ai-agent-book 精选快照（<2MB 代码与文档，来自 github.com/bojieli/ai-agent-book）

2026-08-20 13:12:50 +00:00
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+## Code map
+
+- **Run first:** python generate_data.py --max_problems 2 (a bounded teacher-trajectory smoke).
+- **Start here:** generate_data.py produces raw and verified trajectories.
+- **Core behavior:** train_student.py masks prompt tokens and updates the student; evaluate_student.py runs the paired comparison.
+- **State / protocol:** JSONL messages, answer-validation fields, checkpoint directory and training manifest.
+- **Verifier:** exact answer validator, paired sign test and reflection/backtracking audit.
+- **Experiment variable:** teacher endpoint/model, sampling budget, student base model and LoRA settings.
+- **Skip on first pass:** provider retry code, tokenizer diagnostics and long raw trajectory files.
+
+## English
+
+# CoT Distillation: Collecting SFT Data from Frontier Cloud Models
+
+This experiment implements all three manuscript stages: auditable collection of
+verified teacher trajectories, a real student parameter-update run, and a
+paired baseline/student/teacher evaluation with explicit completion gates.
+
+## Method
+
+The pipeline loads problems, asks a selected teacher model to produce worked solutions, extracts the final answer, and keeps only samples that pass answer validation. `train_student.py` then masks prompt tokens and trains only on the teacher assistant trajectory; it emits a real checkpoint and a content-hashed training manifest. `evaluate_student.py` runs the baseline and trained student on the same problems, reuses the saved real teacher trajectories, computes an exact paired sign test, and audits reflection/backtracking/verification behavior.
+
+## Choosing a teacher model
+
+The default does not have to be a closed-source model. A strong open model served through an OpenAI-compatible endpoint is often cheaper, easier to reproduce, and sufficient for data collection. Closed models remain useful as comparison teachers.
+
+## Run
+
+```bash
+# From the repository root: use the shared Chapter 7 environment
+uv sync --locked --python 3.12 --extra ch7
+
+# Activate it before changing directories:
+# macOS/Linux:
+source .venv/bin/activate
+# Windows PowerShell: .\.venv\Scripts\Activate.ps1
+# Windows cmd: .venv\Scripts\activate.bat
+
+# pip fallback when uv is not installed:
+# python -m pip install -e ".[ch7]"
+
+cd chapter8/cot-distillation
+
+# Single-project compatibility path, still supported during migration:
+# python -m pip install -r requirements.txt
+
+cp env.example .env
+
+# Small smoke test with two problems
+python generate_data.py --max_problems 2
+
+# Collect the full set of 24 AIME problems
+python generate_data.py
+
+# Inspect dataset statistics
+python analyze_data.py
+
+# Real parameter update (CUDA; no mock/CPU success fallback)
+python train_student.py --preflight
+python train_student.py --train-data data/sft_cot_distill_aime_kimi_k3.jsonl \
+  --output-dir checkpoints/cot-student
+
+# Same-problem baseline/student/teacher comparison
+python evaluate_student.py --student-model checkpoints/cot-student \
+  --teacher-data data/raw_trajectories_aime_kimi_k3.jsonl
+```
+
+Provider, model, concurrency, retry, and output settings can be configured through command-line arguments and environment variables. See `python generate_data.py --help` for the complete list.
+
+## Output
+
+Each accepted JSONL row contains the problem, messages, teacher metadata, extracted answer, reference answer, and validation result. Failed or invalid generations are recorded separately so collection runs remain auditable.
+
+## AIME comparison
+
+The included notes compare three teachers over 24 AIME problems, covering answer accuracy, accepted sample counts, token use, latency, and estimated cost. Results depend on model versions and provider conditions, so rerun the experiment before making a production choice.
+
+## Archived simple Chinese problems
+
+Earlier easy Chinese arithmetic samples are retained as historical artifacts. They are useful for smoke testing but are not a meaningful reasoning benchmark.
+
+---
+
+## 中文
+
+# CoT 蒸馏：从前沿云模型采集 SFT 数据
+
+配套书中**实验 8-9（思维链蒸馏）**。SFT 的第一步是拿到高质量示范数据，而获取
+SFT 数据最高效的方式就是**蒸馏前沿模型**：通过大规模 API 调用，把教师模型的
+"思考 + 答案"轨迹采集下来，经规则验证器过滤后作为学生模型的训练数据
+（DeepSeek-R1 蒸馏小模型走的就是这条路线）。
+
+## 方法
+
+三步流程（完整对应实验 8-9，而非只停在采集轨迹）：
+
+1. **采样任务**：`problems.jsonl` 内置 24 道 AIME 真题（1986–2024 年，按题号
+   难度分层抽样：P1–5/P6–10/P11–15 各 8 道，已剔除含图形的题），答案是
+   0–999 的整数，可以用规则验证器自动判对错。`problems_zh.jsonl` 另附 24 道
+   简单中文数学题（鸡兔同笼、工程问题等），适合低成本冒烟测试。
+2. **采集轨迹**：`generate_data.py` 通过 OpenRouter 调用教师模型
+   （默认 `anthropic/claude-opus-4.8`），开启 `reasoning` 参数获取思维链。
+   注意：Claude API 返回的是 **summarized thinking**（由单独的摘要模型改写，
+   逐 token 的原始思维链只存在于加密的 `signature` 字段中，API 不暴露），
+   且模型越新摘要越激进（见文末实测）。若需要逐 token 原文，
+   推荐直接用开放模型原生 API，例如 Kimi K3（见下文对照实验的运行参数）。
+3. **验证过滤**：用规则验证器核对 `Final Answer` 数值，只保留答对的轨迹，
+   写成 `问题 → <think>思考</think> + 最终答案` 的 messages 格式 SFT 数据。
+   错误的思考过程会被学生一并模仿，所以这一步不能省。
+4. **学生 SFT**：`train_student.py` 对提示 token 做 loss mask，只在教师的
+   `<think>…</think> + 最终答案` 上回传梯度；真实 CUDA 训练后写出 checkpoint、
+   数据 SHA、基模、GPU、超参数和训练指标。脚本没有 mock 或 CPU 假成功路径。
+5. **同集对照验收**：`evaluate_student.py` 在同一批题上运行未训练基线与学生，
+   并复用保存的真实 API 教师轨迹；报告三臂准确率、配对胜负、精确双侧检验、
+   教师能力恢复比例，以及反思/回溯/验算行为。只有学生显著优于基线且这些
+   行为在真实输出中出现时，机器可读结果才标为 `complete`。
+
+## 教师模型怎么选：默认开源 SOTA，不必盯着闭源
+
+对绝大多数做后训练的人来说，**不需要**去蒸馏闭源模型的思维链。当前最先进的
+开源模型（DeepSeek V4、Kimi K3、GLM 5.2 等）与 SOTA 闭源模型的差距并没有
+想象中大；如果你要后训练的是 200B 及以下规模的模型，用开源 SOTA 模型当教师
+已经完全够用——教师的水平只需要"明显高于学生"，不需要"全球第一"。
+
+本目录保留 Claude 的采集结果，目的是做一个对照：**闭源 API 的
+summarized thinking 和开源模型的原始思维链，作为 SFT 数据到底有什么差别**。
+
+> 合规说明：本实验只使用各厂商官方 API 提供的 reasoning/thinking 能力获取思维链
+> （Claude 在 API 中返回 summarized thinking，Kimi K3、DeepSeek 等开放模型直接
+> 返回原始思维链），不涉及任何绕过厂商安全机制的手段。对闭源模型，
+> 蒸馏产物的使用需遵守对应服务商的条款。
+
+## 运行
+
+```bash
+# 在仓库根目录使用统一的第 7 章环境
+uv sync --locked --python 3.12 --extra ch7
+
+# 切换目录前先激活环境：
+# macOS/Linux：
+source .venv/bin/activate
+# Windows PowerShell：.\.venv\Scripts\Activate.ps1
+# Windows cmd：.venv\Scripts\activate.bat
+
+# 未安装 uv 时可用 pip 兜底：
+# python -m pip install -e ".[ch7]"
+
+cd chapter8/cot-distillation
+
+# 迁移期间仍支持单项目兼容路径：
+# python -m pip install -r requirements.txt
+
+export OPENROUTER_API_KEY=your-openrouter-api-key
+
+# 小规模冒烟（2 道题）
+python generate_data.py --max_problems 2 \
+  --sft_output /tmp/smoke_sft.jsonl --raw_output /tmp/smoke_raw.jsonl
+
+# 全量采集（24 道 AIME 题；Opus 4.8 输出约 4 万 token，Kimi K3 约 6 万）
+python generate_data.py
+
+# 数据统计
+python analyze_data.py
+
+# 第二步：真实学生参数训练（需要 CUDA）
+python train_student.py --preflight
+python train_student.py \
+  --train-data data/sft_cot_distill_aime_kimi_k3.jsonl \
+  --base-model Qwen/Qwen2.5-1.5B-Instruct \
+  --output-dir checkpoints/cot-student
+
+# 第三步：基线 / 学生 / 教师同题对照
+python evaluate_student.py \
+  --baseline-model Qwen/Qwen2.5-1.5B-Instruct \
+  --student-model checkpoints/cot-student \
+  --teacher-data data/raw_trajectories_aime_kimi_k3.jsonl \
+  --output validation/experiment_8_9.json
+```
+
+常用参数：`--model` 换教师模型、`--base_url`/`--api_key_env` 换端点、
+`--reasoning_effort`（Opus 4.8 等自适应思考模型）与 `--reasoning_max_tokens`
+（Sonnet 4.5 等手动预算模型）控制思维链、`--concurrency` 并发数、
+`--max_retries` 失败重试次数（重试时自动升温换取不同轨迹）、
+`--request_timeout` 单请求硬超时（采集长思考模型时必备，见文末工程教训）。
+
+## 输出
+
+| 文件 | 内容 |
+| --- | --- |
+| `data/sft_cot_distill_aime.jsonl` | Claude Opus 4.8 的 SFT 训练数据（messages 格式，思维链包在 `<think>` 标签内） |
+| `data/sft_cot_distill_aime_kimi_k3.jsonl` | Kimi K3 的 SFT 训练数据 |
+| `data/raw_trajectories_*.jsonl` | 全部原始轨迹（含未通过验证的），用于分析教师错误模式 |
+| `data/*_zh*.jsonl` | 中文简单题（`problems_zh.jsonl`）的归档采集结果 |
+| `train_student.py` | 真实 SFT 参数更新；提示 token mask、LoRA/全参训练和训练 manifest |
+| `evaluate_student.py` | 同题三臂评测、配对显著性与教师式行为验收 |
+
+当前仓库保存了 24/24 Kimi K3 AIME 完整轨迹。规则验证器接受其中 23 条进入 SFT；
+`aime-2016-9-I` 在原生 low-reasoning 控制下完成，但答案错误，因此被正确拒绝。
+第二步与第三步已在 RTX PRO 6000 Blackwell Workstation Edition 上完成
+真实 CUDA 训练：[`student_sft_preflight_20260801_gpu.json`](validation/student_sft_preflight_20260801_gpu.json)
+证明训练栈可用；[`training_manifest.json`](checkpoints/exp8-9-qwen25-1.5b-kimi-k3-20260801-v1/training_manifest.json)
+记录了 Qwen2.5-1.5B-Instruct + LoRA 的真实参数更新（3 epochs，约 27 秒，最终 loss
+2.17）；[`experiment_8_9_complete_20260803_v2.json`](validation/experiment_8_9_complete_20260803_v2.json)
+给出同题三臂对照：基线 1/24、学生 2/24、教师 23/24，配对检验 p=1.0 不显著，能力
+恢复比例约 4.5%。学生输出中确实出现了少量反思/验算行为，但总体与基线接近。
+实验执行与证据状态为 **complete**；配对提升不显著是本次实验的负结果，而不是缺失门禁。
+若要检验更强的蒸馏效果，下一轮应扩大已验证训练集后重新训练，而不是把负结果改写成成功。
+
+## AIME 实测：三位教师的对照（24 题）
+
+| | Claude Sonnet 4.5 | Claude Opus 4.8 | Kimi K3 |
+| --- | --- | --- | --- |
+| 验证通过率 | 22/24 | 24/24 | 23/24 |
+| 思维链性质 | 摘要，近 1:1 保真 | 摘要，激进压缩 | 原始思维链直出 |
+| 原始/可见 token 比（精确对账） | 1.03–1.09 | **2.41** | **1.007** |
+| 可见思维链规模 | 中位 6.2k 字符 | 均值 536 token | 均值 2.5k token |
+| 无思维链的题 | 0 | 3（自适应思考跳过） | 0 |
+| `reasoning_tokens` 字段可信度 | 虚低至 55%–75%（OpenRouter 侧） | 同样虚低 | 准确（1.001） |
+
+token 对账方法：用模型自身 tokenizer（`max_tokens=1` 探针读 `prompt_tokens`）
+数出可见思维链与正文的 token 数，`completion_tokens − 正文 token` 即为计费的
+原始思考量。OpenRouter 返回的 `reasoning_tokens` 详情字段对 Claude 系统性虚低，
+做成本核算时不可直接采信。
+
+三个对后训练有直接意义的观察：
+
+1. **模型越新，思维链围墙越高。** 同是 Claude，Sonnet 4.5 的摘要还接近逐字，
+   Opus 4.8 已压到不足一半、且 3 道题完全不给思维链。思维链透明度在持续收紧。
+2. **教师能力 ≠ 可蒸馏性。** Opus 4.8 答对率最高，给出的蒸馏材料却最差
+   （摘要稀疏、截断、缺失）；Kimi K3 少对 1 题，但每条轨迹都是完整原文。
+   选教师要同时看"会不会做"和"给不给看"。
+3. **原始思维链含元噪声。** Kimi K3 的原文里有英文元思考、输出格式纠结
+   （曾在简单题上用 700+ token 争论该写 `16` 还是 `16%`）、中途自我打断。
+   答案验证器滤不掉这类噪声，用它做 SFT 前值得加一道清洗或重写。
+
+工程教训（也是书中"数据管线健壮性"的实例）：Kimi K3 在个别 AIME 题上思考
+超过 15 分钟（aime-2016-9-I 三次尝试均超 900 秒，最终放弃该题），且
+Moonshot 端会出现"停止发送但不关闭连接"的半开状态。采集 pipeline 必须：
+每题完成即落盘（本脚本增量写入 `raw_trajectories`）、用 `asyncio.wait_for`
+做硬超时（httpx 读超时对半开连接无效）、失败可重试。
+
+## 中文简单题归档结果
+
+`problems_zh.jsonl`（24 道中学数学题）上两位教师均 24/24 通过，思维链短
+（Claude 摘要均值约 290 字符，Kimi 原文均值约 165 token），适合几分钟、
+几分钱成本验证 pipeline 是否工作，再切换到 AIME 或自己的目标分布。
+
+规模化的做法是把 `problems.jsonl` 换成目标分布的题目来源（如 GSM8K、MATH
+训练集），提高并发，并按书中"数据质量三维度"控制覆盖面、多样性与标注准确性。
@@ -0,0 +1,71 @@
+"""统计蒸馏得到的 SFT 数据：规模、token/字符分布、思考链特征。"""
+
+import argparse
+import json
+import re
+
+
+def main():
+    parser = argparse.ArgumentParser(description="统计 CoT 蒸馏 SFT 数据")
+    parser.add_argument("--sft", default="./data/sft_cot_distill_aime.jsonl")
+    parser.add_argument("--raw", default="./data/raw_trajectories_aime.jsonl")
+    args = parser.parse_args()
+
+    with open(args.sft, encoding="utf-8") as f:
+        samples = [json.loads(line) for line in f if line.strip()]
+    print(f"SFT 样本数：{len(samples)}")
+
+    think_lens, answer_lens = [], []
+    n_reflect = 0
+    n_skipped_short = 0
+    for s in samples:
+        messages = s.get("messages") or []
+        # Incomplete SFT rows (user-only / truncated export) must not IndexError.
+        if len(messages) < 2:
+            n_skipped_short += 1
+            continue
+        assistant_msg = messages[1]
+        if not isinstance(assistant_msg, dict):
+            n_skipped_short += 1
+            continue
+        assistant = assistant_msg.get("content")
+        if not isinstance(assistant, str):
+            n_skipped_short += 1
+            continue
+        m = re.search(r"<think>\n?(.*?)\n?</think>", assistant, re.DOTALL)
+        think = m.group(1) if m else ""
+        think_lens.append(len(think))
+        answer_lens.append(len(assistant))
+        # 教师式的反思/验算行为（实验 8-9 验收标准之一）
+        if re.search(r"(验算|检查|重新|等等|不对|再算|反思|verify|check|wait)", think, re.IGNORECASE):
+            n_reflect += 1
+
+    def stats(xs, name):
+        if not xs:
+            print(f"{name}：无数据")
+            return
+        xs = sorted(xs)
+        n = len(xs)
+        print(f"{name}：均值 {sum(xs)/n:.0f}，中位 {xs[n//2]}，最小 {xs[0]}，最大 {xs[-1]}")
+
+    stats(think_lens, "思考链长度（字符）")
+    stats(answer_lens, "完整回答长度（字符）")
+    n_scored = len(samples) - n_skipped_short
+    print(f"含反思/验算行为的样本：{n_reflect}/{n_scored}")
+    if n_skipped_short:
+        print(f"跳过 messages 不足 2 条的样本：{n_skipped_short}")
+
+    try:
+        with open(args.raw, encoding="utf-8") as f:
+            raw = [json.loads(line) for line in f if line.strip()]
+        failed = [r for r in raw if not r["verified"]]
+        print(f"\n原始轨迹 {len(raw)} 条，未通过验证 {len(failed)} 条：")
+        for r in failed:
+            pred = r["content"][-80:].replace("\n", " ") if r["content"] else "(无输出)"
+            print(f"  {r['id']}: gold={r['gold_answer']}  输出末尾: …{pred}  error={r['error']}")
+    except FileNotFoundError:
+        pass
+
+
+if __name__ == "__main__":
+    main()
@@ -0,0 +1,207 @@
+---
+base_model: Qwen/Qwen2.5-1.5B-Instruct
+library_name: peft
+pipeline_tag: text-generation
+tags:
+- base_model:adapter:Qwen/Qwen2.5-1.5B-Instruct
+- lora
+- transformers
+---
+
+# Model Card for Model ID
+
+<!-- Provide a quick summary of what the model is/does. -->
+
+
+
+## Model Details
+
+### Model Description
+
+<!-- Provide a longer summary of what this model is. -->
+
+
+
+- **Developed by:** [More Information Needed]
+- **Funded by [optional]:** [More Information Needed]
+- **Shared by [optional]:** [More Information Needed]
+- **Model type:** [More Information Needed]
+- **Language(s) (NLP):** [More Information Needed]
+- **License:** [More Information Needed]
+- **Finetuned from model [optional]:** [More Information Needed]
+
+### Model Sources [optional]
+
+<!-- Provide the basic links for the model. -->
+
+- **Repository:** [More Information Needed]
+- **Paper [optional]:** [More Information Needed]
+- **Demo [optional]:** [More Information Needed]
+
+## Uses
+
+<!-- Address questions around how the model is intended to be used, including the foreseeable users of the model and those affected by the model. -->
+
+### Direct Use
+
+<!-- This section is for the model use without fine-tuning or plugging into a larger ecosystem/app. -->
+
+[More Information Needed]
+
+### Downstream Use [optional]
+
+<!-- This section is for the model use when fine-tuned for a task, or when plugged into a larger ecosystem/app -->
+
+[More Information Needed]
+
+### Out-of-Scope Use
+
+<!-- This section addresses misuse, malicious use, and uses that the model will not work well for. -->
+
+[More Information Needed]
+
+## Bias, Risks, and Limitations
+
+<!-- This section is meant to convey both technical and sociotechnical limitations. -->
+
+[More Information Needed]
+
+### Recommendations
+
+<!-- This section is meant to convey recommendations with respect to the bias, risk, and technical limitations. -->
+
+Users (both direct and downstream) should be made aware of the risks, biases and limitations of the model. More information needed for further recommendations.
+
+## How to Get Started with the Model
+
+Use the code below to get started with the model.
+
+[More Information Needed]
+
+## Training Details
+
+### Training Data
+
+<!-- This should link to a Dataset Card, perhaps with a short stub of information on what the training data is all about as well as documentation related to data pre-processing or additional filtering. -->
+
+[More Information Needed]
+
+### Training Procedure
+
+<!-- This relates heavily to the Technical Specifications. Content here should link to that section when it is relevant to the training procedure. -->
+
+#### Preprocessing [optional]
+
+[More Information Needed]
+
+
+#### Training Hyperparameters
+
+- **Training regime:** [More Information Needed] <!--fp32, fp16 mixed precision, bf16 mixed precision, bf16 non-mixed precision, fp16 non-mixed precision, fp8 mixed precision -->
+
+#### Speeds, Sizes, Times [optional]
+
+<!-- This section provides information about throughput, start/end time, checkpoint size if relevant, etc. -->
+
+[More Information Needed]
+
+## Evaluation
+
+<!-- This section describes the evaluation protocols and provides the results. -->
+
+### Testing Data, Factors & Metrics
+
+#### Testing Data
+
+<!-- This should link to a Dataset Card if possible. -->
+
+[More Information Needed]
+
+#### Factors
+
+<!-- These are the things the evaluation is disaggregating by, e.g., subpopulations or domains. -->
+
+[More Information Needed]
+
+#### Metrics
+
+<!-- These are the evaluation metrics being used, ideally with a description of why. -->
+
+[More Information Needed]
+
+### Results
+
+[More Information Needed]
+
+#### Summary
+
+
+
+## Model Examination [optional]
+
+<!-- Relevant interpretability work for the model goes here -->
+
+[More Information Needed]
+
+## Environmental Impact
+
+<!-- Total emissions (in grams of CO2eq) and additional considerations, such as electricity usage, go here. Edit the suggested text below accordingly -->
+
+Carbon emissions can be estimated using the [Machine Learning Impact calculator](https://mlco2.github.io/impact#compute) presented in [Lacoste et al. (2019)](https://arxiv.org/abs/1910.09700).
+
+- **Hardware Type:** [More Information Needed]
+- **Hours used:** [More Information Needed]
+- **Cloud Provider:** [More Information Needed]
+- **Compute Region:** [More Information Needed]
+- **Carbon Emitted:** [More Information Needed]
+
+## Technical Specifications [optional]
+
+### Model Architecture and Objective
+
+[More Information Needed]
+
+### Compute Infrastructure
+
+[More Information Needed]
+
+#### Hardware
+
+[More Information Needed]
+
+#### Software
+
+[More Information Needed]
+
+## Citation [optional]
+
+<!-- If there is a paper or blog post introducing the model, the APA and Bibtex information for that should go in this section. -->
+
+**BibTeX:**
+
+[More Information Needed]
+
+**APA:**
+
+[More Information Needed]
+
+## Glossary [optional]
+
+<!-- If relevant, include terms and calculations in this section that can help readers understand the model or model card. -->
+
+[More Information Needed]
+
+## More Information [optional]
+
+[More Information Needed]
+
+## Model Card Authors [optional]
+
+[More Information Needed]
+
+## Model Card Contact
+
+[More Information Needed]
+### Framework versions
+
+- PEFT 0.19.1
@@ -0,0 +1,48 @@
+{
+  "alora_invocation_tokens": null,
+  "alpha_pattern": {},
+  "arrow_config": null,
+  "auto_mapping": null,
+  "base_model_name_or_path": "Qwen/Qwen2.5-1.5B-Instruct",
+  "bias": "none",
+  "corda_config": null,
+  "ensure_weight_tying": false,
+  "eva_config": null,
+  "exclude_modules": null,
+  "fan_in_fan_out": false,
+  "inference_mode": true,
+  "init_lora_weights": true,
+  "layer_replication": null,
+  "layers_pattern": null,
+  "layers_to_transform": null,
+  "loftq_config": {},
+  "lora_alpha": 64,
+  "lora_bias": false,
+  "lora_dropout": 0.05,
+  "lora_ga_config": null,
+  "megatron_config": null,
+  "megatron_core": "megatron.core",
+  "modules_to_save": null,
+  "peft_type": "LORA",
+  "peft_version": "0.19.1",
+  "qalora_group_size": 16,
+  "r": 32,
+  "rank_pattern": {},
+  "revision": null,
+  "target_modules": [
+    "up_proj",
+    "down_proj",
+    "k_proj",
+    "gate_proj",
+    "q_proj",
+    "v_proj",
+    "o_proj"
+  ],
+  "target_parameters": null,
+  "task_type": "CAUSAL_LM",
+  "trainable_token_indices": null,
+  "use_bdlora": null,
+  "use_dora": false,
+  "use_qalora": false,
+  "use_rslora": false
+}
@@ -0,0 +1,54 @@
+{%- if tools %}
+    {{- '<|im_start|>system\n' }}
+    {%- if messages[0]['role'] == 'system' %}
+        {{- messages[0]['content'] }}
+    {%- else %}
+        {{- 'You are Qwen, created by Alibaba Cloud. You are a helpful assistant.' }}
+    {%- endif %}
+    {{- "\n\n# Tools\n\nYou may call one or more functions to assist with the user query.\n\nYou are provided with function signatures within <tools></tools> XML tags:\n<tools>" }}
+    {%- for tool in tools %}
+        {{- "\n" }}
+        {{- tool | tojson }}
+    {%- endfor %}
+    {{- "\n</tools>\n\nFor each function call, return a json object with function name and arguments within <tool_call></tool_call> XML tags:\n<tool_call>\n{\"name\": <function-name>, \"arguments\": <args-json-object>}\n</tool_call><|im_end|>\n" }}
+{%- else %}
+    {%- if messages[0]['role'] == 'system' %}
+        {{- '<|im_start|>system\n' + messages[0]['content'] + '<|im_end|>\n' }}
+    {%- else %}
+        {{- '<|im_start|>system\nYou are Qwen, created by Alibaba Cloud. You are a helpful assistant.<|im_end|>\n' }}
+    {%- endif %}
+{%- endif %}
+{%- for message in messages %}
+    {%- if (message.role == "user") or (message.role == "system" and not loop.first) or (message.role == "assistant" and not message.tool_calls) %}
+        {{- '<|im_start|>' + message.role + '\n' + message.content + '<|im_end|>' + '\n' }}
+    {%- elif message.role == "assistant" %}
+        {{- '<|im_start|>' + message.role }}
+        {%- if message.content %}
+            {{- '\n' + message.content }}
+        {%- endif %}
+        {%- for tool_call in message.tool_calls %}
+            {%- if tool_call.function is defined %}
+                {%- set tool_call = tool_call.function %}
+            {%- endif %}
+            {{- '\n<tool_call>\n{"name": "' }}
+            {{- tool_call.name }}
+            {{- '", "arguments": ' }}
+            {{- tool_call.arguments | tojson }}
+            {{- '}\n</tool_call>' }}
+        {%- endfor %}
+        {{- '<|im_end|>\n' }}
+    {%- elif message.role == "tool" %}
+        {%- if (loop.index0 == 0) or (messages[loop.index0 - 1].role != "tool") %}
+            {{- '<|im_start|>user' }}
+        {%- endif %}
+        {{- '\n<tool_response>\n' }}
+        {{- message.content }}
+        {{- '\n</tool_response>' }}
+        {%- if loop.last or (messages[loop.index0 + 1].role != "tool") %}
+            {{- '<|im_end|>\n' }}
+        {%- endif %}
+    {%- endif %}
+{%- endfor %}
+{%- if add_generation_prompt %}
+    {{- '<|im_start|>assistant\n' }}
+{%- endif %}
@@ -0,0 +1,207 @@
+---
+base_model: Qwen/Qwen2.5-1.5B-Instruct
+library_name: peft
+pipeline_tag: text-generation
+tags:
+- base_model:adapter:Qwen/Qwen2.5-1.5B-Instruct
+- lora
+- transformers
+---
+
+# Model Card for Model ID
+
+<!-- Provide a quick summary of what the model is/does. -->
+
+
+
+## Model Details
+
+### Model Description
+
+<!-- Provide a longer summary of what this model is. -->
+
+
+
+- **Developed by:** [More Information Needed]
+- **Funded by [optional]:** [More Information Needed]
+- **Shared by [optional]:** [More Information Needed]
+- **Model type:** [More Information Needed]
+- **Language(s) (NLP):** [More Information Needed]
+- **License:** [More Information Needed]
+- **Finetuned from model [optional]:** [More Information Needed]
+
+### Model Sources [optional]
+
+<!-- Provide the basic links for the model. -->
+
+- **Repository:** [More Information Needed]
+- **Paper [optional]:** [More Information Needed]
+- **Demo [optional]:** [More Information Needed]
+
+## Uses
+
+<!-- Address questions around how the model is intended to be used, including the foreseeable users of the model and those affected by the model. -->
+
+### Direct Use
+
+<!-- This section is for the model use without fine-tuning or plugging into a larger ecosystem/app. -->
+
+[More Information Needed]
+
+### Downstream Use [optional]
+
+<!-- This section is for the model use when fine-tuned for a task, or when plugged into a larger ecosystem/app -->
+
+[More Information Needed]
+
+### Out-of-Scope Use
+
+<!-- This section addresses misuse, malicious use, and uses that the model will not work well for. -->
+
+[More Information Needed]
+
+## Bias, Risks, and Limitations
+
+<!-- This section is meant to convey both technical and sociotechnical limitations. -->
+
+[More Information Needed]
+
+### Recommendations
+
+<!-- This section is meant to convey recommendations with respect to the bias, risk, and technical limitations. -->
+
+Users (both direct and downstream) should be made aware of the risks, biases and limitations of the model. More information needed for further recommendations.
+
+## How to Get Started with the Model
+
+Use the code below to get started with the model.
+
+[More Information Needed]
+
+## Training Details
+
+### Training Data
+
+<!-- This should link to a Dataset Card, perhaps with a short stub of information on what the training data is all about as well as documentation related to data pre-processing or additional filtering. -->
+
+[More Information Needed]
+
+### Training Procedure
+
+<!-- This relates heavily to the Technical Specifications. Content here should link to that section when it is relevant to the training procedure. -->
+
+#### Preprocessing [optional]
+
+[More Information Needed]
+
+
+#### Training Hyperparameters
+
+- **Training regime:** [More Information Needed] <!--fp32, fp16 mixed precision, bf16 mixed precision, bf16 non-mixed precision, fp16 non-mixed precision, fp8 mixed precision -->
+
+#### Speeds, Sizes, Times [optional]
+
+<!-- This section provides information about throughput, start/end time, checkpoint size if relevant, etc. -->
+
+[More Information Needed]
+
+## Evaluation
+
+<!-- This section describes the evaluation protocols and provides the results. -->
+
+### Testing Data, Factors & Metrics
+
+#### Testing Data
+
+<!-- This should link to a Dataset Card if possible. -->
+
+[More Information Needed]
+
+#### Factors
+
+<!-- These are the things the evaluation is disaggregating by, e.g., subpopulations or domains. -->
+
+[More Information Needed]
+
+#### Metrics
+
+<!-- These are the evaluation metrics being used, ideally with a description of why. -->
+
+[More Information Needed]
+
+### Results
+
+[More Information Needed]
+
+#### Summary
+
+
+
+## Model Examination [optional]
+
+<!-- Relevant interpretability work for the model goes here -->
+
+[More Information Needed]
+
+## Environmental Impact
+
+<!-- Total emissions (in grams of CO2eq) and additional considerations, such as electricity usage, go here. Edit the suggested text below accordingly -->
+
+Carbon emissions can be estimated using the [Machine Learning Impact calculator](https://mlco2.github.io/impact#compute) presented in [Lacoste et al. (2019)](https://arxiv.org/abs/1910.09700).
+
+- **Hardware Type:** [More Information Needed]
+- **Hours used:** [More Information Needed]
+- **Cloud Provider:** [More Information Needed]
+- **Compute Region:** [More Information Needed]
+- **Carbon Emitted:** [More Information Needed]
+
+## Technical Specifications [optional]
+
+### Model Architecture and Objective
+
+[More Information Needed]
+
+### Compute Infrastructure
+
+[More Information Needed]
+
+#### Hardware
+
+[More Information Needed]
+
+#### Software
+
+[More Information Needed]
+
+## Citation [optional]
+
+<!-- If there is a paper or blog post introducing the model, the APA and Bibtex information for that should go in this section. -->
+
+**BibTeX:**
+
+[More Information Needed]
+
+**APA:**
+
+[More Information Needed]
+
+## Glossary [optional]
+
+<!-- If relevant, include terms and calculations in this section that can help readers understand the model or model card. -->
+
+[More Information Needed]
+
+## More Information [optional]
+
+[More Information Needed]
+
+## Model Card Authors [optional]
+
+[More Information Needed]
+
+## Model Card Contact
+
+[More Information Needed]
+### Framework versions
+
+- PEFT 0.19.1
@@ -0,0 +1,48 @@
+{
+  "alora_invocation_tokens": null,
+  "alpha_pattern": {},
+  "arrow_config": null,
+  "auto_mapping": null,
+  "base_model_name_or_path": "Qwen/Qwen2.5-1.5B-Instruct",
+  "bias": "none",
+  "corda_config": null,
+  "ensure_weight_tying": false,
+  "eva_config": null,
+  "exclude_modules": null,
+  "fan_in_fan_out": false,
+  "inference_mode": true,
+  "init_lora_weights": true,
+  "layer_replication": null,
+  "layers_pattern": null,
+  "layers_to_transform": null,
+  "loftq_config": {},
+  "lora_alpha": 64,
+  "lora_bias": false,
+  "lora_dropout": 0.05,
+  "lora_ga_config": null,
+  "megatron_config": null,
+  "megatron_core": "megatron.core",
+  "modules_to_save": null,
+  "peft_type": "LORA",
+  "peft_version": "0.19.1",
+  "qalora_group_size": 16,
+  "r": 32,
+  "rank_pattern": {},
+  "revision": null,
+  "target_modules": [
+    "up_proj",
+    "down_proj",
+    "k_proj",
+    "gate_proj",
+    "q_proj",
+    "v_proj",
+    "o_proj"
+  ],
+  "target_parameters": null,
+  "task_type": "CAUSAL_LM",
+  "trainable_token_indices": null,
+  "use_bdlora": null,
+  "use_dora": false,
+  "use_qalora": false,
+  "use_rslora": false
+}
@@ -0,0 +1,48 @@
+{
+  "best_global_step": null,
+  "best_metric": null,
+  "best_model_checkpoint": null,
+  "epoch": 1.0,
+  "eval_steps": 500,
+  "global_step": 2,
+  "is_hyper_param_search": false,
+  "is_local_process_zero": true,
+  "is_world_process_zero": true,
+  "log_history": [
+    {
+      "epoch": 0.6956521739130435,
+      "grad_norm": 0.45245930552482605,
+      "learning_rate": 2e-05,
+      "loss": 2.1483168601989746,
+      "step": 1
+    },
+    {
+      "epoch": 1.0,
+      "grad_norm": 0.5142671465873718,
+      "learning_rate": 1.6666666666666667e-05,
+      "loss": 2.2101056575775146,
+      "step": 2
+    }
+  ],
+  "logging_steps": 1,
+  "max_steps": 6,
+  "num_input_tokens_seen": 0,
+  "num_train_epochs": 3,
+  "save_steps": 500,
+  "stateful_callbacks": {
+    "TrainerControl": {
+      "args": {
+        "should_epoch_stop": false,
+        "should_evaluate": false,
+        "should_log": false,
+        "should_save": true,
+        "should_training_stop": false
+      },
+      "attributes": {}
+    }
+  },
+  "total_flos": 758785239641088.0,
+  "train_batch_size": 1,
+  "trial_name": null,
+  "trial_params": null
+}
@@ -0,0 +1,207 @@
+---
+base_model: Qwen/Qwen2.5-1.5B-Instruct
+library_name: peft
+pipeline_tag: text-generation
+tags:
+- base_model:adapter:Qwen/Qwen2.5-1.5B-Instruct
+- lora
+- transformers
+---
+
+# Model Card for Model ID
+
+<!-- Provide a quick summary of what the model is/does. -->
+
+
+
+## Model Details
+
+### Model Description
+
+<!-- Provide a longer summary of what this model is. -->
+
+
+
+- **Developed by:** [More Information Needed]
+- **Funded by [optional]:** [More Information Needed]
+- **Shared by [optional]:** [More Information Needed]
+- **Model type:** [More Information Needed]
+- **Language(s) (NLP):** [More Information Needed]
+- **License:** [More Information Needed]
+- **Finetuned from model [optional]:** [More Information Needed]
+
+### Model Sources [optional]
+
+<!-- Provide the basic links for the model. -->
+
+- **Repository:** [More Information Needed]
+- **Paper [optional]:** [More Information Needed]
+- **Demo [optional]:** [More Information Needed]
+
+## Uses
+
+<!-- Address questions around how the model is intended to be used, including the foreseeable users of the model and those affected by the model. -->
+
+### Direct Use
+
+<!-- This section is for the model use without fine-tuning or plugging into a larger ecosystem/app. -->
+
+[More Information Needed]
+
+### Downstream Use [optional]
+
+<!-- This section is for the model use when fine-tuned for a task, or when plugged into a larger ecosystem/app -->
+
+[More Information Needed]
+
+### Out-of-Scope Use
+
+<!-- This section addresses misuse, malicious use, and uses that the model will not work well for. -->
+
+[More Information Needed]
+
+## Bias, Risks, and Limitations
+
+<!-- This section is meant to convey both technical and sociotechnical limitations. -->
+
+[More Information Needed]
+
+### Recommendations
+
+<!-- This section is meant to convey recommendations with respect to the bias, risk, and technical limitations. -->
+
+Users (both direct and downstream) should be made aware of the risks, biases and limitations of the model. More information needed for further recommendations.
+
+## How to Get Started with the Model
+
+Use the code below to get started with the model.
+
+[More Information Needed]
+
+## Training Details
+
+### Training Data
+
+<!-- This should link to a Dataset Card, perhaps with a short stub of information on what the training data is all about as well as documentation related to data pre-processing or additional filtering. -->
+
+[More Information Needed]
+
+### Training Procedure
+
+<!-- This relates heavily to the Technical Specifications. Content here should link to that section when it is relevant to the training procedure. -->
+
+#### Preprocessing [optional]
+
+[More Information Needed]
+
+
+#### Training Hyperparameters
+
+- **Training regime:** [More Information Needed] <!--fp32, fp16 mixed precision, bf16 mixed precision, bf16 non-mixed precision, fp16 non-mixed precision, fp8 mixed precision -->
+
+#### Speeds, Sizes, Times [optional]
+
+<!-- This section provides information about throughput, start/end time, checkpoint size if relevant, etc. -->
+
+[More Information Needed]
+
+## Evaluation
+
+<!-- This section describes the evaluation protocols and provides the results. -->
+
+### Testing Data, Factors & Metrics
+
+#### Testing Data
+
+<!-- This should link to a Dataset Card if possible. -->
+
+[More Information Needed]
+
+#### Factors
+
+<!-- These are the things the evaluation is disaggregating by, e.g., subpopulations or domains. -->
+
+[More Information Needed]
+
+#### Metrics
+
+<!-- These are the evaluation metrics being used, ideally with a description of why. -->
+
+[More Information Needed]
+
+### Results
+
+[More Information Needed]
+
+#### Summary
+
+
+
+## Model Examination [optional]
+
+<!-- Relevant interpretability work for the model goes here -->
+
+[More Information Needed]
+
+## Environmental Impact
+
+<!-- Total emissions (in grams of CO2eq) and additional considerations, such as electricity usage, go here. Edit the suggested text below accordingly -->
+
+Carbon emissions can be estimated using the [Machine Learning Impact calculator](https://mlco2.github.io/impact#compute) presented in [Lacoste et al. (2019)](https://arxiv.org/abs/1910.09700).
+
+- **Hardware Type:** [More Information Needed]
+- **Hours used:** [More Information Needed]
+- **Cloud Provider:** [More Information Needed]
+- **Compute Region:** [More Information Needed]
+- **Carbon Emitted:** [More Information Needed]
+
+## Technical Specifications [optional]
+
+### Model Architecture and Objective
+
+[More Information Needed]
+
+### Compute Infrastructure
+
+[More Information Needed]
+
+#### Hardware
+
+[More Information Needed]
+
+#### Software
+
+[More Information Needed]
+
+## Citation [optional]
+
+<!-- If there is a paper or blog post introducing the model, the APA and Bibtex information for that should go in this section. -->
+
+**BibTeX:**
+
+[More Information Needed]
+
+**APA:**
+
+[More Information Needed]
+
+## Glossary [optional]
+
+<!-- If relevant, include terms and calculations in this section that can help readers understand the model or model card. -->
+
+[More Information Needed]
+
+## More Information [optional]
+
+[More Information Needed]
+
+## Model Card Authors [optional]
+
+[More Information Needed]
+
+## Model Card Contact
+
+[More Information Needed]
+### Framework versions
+
+- PEFT 0.19.1
@@ -0,0 +1,48 @@
+{
+  "alora_invocation_tokens": null,
+  "alpha_pattern": {},
+  "arrow_config": null,
+  "auto_mapping": null,
+  "base_model_name_or_path": "Qwen/Qwen2.5-1.5B-Instruct",
+  "bias": "none",
+  "corda_config": null,
+  "ensure_weight_tying": false,
+  "eva_config": null,
+  "exclude_modules": null,
+  "fan_in_fan_out": false,
+  "inference_mode": true,
+  "init_lora_weights": true,
+  "layer_replication": null,
+  "layers_pattern": null,
+  "layers_to_transform": null,
+  "loftq_config": {},
+  "lora_alpha": 64,
+  "lora_bias": false,
+  "lora_dropout": 0.05,
+  "lora_ga_config": null,
+  "megatron_config": null,
+  "megatron_core": "megatron.core",
+  "modules_to_save": null,
+  "peft_type": "LORA",
+  "peft_version": "0.19.1",
+  "qalora_group_size": 16,
+  "r": 32,
+  "rank_pattern": {},
+  "revision": null,
+  "target_modules": [
+    "up_proj",
+    "down_proj",
+    "k_proj",
+    "gate_proj",
+    "q_proj",
+    "v_proj",
+    "o_proj"
+  ],
+  "target_parameters": null,
+  "task_type": "CAUSAL_LM",
+  "trainable_token_indices": null,
+  "use_bdlora": null,
+  "use_dora": false,
+  "use_qalora": false,
+  "use_rslora": false
+}
@@ -0,0 +1,62 @@
+{
+  "best_global_step": null,
+  "best_metric": null,
+  "best_model_checkpoint": null,
+  "epoch": 2.0,
+  "eval_steps": 500,
+  "global_step": 4,
+  "is_hyper_param_search": false,
+  "is_local_process_zero": true,
+  "is_world_process_zero": true,
+  "log_history": [
+    {
+      "epoch": 0.6956521739130435,
+      "grad_norm": 0.45245930552482605,
+      "learning_rate": 2e-05,
+      "loss": 2.1483168601989746,
+      "step": 1
+    },
+    {
+      "epoch": 1.0,
+      "grad_norm": 0.5142671465873718,
+      "learning_rate": 1.6666666666666667e-05,
+      "loss": 2.2101056575775146,
+      "step": 2
+    },
+    {
+      "epoch": 1.6956521739130435,
+      "grad_norm": 0.39382970333099365,
+      "learning_rate": 1.3333333333333333e-05,
+      "loss": 2.0590713024139404,
+      "step": 3
+    },
+    {
+      "epoch": 2.0,
+      "grad_norm": 0.45731329917907715,
+      "learning_rate": 1e-05,
+      "loss": 2.369413137435913,
+      "step": 4
+    }
+  ],
+  "logging_steps": 1,
+  "max_steps": 6,
+  "num_input_tokens_seen": 0,
+  "num_train_epochs": 3,
+  "save_steps": 500,
+  "stateful_callbacks": {
+    "TrainerControl": {
+      "args": {
+        "should_epoch_stop": false,
+        "should_evaluate": false,
+        "should_log": false,
+        "should_save": true,
+        "should_training_stop": false
+      },
+      "attributes": {}
+    }
+  },
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+  "train_batch_size": 1,
+  "trial_name": null,
+  "trial_params": null
+}
@@ -0,0 +1,207 @@
+---
+base_model: Qwen/Qwen2.5-1.5B-Instruct
+library_name: peft
+pipeline_tag: text-generation
+tags:
+- base_model:adapter:Qwen/Qwen2.5-1.5B-Instruct
+- lora
+- transformers
+---
+
+# Model Card for Model ID
+
+<!-- Provide a quick summary of what the model is/does. -->
+
+
+
+## Model Details
+
+### Model Description
+
+<!-- Provide a longer summary of what this model is. -->
+
+
+
+- **Developed by:** [More Information Needed]
+- **Funded by [optional]:** [More Information Needed]
+- **Shared by [optional]:** [More Information Needed]
+- **Model type:** [More Information Needed]
+- **Language(s) (NLP):** [More Information Needed]
+- **License:** [More Information Needed]
+- **Finetuned from model [optional]:** [More Information Needed]
+
+### Model Sources [optional]
+
+<!-- Provide the basic links for the model. -->
+
+- **Repository:** [More Information Needed]
+- **Paper [optional]:** [More Information Needed]
+- **Demo [optional]:** [More Information Needed]
+
+## Uses
+
+<!-- Address questions around how the model is intended to be used, including the foreseeable users of the model and those affected by the model. -->
+
+### Direct Use
+
+<!-- This section is for the model use without fine-tuning or plugging into a larger ecosystem/app. -->
+
+[More Information Needed]
+
+### Downstream Use [optional]
+
+<!-- This section is for the model use when fine-tuned for a task, or when plugged into a larger ecosystem/app -->
+
+[More Information Needed]
+
+### Out-of-Scope Use
+
+<!-- This section addresses misuse, malicious use, and uses that the model will not work well for. -->
+
+[More Information Needed]
+
+## Bias, Risks, and Limitations
+
+<!-- This section is meant to convey both technical and sociotechnical limitations. -->
+
+[More Information Needed]
+
+### Recommendations
+
+<!-- This section is meant to convey recommendations with respect to the bias, risk, and technical limitations. -->
+
+Users (both direct and downstream) should be made aware of the risks, biases and limitations of the model. More information needed for further recommendations.
+
+## How to Get Started with the Model
+
+Use the code below to get started with the model.
+
+[More Information Needed]
+
+## Training Details
+
+### Training Data
+
+<!-- This should link to a Dataset Card, perhaps with a short stub of information on what the training data is all about as well as documentation related to data pre-processing or additional filtering. -->
+
+[More Information Needed]
+
+### Training Procedure
+
+<!-- This relates heavily to the Technical Specifications. Content here should link to that section when it is relevant to the training procedure. -->
+
+#### Preprocessing [optional]
+
+[More Information Needed]
+
+
+#### Training Hyperparameters
+
+- **Training regime:** [More Information Needed] <!--fp32, fp16 mixed precision, bf16 mixed precision, bf16 non-mixed precision, fp16 non-mixed precision, fp8 mixed precision -->
+
+#### Speeds, Sizes, Times [optional]
+
+<!-- This section provides information about throughput, start/end time, checkpoint size if relevant, etc. -->
+
+[More Information Needed]
+
+## Evaluation
+
+<!-- This section describes the evaluation protocols and provides the results. -->
+
+### Testing Data, Factors & Metrics
+
+#### Testing Data
+
+<!-- This should link to a Dataset Card if possible. -->
+
+[More Information Needed]
+
+#### Factors
+
+<!-- These are the things the evaluation is disaggregating by, e.g., subpopulations or domains. -->
+
+[More Information Needed]
+
+#### Metrics
+
+<!-- These are the evaluation metrics being used, ideally with a description of why. -->
+
+[More Information Needed]
+
+### Results
+
+[More Information Needed]
+
+#### Summary
+
+
+
+## Model Examination [optional]
+
+<!-- Relevant interpretability work for the model goes here -->
+
+[More Information Needed]
+
+## Environmental Impact
+
+<!-- Total emissions (in grams of CO2eq) and additional considerations, such as electricity usage, go here. Edit the suggested text below accordingly -->
+
+Carbon emissions can be estimated using the [Machine Learning Impact calculator](https://mlco2.github.io/impact#compute) presented in [Lacoste et al. (2019)](https://arxiv.org/abs/1910.09700).
+
+- **Hardware Type:** [More Information Needed]
+- **Hours used:** [More Information Needed]
+- **Cloud Provider:** [More Information Needed]
+- **Compute Region:** [More Information Needed]
+- **Carbon Emitted:** [More Information Needed]
+
+## Technical Specifications [optional]
+
+### Model Architecture and Objective
+
+[More Information Needed]
+
+### Compute Infrastructure
+
+[More Information Needed]
+
+#### Hardware
+
+[More Information Needed]
+
+#### Software
+
+[More Information Needed]
+
+## Citation [optional]
+
+<!-- If there is a paper or blog post introducing the model, the APA and Bibtex information for that should go in this section. -->
+
+**BibTeX:**
+
+[More Information Needed]
+
+**APA:**
+
+[More Information Needed]
+
+## Glossary [optional]
+
+<!-- If relevant, include terms and calculations in this section that can help readers understand the model or model card. -->
+
+[More Information Needed]
+
+## More Information [optional]
+
+[More Information Needed]
+
+## Model Card Authors [optional]
+
+[More Information Needed]
+
+## Model Card Contact
+
+[More Information Needed]
+### Framework versions
+
+- PEFT 0.19.1
@@ -0,0 +1,48 @@
+{
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+  "alpha_pattern": {},
+  "arrow_config": null,
+  "auto_mapping": null,
+  "base_model_name_or_path": "Qwen/Qwen2.5-1.5B-Instruct",
+  "bias": "none",
+  "corda_config": null,
+  "ensure_weight_tying": false,
+  "eva_config": null,
+  "exclude_modules": null,
+  "fan_in_fan_out": false,
+  "inference_mode": true,
+  "init_lora_weights": true,
+  "layer_replication": null,
+  "layers_pattern": null,
+  "layers_to_transform": null,
+  "loftq_config": {},
+  "lora_alpha": 64,
+  "lora_bias": false,
+  "lora_dropout": 0.05,
+  "lora_ga_config": null,
+  "megatron_config": null,
+  "megatron_core": "megatron.core",
+  "modules_to_save": null,
+  "peft_type": "LORA",
+  "peft_version": "0.19.1",
+  "qalora_group_size": 16,
+  "r": 32,
+  "rank_pattern": {},
+  "revision": null,
+  "target_modules": [
+    "up_proj",
+    "down_proj",
+    "k_proj",
+    "gate_proj",
+    "q_proj",
+    "v_proj",
+    "o_proj"
+  ],
+  "target_parameters": null,
+  "task_type": "CAUSAL_LM",
+  "trainable_token_indices": null,
+  "use_bdlora": null,
+  "use_dora": false,
+  "use_qalora": false,
+  "use_rslora": false
+}
@@ -0,0 +1,76 @@
+{
+  "best_global_step": null,
+  "best_metric": null,
+  "best_model_checkpoint": null,
+  "epoch": 3.0,
+  "eval_steps": 500,
+  "global_step": 6,
+  "is_hyper_param_search": false,
+  "is_local_process_zero": true,
+  "is_world_process_zero": true,
+  "log_history": [
+    {
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+      "grad_norm": 0.45245930552482605,
+      "learning_rate": 2e-05,
+      "loss": 2.1483168601989746,
+      "step": 1
+    },
+    {
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+      "grad_norm": 0.5142671465873718,
+      "learning_rate": 1.6666666666666667e-05,
+      "loss": 2.2101056575775146,
+      "step": 2
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+    {
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+      "grad_norm": 0.39382970333099365,
+      "learning_rate": 1.3333333333333333e-05,
+      "loss": 2.0590713024139404,
+      "step": 3
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+    {
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+      "grad_norm": 0.45731329917907715,
+      "learning_rate": 1e-05,
+      "loss": 2.369413137435913,
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+      "grad_norm": 0.3741995692253113,
+      "learning_rate": 6.666666666666667e-06,
+      "loss": 2.152299404144287,
+      "step": 5
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+    {
+      "epoch": 3.0,
+      "grad_norm": 0.38987115025520325,
+      "learning_rate": 3.3333333333333333e-06,
+      "loss": 2.0988121032714844,
+      "step": 6
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+  ],
+  "logging_steps": 1,
+  "max_steps": 6,
+  "num_input_tokens_seen": 0,
+  "num_train_epochs": 3,
+  "save_steps": 500,
+  "stateful_callbacks": {
+    "TrainerControl": {
+      "args": {
+        "should_epoch_stop": false,
+        "should_evaluate": false,
+        "should_log": false,
+        "should_save": true,
+        "should_training_stop": true
+      },
+      "attributes": {}
+    }
+  },
+  "total_flos": 2276355718923264.0,
+  "train_batch_size": 1,
+  "trial_name": null,
+  "trial_params": null
+}
@@ -0,0 +1,30 @@
+{
+  "add_prefix_space": false,
+  "backend": "tokenizers",
+  "bos_token": null,
+  "clean_up_tokenization_spaces": false,
+  "eos_token": "<|im_end|>",
+  "errors": "replace",
+  "extra_special_tokens": [
+    "<|im_start|>",
+    "<|im_end|>",
+    "<|object_ref_start|>",
+    "<|object_ref_end|>",
+    "<|box_start|>",
+    "<|box_end|>",
+    "<|quad_start|>",
+    "<|quad_end|>",
+    "<|vision_start|>",
+    "<|vision_end|>",
+    "<|vision_pad|>",
+    "<|image_pad|>",
+    "<|video_pad|>"
+  ],
+  "is_local": false,
+  "local_files_only": false,
+  "model_max_length": 131072,
+  "pad_token": "<|endoftext|>",
+  "split_special_tokens": false,
+  "tokenizer_class": "Qwen2Tokenizer",
+  "unk_token": null
+}
@@ -0,0 +1,45 @@
+{
+  "schema_version": 1,
+  "experiment": "8-9",
+  "stage": "student_sft",
+  "status": "complete",
+  "generated_at_utc": "2026-08-01T06:30:16.654563+00:00",
+  "book_git_commit": "88a5c5e86da9f9b0f97a46b4d501d715dc741e5a",
+  "training_data": {
+    "path": "/home/ubuntu/ai-agent-book/chapter8/cot-distillation/data/sft_cot_distill_aime_kimi_k3.jsonl",
+    "sha256": "0c6cab7cb8e0bd13671eda4e2bd0dc2530f7483f09ae42e946b87600a979968e",
+    "samples": 23
+  },
+  "base_model": "Qwen/Qwen2.5-1.5B-Instruct",
+  "output_dir": "/home/ubuntu/ai-agent-book/chapter8/cot-distillation/checkpoints/exp8-9-qwen25-1.5b-kimi-k3-20260801-v1",
+  "host": {
+    "platform": "Linux-6.8.0-111-generic-x86_64-with-glibc2.35",
+    "gpu_names": [
+      "NVIDIA RTX PRO 6000 Blackwell Workstation Edition"
+    ],
+    "torch": "2.11.0+cu130"
+  },
+  "dependency_versions": {
+    "torch": "2.11.0",
+    "transformers": "5.14.1",
+    "accelerate": "1.14.0",
+    "peft": "0.19.1"
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+    "learning_rate": 2e-05,
+    "max_length": 20000,
+    "batch_size": 1,
+    "gradient_accumulation": 16,
+    "lora_rank": 32,
+    "seed": 42,
+    "metrics": {
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+      "total_flos": 2276355718923264.0,
+      "train_loss": 2.173003077507019,
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+  }
+}
@@ -0,0 +1,24 @@
+{"id": "m01", "question": "一个书架有 5 层，每层放 28 本书。已经借出了 37 本，书架上还剩多少本书？", "gold_answer": 103, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步计算：\n\n**第一步：计算书架原本有多少本书**\n- 书架有5层\n- 每层放28本书\n- 总共的书：5 × 28 = 140本\n\n**第二步：计算借出后还剩多少本**\n- 原本有140本书\n- 借出了37本\n- 还剩：140 - 37 = 103本\n\nFinal Answer: 103", "reasoning": "让我一步步解决这个问题。\n\n1. 首先计算书架原本有多少本书：\n   - 书架有5层\n   - 每层放28本书\n   - 总共有：5 × 28 = 140本书\n\n2. 已经借出了37本书\n\n3. 书架上还剩多少本书：\n   - 140 - 37 = 103本书\n\n所以答案是103。", "verified": true, "usage": {"completion_tokens": 248, "prompt_tokens": 121, "total_tokens": 369, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 89, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.004083, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.004083, "upstream_inference_prompt_cost": 0.000363, "upstream_inference_completions_cost": 0.00372}}, "error": null}
+{"id": "m02", "question": "小明买了 3 支单价 4.5 元的笔和 2 本单价 12 元的笔记本，付了 50 元，应找回多少元？", "gold_answer": 12.5, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步计算小明应找回多少钱：\n\n**第一步：计算3支笔的总价**\n- 单价：4.5元/支\n- 数量：3支\n- 总价：3 × 4.5 = 13.5元\n\n**第二步：计算2本笔记本的总价**\n- 单价：12元/本\n- 数量：2本\n- 总价：2 × 12 = 24元\n\n**第三步：计算所有商品的总价**\n- 笔的总价：13.5元\n- 笔记本的总价：24元\n- 合计：13.5 + 24 = 37.5元\n\n**第四步：计算应找回的钱**\n- 付款：50元\n- 实际花费：37.5元\n- 应找回：50 - 37.5 = 12.5元\n\nFinal Answer: 12.5", "reasoning": "让我一步步计算：\n\n1. 首先计算3支笔的总价：\n   3 × 4.5 = 13.5元\n\n2. 然后计算2本笔记本的总价：\n   2 × 12 = 24元\n\n3. 计算总共花费：\n   13.5 + 24 = 37.5元\n\n4. 计算应找回的钱：\n   50 - 37.5 = 12.5元\n\n所以应该找回12.5元。", "verified": true, "usage": {"completion_tokens": 400, "prompt_tokens": 131, "total_tokens": 531, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 87, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.006393, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.006393, "upstream_inference_prompt_cost": 0.000393, "upstream_inference_completions_cost": 0.006}}, "error": null}
+{"id": "m03", "question": "一辆汽车以每小时 60 公里的速度行驶了 2.5 小时，又以每小时 80 公里的速度行驶了 1.5 小时，一共行驶了多少公里？", "gold_answer": 270, "model": "anthropic/claude-sonnet-4.5", "content": "好的，让我一步步计算这道题。\n\n**第一段行驶：**\n- 速度：60 公里/小时\n- 时间：2.5 小时\n- 路程 = 速度 × 时间 = 60 × 2.5 = 150 公里\n\n**第二段行驶：**\n- 速度：80 公里/小时\n- 时间：1.5 小时\n- 路程 = 速度 × 时间 = 80 × 1.5 = 120 公里\n\n**总路程：**\n总路程 = 第一段路程 + 第二段路程 = 150 + 120 = 270 公里\n\nFinal Answer: 270", "reasoning": "这道题需要计算两段行驶的总路程。\n\n第一段：\n速度：60 公里/小时\n时间：2.5 小时\n路程 = 速度 × 时间 = 60 × 2.5 = 150 公里\n\n第二段：\n速度：80 公里/小时\n时间：1.5 小时\n路程 = 速度 × 时间 = 80 × 1.5 = 120 公里\n\n总路程 = 第一段路程 + 第二段路程 = 150 + 120 = 270 公里", "verified": true, "usage": {"completion_tokens": 371, "prompt_tokens": 147, "total_tokens": 518, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 109, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.006006, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.006006, "upstream_inference_prompt_cost": 0.000441, "upstream_inference_completions_cost": 0.005565}}, "error": null}
+{"id": "m04", "question": "某班有 48 名学生，其中 3/8 参加了数学兴趣小组，参加数学兴趣小组的有多少人？", "gold_answer": 18, "model": "anthropic/claude-sonnet-4.5", "content": "让我来解决这个问题：\n\n**已知条件：**\n- 全班学生总数：48名\n- 参加数学兴趣小组的比例：3/8\n\n**计算过程：**\n参加数学兴趣小组的人数 = 总人数 × 参加比例\n\n参加人数 = 48 × 3/8\n\n参加人数 = (48 × 3) ÷ 8\n\n参加人数 = 144 ÷ 8\n\n参加人数 = 18\n\n**验证：**\n18 ÷ 48 = 3/8 ✓\n\nFinal Answer: 18", "reasoning": "这道题是一个简单的分数乘法问题。\n\n总人数：48名学生\n参加数学兴趣小组的比例：3/8\n\n参加数学兴趣小组的人数 = 总人数 × 比例\n= 48 × 3/8\n= (48 × 3) / 8\n= 144 / 8\n= 18\n\n所以参加数学兴趣小组的有18人。", "verified": true, "usage": {"completion_tokens": 308, "prompt_tokens": 126, "total_tokens": 434, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 84, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.004998, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.004998, "upstream_inference_prompt_cost": 0.000378, "upstream_inference_completions_cost": 0.00462}}, "error": null}
+{"id": "m05", "question": "一个长方形的长是宽的 3 倍，周长是 64 厘米，它的面积是多少平方厘米？", "gold_answer": 192, "model": "anthropic/claude-sonnet-4.5", "content": "设长方形的宽为 $x$ 厘米，则长为 $3x$ 厘米。\n\n根据周长公式求宽：\n- 周长 = 2(长 + 宽)\n- $64 = 2(3x + x)$\n- $64 = 2 \\times 4x$\n- $64 = 8x$\n- $x = 8$\n\n所以：\n- 宽 = 8 厘米\n- 长 = 3 × 8 = 24 厘米\n\n计算面积：\n- 面积 = 长 × 宽\n- 面积 = 24 × 8 = 192 平方厘米\n\nFinal Answer: 192", "reasoning": "设长方形的宽为 $x$ 厘米，那么长为 $3x$ 厘米。\n\n根据周长公式：\n周长 = 2(长 + 宽)\n\n所以：\n$2(3x + x) = 64$\n$2(4x) = 64$\n$8x = 64$\n$x = 8$\n\n所以宽为 8 厘米，长为 $3 \\times 8 = 24$ 厘米。\n\n面积 = 长 × 宽 = $24 \\times 8 = 192$ 平方厘米。", "verified": true, "usage": {"completion_tokens": 366, "prompt_tokens": 121, "total_tokens": 487, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 88, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.005853, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.005853, "upstream_inference_prompt_cost": 0.000363, "upstream_inference_completions_cost": 0.00549}}, "error": null}
+{"id": "m06", "question": "商店把一件商品先提价 20%，再降价 20%，现价是 96 元。这件商品的原价是多少元？", "gold_answer": 100, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步推理这道题。\n\n设这件商品的原价为 $x$ 元。\n\n**第一步：提价 20%**\n- 提价后的价格 = $x \\times (1 + 20\\%)$\n- 提价后的价格 = $x \\times 1.2$\n\n**第二步：再降价 20%**\n- 注意：这里的降价 20% 是基于提价后的价格\n- 降价后的价格 = $x \\times 1.2 \\times (1 - 20\\%)$\n- 降价后的价格 = $x \\times 1.2 \\times 0.8$\n- 降价后的价格 = $x \\times 0.96$\n\n**第三步：列方程**\n- 根据题意，现价是 96 元，所以：\n$$x \\times 0.96 = 96$$\n\n**第四步：解方程**\n$$x = \\frac{96}{0.96} = \\frac{96}{\\frac{96}{100}} = \\frac{96 \\times 100}{96} = 100$$\n\n**验证：**\n- 原价：100 元\n- 提价 20%：100 × 1.2 = 120 元\n- 降价 20%：120 × 0.8 = 96 元 ✓\n\nFinal Answer: 100", "reasoning": "设这件商品的原价是 $x$ 元。\n\n第一步：先提价 20%\n提价后的价格 = $x \\times (1 + 20\\%) = x \\times 1.2$\n\n第二步：再降价 20%\n注意这里的降价 20% 是基于提价后的价格，不是原价。\n降价后的价格 = $x \\times 1.2 \\times (1 - 20\\%) = x \\times 1.2 \\times 0.8$\n\n第三步：列方程\n根据题意，降价后的现价是 96 元，所以：\n$x \\times 1.2 \\times 0.8 = 96$\n\n第四步：解方程\n$x \\times 0.96 = 96$\n$x = 96 \\div 0.96$\n$x = 100$\n\n验证：\n原价 100 元\n提价 20%：100 × 1.2 = 120 元\n降价 20%：120 × 0.8 = 96 元 ✓\n\n所以原价是 100 元。", "verified": true, "usage": {"completion_tokens": 677, "prompt_tokens": 122, "total_tokens": 799, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 187, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010521, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010521, "upstream_inference_prompt_cost": 0.000366, "upstream_inference_completions_cost": 0.010155}}, "error": null}
+{"id": "m07", "question": "5 台同样的机器 8 小时可以生产 600 个零件。照这样计算，8 台机器 10 小时可以生产多少个零件？", "gold_answer": 1200, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步推理这道工程问题：\n\n**第一步：求单台机器每小时的生产效率**\n\n5台机器8小时生产600个零件\n\n1台机器8小时生产：600 ÷ 5 = 120（个）\n\n1台机器1小时生产：120 ÷ 8 = 15（个）\n\n**第二步：计算8台机器10小时的产量**\n\n8台机器1小时生产：15 × 8 = 120（个）\n\n8台机器10小时生产：120 × 10 = 1200（个）\n\n**验证：**\n- 5台机器8小时 = 40个\"机器·小时\"，生产600个零件\n- 单位效率 = 600 ÷ 40 = 15个/（机器·小时）\n- 8台机器10小时 = 80个\"机器·小时\"\n- 产量 = 80 × 15 = 1200个 ✓\n\nFinal Answer: 1200", "reasoning": "让我一步步分析这道题：\n\n已知：5台机器8小时生产600个零件\n求：8台机器10小时生产多少个零件\n\n第一步：计算1台机器1小时生产多少个零件（即单位效率）\n5台机器8小时生产600个零件\n1台机器8小时生产：600 ÷ 5 = 120个零件\n1台机器1小时生产：120 ÷ 8 = 15个零件\n\n第二步：计算8台机器10小时生产多少个零件\n1台机器1小时生产15个零件\n8台机器1小时生产：15 × 8 = 120个零件\n8台机器10小时生产：120 × 10 = 1200个零件\n\n让我验证一下：\n- 5台机器8小时 = 5 × 8 = 40 台·小时，生产600个零件\n- 单位效率 = 600 ÷ 40 = 15 个零件/(台·小时)\n- 8台机器10小时 = 8 × 10 = 80 台·小时\n- 生产零件数 = 80 × 15 = 1200个零件\n\n答案是1200个零件。", "verified": true, "usage": {"completion_tokens": 651, "prompt_tokens": 131, "total_tokens": 782, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 256, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010158, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010158, "upstream_inference_prompt_cost": 0.000393, "upstream_inference_completions_cost": 0.009765}}, "error": null}
+{"id": "m08", "question": "某数加上它的 1/4 等于 35，这个数是多少？", "gold_answer": 28, "model": "anthropic/claude-sonnet-4.5", "content": "我来一步步解决这个问题。\n\n**设这个数为 x**\n\n根据题意可以列出方程：\n- 某数加上它的 1/4 等于 35\n- 即：x + (1/4)x = 35\n\n**化简方程：**\n- x + (1/4)x = 35\n- 可以写成：x + x/4 = 35\n- 通分：(4x + x)/4 = 35\n- 合并：5x/4 = 35\n\n**求解 x：**\n- 两边同时乘以 4：5x = 140\n- 两边同时除以 5：x = 28\n\n**验证：**\n- 28 + 28×(1/4) = 28 + 7 = 35 ✓\n\nFinal Answer: 28", "reasoning": "设这个数为 x。\n\n根据题意：某数加上它的 1/4 等于 35\n\n用数学表达式写出来就是：\nx + (1/4)x = 35\n\n化简左边：\nx + x/4 = 35\n(4x + x)/4 = 35\n5x/4 = 35\n\n两边同时乘以 4：\n5x = 140\n\n两边同时除以 5：\nx = 28\n\n验证：28 + 28×(1/4) = 28 + 7 = 35 ✓\n\n所以这个数是 28。", "verified": true, "usage": {"completion_tokens": 394, "prompt_tokens": 104, "total_tokens": 498, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 100, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.006222, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.006222, "upstream_inference_prompt_cost": 0.000312, "upstream_inference_completions_cost": 0.00591}}, "error": null}
+{"id": "m09", "question": "甲、乙两人从相距 240 公里的两地同时出发相向而行，甲每小时走 14 公里，乙每小时走 10 公里，几小时后两人相遇？", "gold_answer": 10, "model": "anthropic/claude-sonnet-4.5", "content": "让我来一步步分析这个相遇问题。\n\n**已知条件：**\n- 两地距离：240公里\n- 甲的速度：14公里/小时\n- 乙的速度：10公里/小时\n- 两人同时出发，相向而行\n\n**解题步骤：**\n\n第一步：理解相向而行的概念\n相向而行意味着两人朝对方方向前进，他们之间的距离会越来越近。\n\n第二步：计算相对速度\n由于两人相向而行，他们的相对速度等于两人速度之和：\n相对速度 = 14 + 10 = 24公里/小时\n\n第三步：计算相遇时间\n相遇时间 = 总距离 ÷ 相对速度\n相遇时间 = 240 ÷ 24 = 10小时\n\n**验证：**\n- 甲走的距离：14 × 10 = 140公里\n- 乙走的距离：10 × 10 = 100公里\n- 两人走的总距离：140 + 100 = 240公里 ✓\n\nFinal Answer: 10", "reasoning": "这是一个相遇问题。\n\n甲和乙从相距240公里的两地同时出发相向而行。\n- 甲的速度：14公里/小时\n- 乙的速度：10公里/小时\n- 两地距离：240公里\n\n相向而行时，两人的相对速度是两人速度之和。\n\n相对速度 = 甲的速度 + 乙的速度 = 14 + 10 = 24公里/小时\n\n相遇时间 = 总距离 ÷ 相对速度 = 240 ÷ 24 = 10小时\n\n验算：\n- 甲走的距离：14 × 10 = 140公里\n- 乙走的距离：10 × 10 = 100公里\n- 总距离：140 + 100 = 240公里 ✓\n\n所以两人将在10小时后相遇。", "verified": true, "usage": {"completion_tokens": 598, "prompt_tokens": 142, "total_tokens": 740, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 180, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.009396, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.009396, "upstream_inference_prompt_cost": 0.000426, "upstream_inference_completions_cost": 0.00897}}, "error": null}
+{"id": "m10", "question": "一个等差数列的第 3 项是 11，第 7 项是 27，它的第 10 项是多少？", "gold_answer": 39, "model": "anthropic/claude-sonnet-4.5", "content": "我需要求等差数列的第10项，已知第3项和第7项。\n\n设等差数列的首项为 $a_1$，公差为 $d$。\n\n**根据等差数列通项公式：**\n$$a_n = a_1 + (n-1)d$$\n\n**列出已知条件的方程：**\n\n第3项：$a_3 = a_1 + 2d = 11$ ... (1)\n\n第7项：$a_7 = a_1 + 6d = 27$ ... (2)\n\n**求公差 $d$：**\n\n方程(2) - 方程(1)：\n$$(a_1 + 6d) - (a_1 + 2d) = 27 - 11$$\n$$4d = 16$$\n$$d = 4$$\n\n**求首项 $a_1$：**\n\n将 $d = 4$ 代入方程(1)：\n$$a_1 + 2 \\times 4 = 11$$\n$$a_1 + 8 = 11$$\n$$a_1 = 3$$\n\n**求第10项：**\n$$a_{10} = a_1 + 9d = 3 + 9 \\times 4 = 3 + 36 = 39$$\n\n**验证：**\n- $a_3 = 3 + 2 \\times 4 = 11$ ✓\n- $a_7 = 3 + 6 \\times 4 = 27$ ✓\n\nFinal Answer: 39", "reasoning": "等差数列的通项公式是：$a_n = a_1 + (n-1)d$\n\n其中 $a_1$ 是首项，$d$ 是公差。\n\n已知：\n- 第3项：$a_3 = 11$\n- 第7项：$a_7 = 27$\n\n根据通项公式：\n- $a_3 = a_1 + 2d = 11$ ... (1)\n- $a_7 = a_1 + 6d = 27$ ... (2)\n\n从方程(2)减去方程(1)：\n$(a_1 + 6d) - (a_1 + 2d) = 27 - 11$\n$4d = 16$\n$d = 4$\n\n将 $d = 4$ 代入方程(1)：\n$a_1 + 2(4) = 11$\n$a_1 + 8 = 11$\n$a_1 = 3$\n\n现在求第10项：\n$a_{10} = a_1 + 9d = 3 + 9(4) = 3 + 36 = 39$\n\n让我验证一下：\n- $a_3 = 3 + 2(4) = 3 + 8 = 11$ ✓\n- $a_7 = 3 + 6(4) = 3 + 24 = 27$ ✓\n- $a_{10} = 3 + 9(4) = 3 + 36 = 39$\n\n答案是39。", "verified": true, "usage": {"completion_tokens": 829, "prompt_tokens": 117, "total_tokens": 946, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 179, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.012786, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.012786, "upstream_inference_prompt_cost": 0.000351, "upstream_inference_completions_cost": 0.012435}}, "error": null}
+{"id": "m11", "question": "农场里鸡和兔共有 35 个头、94 只脚，兔有多少只？", "gold_answer": 12, "model": "anthropic/claude-sonnet-4.5", "content": "我来用方程法解决这个鸡兔同笼问题。\n\n设鸡有 $x$ 只，兔有 $y$ 只。\n\n**建立方程组：**\n\n根据头的数量：\n$$x + y = 35 \\quad \\text{(方程1)}$$\n\n根据脚的数量（鸡有2只脚，兔有4只脚）：\n$$2x + 4y = 94 \\quad \\text{(方程2)}$$\n\n**求解方程组：**\n\n从方程1得到：\n$$x = 35 - y$$\n\n将其代入方程2：\n$$2(35 - y) + 4y = 94$$\n\n展开：\n$$70 - 2y + 4y = 94$$\n\n合并同类项：\n$$70 + 2y = 94$$\n\n解得：\n$$2y = 24$$\n$$y = 12$$\n\n**验证：**\n- 兔有 12 只，鸡有 $35 - 12 = 23$ 只\n- 头数：$23 + 12 = 35$ ✓\n- 脚数：$23 \\times 2 + 12 \\times 4 = 46 + 48 = 94$ ✓\n\nFinal Answer: 12", "reasoning": "设鸡有 $x$ 只，兔有 $y$ 只。\n\n根据题意：\n- 鸡和兔共有 35 个头：$x + y = 35$\n- 鸡和兔共有 94 只脚：$2x + 4y = 94$（鸡有2只脚，兔有4只脚）\n\n从第一个方程得：$x = 35 - y$\n\n代入第二个方程：\n$2(35 - y) + 4y = 94$\n$70 - 2y + 4y = 94$\n$70 + 2y = 94$\n$2y = 24$\n$y = 12$\n\n验证：\n鸡有 $35 - 12 = 23$ 只\n脚数：$23 \\times 2 + 12 \\times 4 = 46 + 48 = 94$ ✓\n头数：$23 + 12 = 35$ ✓\n\n所以兔有 12 只。", "verified": true, "usage": {"completion_tokens": 629, "prompt_tokens": 111, "total_tokens": 740, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 140, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.009768, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.009768, "upstream_inference_prompt_cost": 0.000333, "upstream_inference_completions_cost": 0.009435}}, "error": null}
+{"id": "m12", "question": "一项工程，甲单独做 12 天完成，乙单独做 18 天完成。两人合作 4 天后，剩下的工程由乙单独完成，还需要多少天？", "gold_answer": 8, "model": "anthropic/claude-sonnet-4.5", "content": "让我逐步分析这道工程问题。\n\n**第一步：确定工作效率**\n- 甲单独做12天完成，甲的工作效率 = 1/12\n- 乙单独做18天完成，乙的工作效率 = 1/18\n\n**第二步：计算两人合作的工作效率**\n两人合作的效率 = 1/12 + 1/18\n\n找公分母：\n- 1/12 = 3/36\n- 1/18 = 2/36\n- 合作效率 = 3/36 + 2/36 = 5/36\n\n**第三步：计算合作4天完成的工作量**\n合作4天完成 = 4 × 5/36 = 20/36 = 5/9\n\n**第四步：计算剩余工作量**\n剩余工作量 = 1 - 5/9 = 4/9\n\n**第五步：计算乙单独完成剩余工作需要的天数**\n乙单独完成剩余工作需要的天数 = (4/9) ÷ (1/18)\n= 4/9 × 18\n= 72/9\n= 8天\n\nFinal Answer: 8", "reasoning": "让我一步步解决这个问题。\n\n设这项工程的总工作量为1。\n\n甲单独做12天完成，所以甲每天完成 1/12 的工作量。\n乙单独做18天完成，所以乙每天完成 1/18 的工作量。\n\n两人合作4天，完成的工作量为：\n4 × (1/12 + 1/18)\n\n先计算 1/12 + 1/18：\n1/12 + 1/18 = 3/36 + 2/36 = 5/36\n\n所以两人合作4天完成：\n4 × 5/36 = 20/36 = 5/9\n\n剩下的工程量为：\n1 - 5/9 = 4/9\n\n剩下的工程由乙单独完成，乙每天完成 1/18，所以需要的天数为：\n(4/9) ÷ (1/18) = (4/9) × 18 = 4 × 18/9 = 4 × 2 = 8\n\n所以还需要8天。", "verified": true, "usage": {"completion_tokens": 658, "prompt_tokens": 141, "total_tokens": 799, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 190, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010293, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010293, "upstream_inference_prompt_cost": 0.000423, "upstream_inference_completions_cost": 0.00987}}, "error": null}
+{"id": "m13", "question": "某商品进价 80 元，按标价卖出可赚 25%。若按标价的 9 折出售，每件可赚多少元？", "gold_answer": 10, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步来解决这个问题。\n\n**已知条件：**\n- 进价：80元\n- 按标价卖出可赚25%\n- 求：按标价的9折出售，每件可赚多少元\n\n**第一步：计算标价**\n\n按标价卖出可赚25%，意思是利润率为25%。\n\n利润 = 进价 × 25% = 80 × 0.25 = 20元\n\n标价 = 进价 + 利润 = 80 + 20 = 100元\n\n**第二步：计算9折售价**\n\n9折售价 = 标价 × 0.9 = 100 × 0.9 = 90元\n\n**第三步：计算9折出售时的利润**\n\n利润 = 售价 - 进价 = 90 - 80 = 10元\n\nFinal Answer: 10", "reasoning": "让我来一步步解决这个问题。\n\n已知信息：\n- 进价：80元\n- 按标价卖出可赚25%\n- 问题：按标价的9折出售，每件可赚多少元？\n\n第一步：计算标价\n按标价卖出可赚25%，意思是利润是进价的25%。\n利润 = 80 × 25% = 80 × 0.25 = 20元\n标价 = 进价 + 利润 = 80 + 20 = 100元\n\n第二步：计算9折售价\n9折售价 = 标价 × 0.9 = 100 × 0.9 = 90元\n\n第三步：计算9折出售时的利润\n利润 = 售价 - 进价 = 90 - 80 = 10元\n\n所以按标价的9折出售，每件可赚10元。", "verified": true, "usage": {"completion_tokens": 510, "prompt_tokens": 125, "total_tokens": 635, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 176, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.008025, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.008025, "upstream_inference_prompt_cost": 0.000375, "upstream_inference_completions_cost": 0.00765}}, "error": null}
+{"id": "m14", "question": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？", "gold_answer": 70, "model": "anthropic/claude-sonnet-4.5", "content": "我来一步步解决这个比例分配问题。\n\n**已知条件：**\n- 总共有126本书\n- 按4:5的比例分给甲、乙两个班\n- 其中甲班占4份，乙班占5份\n\n**解题步骤：**\n\n1. 计算总份数\n   - 总份数 = 4 + 5 = 9份\n\n2. 计算每份对应的书本数\n   - 每份 = 126 ÷ 9 = 14本\n\n3. 计算乙班分到的书本数\n   - 乙班占5份\n   - 乙班分到 = 5 × 14 = 70本\n\n**验证：**\n- 甲班：4 × 14 = 56本\n- 乙班：5 × 14 = 70本\n- 合计：56 + 70 = 126本 ✓\n\nFinal Answer: 70", "reasoning": "这道题是关于比例分配的问题。\n\n已知：\n- 总共有126本书\n- 按4:5的比例分给甲、乙两个班\n- 需要求乙班分到多少本\n\n解题步骤：\n1. 首先理解比例4:5的含义：甲班得4份，乙班得5份\n2. 总份数 = 4 + 5 = 9份\n3. 每份对应的书本数 = 126 ÷ 9 = 14本\n4. 乙班分到的书本数 = 5 × 14 = 70本\n\n让我验证一下：\n- 甲班：4 × 14 = 56本\n- 乙班：5 × 14 = 70本\n- 总计：56 + 70 = 126本 ✓\n\n答案是70本。", "verified": true, "usage": {"completion_tokens": 510, "prompt_tokens": 113, "total_tokens": 623, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 154, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.007989, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.007989, "upstream_inference_prompt_cost": 0.000339, "upstream_inference_completions_cost": 0.00765}}, "error": null}
+{"id": "m15", "question": "一个水池，单开进水管 6 小时注满，单开出水管 10 小时放空。两管同时打开，多少小时可以注满空水池？", "gold_answer": 15, "model": "anthropic/claude-sonnet-4.5", "content": "让我逐步分析这道工程问题。\n\n**设定单位：**\n将水池的总容量设为单位\"1\"\n\n**计算各管道的工作效率：**\n- 进水管单独工作6小时注满水池\n  进水管每小时的工作效率 = 1/6\n\n- 出水管单独工作10小时放空水池\n  出水管每小时的工作效率 = 1/10\n\n**计算两管同时打开的净效率：**\n两管同时打开时，每小时的净进水量 = 进水量 - 出水量\n\n净效率 = 1/6 - 1/10\n\n找最小公倍数进行计算：\n- 6和10的最小公倍数是30\n- 1/6 = 5/30\n- 1/10 = 3/30\n- 净效率 = 5/30 - 3/30 = 2/30 = 1/15\n\n**计算注满时间：**\n注满水池所需时间 = 水池总容量 ÷ 净效率\n= 1 ÷ (1/15)\n= 15小时\n\nFinal Answer: 15", "reasoning": "让我来分析这道题目。\n\n设水池的容量为1（单位体积）。\n\n单开进水管6小时注满，那么进水管每小时的进水速度是：1/6\n\n单开出水管10小时放空，那么出水管每小时的出水速度是：1/10\n\n两管同时打开时，每小时的净进水速度是：\n进水速度 - 出水速度 = 1/6 - 1/10\n\n计算 1/6 - 1/10：\n找最小公倍数，6和10的最小公倍数是30\n1/6 = 5/30\n1/10 = 3/30\n1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15\n\n所以每小时净进水是 1/15 的水池容量。\n\n要注满整个水池（容量为1），需要的时间是：\n1 ÷ (1/15) = 15小时\n\n答案是15小时。", "verified": true, "usage": {"completion_tokens": 625, "prompt_tokens": 131, "total_tokens": 756, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 206, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.009768, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.009768, "upstream_inference_prompt_cost": 0.000393, "upstream_inference_completions_cost": 0.009375}}, "error": null}
+{"id": "m16", "question": "小华今年的年龄是小丽的 3 倍，8 年后小华的年龄是小丽的 2 倍。小华今年多少岁？", "gold_answer": 24, "model": "anthropic/claude-sonnet-4.5", "content": "我来逐步分析这道年龄问题。\n\n**设定未知数：**\n- 设小华今年的年龄为 x 岁\n- 设小丽今年的年龄为 y 岁\n\n**根据题意列方程：**\n\n条件1：小华今年的年龄是小丽的 3 倍\n$$x = 3y$$\n\n条件2：8 年后小华的年龄是小丽的 2 倍\n$$x + 8 = 2(y + 8)$$\n\n**求解方程组：**\n\n将第二个方程展开：\n$$x + 8 = 2y + 16$$\n$$x = 2y + 8$$\n\n现在有两个关于 x 的表达式：\n- $x = 3y$\n- $x = 2y + 8$\n\n令两式相等：\n$$3y = 2y + 8$$\n$$y = 8$$\n\n代入第一个方程求 x：\n$$x = 3y = 3 \\times 8 = 24$$\n\n**验证：**\n- 今年：小华 24 岁，小丽 8 岁，24 = 3 × 8 ✓\n- 8 年后：小华 32 岁，小丽 16 岁，32 = 2 × 16 ✓\n\nFinal Answer: 24", "reasoning": "设小华今年的年龄为 $x$ 岁，小丽今年的年龄为 $y$ 岁。\n\n根据题意：\n1. 小华今年的年龄是小丽的 3 倍：$x = 3y$\n2. 8 年后小华的年龄是小丽的 2 倍：$x + 8 = 2(y + 8)$\n\n从第二个方程：\n$x + 8 = 2y + 16$\n$x = 2y + 8$\n\n现在我们有两个方程：\n- $x = 3y$\n- $x = 2y + 8$\n\n因此：\n$3y = 2y + 8$\n$y = 8$\n\n所以：\n$x = 3y = 3 \\times 8 = 24$\n\n验证：\n- 今年小华 24 岁，小丽 8 岁，24 = 3 × 8 ✓\n- 8 年后小华 32 岁，小丽 16 岁，32 = 2 × 16 ✓\n\n所以小华今年 24 岁。", "verified": true, "usage": {"completion_tokens": 697, "prompt_tokens": 130, "total_tokens": 827, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 171, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010845, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010845, "upstream_inference_prompt_cost": 0.00039, "upstream_inference_completions_cost": 0.010455}}, "error": null}
+{"id": "m17", "question": "某工厂第一季度生产零件 2400 个，第二季度比第一季度增产 15%，两个季度一共生产了多少个零件？", "gold_answer": 5160, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步解决这道题：\n\n**第一步：确定第一季度产量**\n- 第一季度生产零件：2400个\n\n**第二步：计算第二季度产量**\n- 第二季度比第一季度增产15%\n- 第二季度产量 = 第一季度产量 × (1 + 15%)\n- 第二季度产量 = 2400 × (1 + 0.15)\n- 第二季度产量 = 2400 × 1.15\n- 第二季度产量 = 2760个\n\n**第三步：计算两个季度总产量**\n- 总产量 = 第一季度产量 + 第二季度产量\n- 总产量 = 2400 + 2760\n- 总产量 = 5160个\n\nFinal Answer: 5160", "reasoning": "让我一步步分析这道题：\n\n第一季度生产零件：2400个\n\n第二季度比第一季度增产15%，意思是第二季度生产的数量是第一季度的(1 + 15%) = 1.15倍\n\n第二季度生产零件数 = 2400 × (1 + 15%)\n= 2400 × 1.15\n= 2760个\n\n两个季度一共生产：\n= 2400 + 2760\n= 5160个\n\n所以答案是5160", "verified": true, "usage": {"completion_tokens": 391, "prompt_tokens": 129, "total_tokens": 520, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 106, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.006252, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.006252, "upstream_inference_prompt_cost": 0.000387, "upstream_inference_completions_cost": 0.005865}}, "error": null}
+{"id": "m18", "question": "一个两位数的个位数字比十位数字大 3，这个两位数等于它的两个数字之和的 4 倍，这个两位数是多少？", "gold_answer": 36, "model": "anthropic/claude-sonnet-4.5", "content": "让我设这个两位数的十位数字为 $a$，个位数字为 $b$。\n\n**根据题意列方程：**\n\n条件1：个位数字比十位数字大 3\n$$b = a + 3$$\n\n条件2：这个两位数等于它的两个数字之和的 4 倍\n$$10a + b = 4(a + b)$$\n\n**解方程：**\n\n将条件2展开：\n$$10a + b = 4a + 4b$$\n\n移项整理：\n$$10a - 4a = 4b - b$$\n$$6a = 3b$$\n$$2a = b$$\n\n**联立两个关系式：**\n\n从 $b = a + 3$ 和 $2a = b$，得：\n$$2a = a + 3$$\n$$a = 3$$\n\n因此：\n$$b = 2 \\times 3 = 6$$\n\n**求两位数：**\n$$10a + b = 10 \\times 3 + 6 = 36$$\n\n**验证：**\n- 个位数字 6 比十位数字 3 大 3 ✓\n- 数字之和：$3 + 6 = 9$，$36 = 4 \\times 9 = 36$ ✓\n\nFinal Answer: 36", "reasoning": "设这个两位数的十位数字为 $a$，个位数字为 $b$。\n\n根据题意：\n1. 个位数字比十位数字大 3：$b = a + 3$\n2. 这个两位数等于它的两个数字之和的 4 倍：$10a + b = 4(a + b)$\n\n从第二个条件：\n$10a + b = 4a + 4b$\n$10a - 4a = 4b - b$\n$6a = 3b$\n$2a = b$\n\n现在我们有两个关系式：\n- $b = a + 3$\n- $2a = b$\n\n将第一个式子代入第二个：\n$2a = a + 3$\n$a = 3$\n\n因此：\n$b = a + 3 = 3 + 3 = 6$\n\n这个两位数是：$10a + b = 10 \\times 3 + 6 = 36$\n\n验证：\n1. 个位数字 6 比十位数字 3 大 3 ✓\n2. 两个数字之和：$3 + 6 = 9$，$9 \\times 4 = 36$ ✓\n\n答案是 36。", "verified": true, "usage": {"completion_tokens": 705, "prompt_tokens": 131, "total_tokens": 836, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 198, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010968, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010968, "upstream_inference_prompt_cost": 0.000393, "upstream_inference_completions_cost": 0.010575}}, "error": null}
+{"id": "m19", "question": "从 1 到 100 的所有整数中，能被 3 整除但不能被 5 整除的数有多少个？", "gold_answer": 27, "model": "anthropic/claude-sonnet-4.5", "content": "好的，让我一步步分析这个问题。\n\n**第一步：找出1到100中能被3整除的数**\n\n能被3整除的数是：3, 6, 9, 12, ..., 99\n\n这些数的个数为：⌊100 ÷ 3⌋ = ⌊33.33...⌋ = 33个\n\n**第二步：找出既能被3整除又能被5整除的数**\n\n如果一个数既能被3整除又能被5整除，那么它必须能被15整除（因为3和5互质，所以最小公倍数是3×5=15）\n\n能被15整除的数是：15, 30, 45, 60, 75, 90\n\n这些数的个数为：⌊100 ÷ 15⌋ = ⌊6.66...⌋ = 6个\n\n**第三步：计算能被3整除但不能被5整除的数**\n\n使用排除法：\n- 能被3整除的数：33个\n- 减去既能被3整除又能被5整除的数：6个\n- 结果：33 - 6 = 27个\n\n**验证：**\n能被3整除但不能被5整除的数包括：\n3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, 99\n共27个 ✓\n\nFinal Answer: 27", "reasoning": "我需要找出从1到100中能被3整除但不能被5整除的数有多少个。\n\n首先，让我找出能被3整除的数有多少个。\n从1到100中，能被3整除的数是：3, 6, 9, 12, ..., 99\n这些数可以表示为3k，其中k = 1, 2, 3, ..., 33\n所以能被3整除的数有：⌊100/3⌋ = 33个\n\n接下来，我需要从这33个数中排除那些也能被5整除的数。\n如果一个数既能被3整除又能被5整除，那么它能被15整除（因为3和5互质）。\n\n从1到100中，能被15整除的数是：15, 30, 45, 60, 75, 90\n这些数可以表示为15k，其中k = 1, 2, 3, 4, 5, 6\n所以能被15整除的数有：⌊100/15⌋ = 6个\n\n因此，能被3整除但不能被5整除的数有：33 - 6 = 27个\n\n让我验证一下：\n能被3整除的数：3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99\n共33个\n\n其中能被5整除的（即能被15整除的）：15, 30, 45, 60, 75, 90\n共6个\n\n排除这些后，剩下的是：3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, 99，共27个。答案确实是27。", "verified": true, "usage": {"completion_tokens": 1045, "prompt_tokens": 119, "total_tokens": 1164, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 359, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.016032, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.016032, "upstream_inference_prompt_cost": 0.000357, "upstream_inference_completions_cost": 0.015675}}, "error": null}
+{"id": "m20", "question": "甲容器中有浓度 20% 的盐水 300 克，乙容器中有浓度 10% 的盐水 200 克，混合后盐水的浓度是多少？（用百分数表示）", "gold_answer": 16, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步计算混合后盐水的浓度。\n\n**第一步：计算甲容器中盐的质量**\n- 甲容器盐水：300 克，浓度 20%\n- 甲容器中盐的质量 = 300 × 20% = 300 × 0.2 = 60 克\n\n**第二步：计算乙容器中盐的质量**\n- 乙容器盐水：200 克，浓度 10%\n- 乙容器中盐的质量 = 200 × 10% = 200 × 0.1 = 20 克\n\n**第三步：计算混合后盐的总质量**\n- 混合后盐的总质量 = 60 + 20 = 80 克\n\n**第四步：计算混合后盐水的总质量**\n- 混合后盐水的总质量 = 300 + 200 = 500 克\n\n**第五步：计算混合后的浓度**\n- 混合后的浓度 = 盐的总质量 / 盐水的总质量 × 100%\n- 混合后的浓度 = 80 / 500 × 100% = 0.16 × 100% = 16%\n\nFinal Answer: 16%", "reasoning": "让我一步步计算混合后盐水的浓度。\n\n甲容器：\n- 盐水总量：300 克\n- 浓度：20%\n- 盐的质量：300 × 20% = 300 × 0.2 = 60 克\n\n乙容器：\n- 盐水总量：200 克\n- 浓度：10%\n- 盐的质量：200 × 10% = 200 × 0.1 = 20 克\n\n混合后：\n- 盐的总质量：60 + 20 = 80 克\n- 盐水的总质量：300 + 200 = 500 克\n- 浓度：80 / 500 = 0.16 = 16%\n\n所以混合后盐水的浓度是 16%。", "verified": true, "usage": {"completion_tokens": 588, "prompt_tokens": 147, "total_tokens": 735, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 133, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.009261, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.009261, "upstream_inference_prompt_cost": 0.000441, "upstream_inference_completions_cost": 0.00882}}, "error": null}
+{"id": "m21", "question": "某快递公司收费标准：首重 1 公斤收 8 元，之后每增加 1 公斤加收 3 元（不足 1 公斤按 1 公斤计）。寄一件 7.3 公斤的包裹要多少元？", "gold_answer": 29, "model": "anthropic/claude-sonnet-4.5", "content": "让我逐步计算这件7.3公斤包裹的快递费用。\n\n**第一步：确定首重费用**\n- 首重1公斤收费：8元\n\n**第二步：计算超出首重的部分**\n- 包裹总重：7.3公斤\n- 超出首重部分：7.3 - 1 = 6.3公斤\n\n**第三步：按照\"不足1公斤按1公斤计\"的规则处理**\n- 6.3公斤不足7公斤，按7公斤计算\n\n**第四步：计算超重部分费用**\n- 超重部分：7公斤\n- 每公斤加收3元\n- 超重费用：7 × 3 = 21元\n\n**第五步：计算总费用**\n- 总费用 = 首重费用 + 超重费用\n- 总费用 = 8 + 21 = 29元\n\nFinal Answer: 29", "reasoning": "让我逐步分析这个快递收费问题。\n\n收费标准：\n- 首重1公斤收8元\n- 之后每增加1公斤加收3元\n- 不足1公斤按1公斤计\n\n包裹重量：7.3公斤\n\n步骤：\n1. 首先，首重1公斤收费：8元\n\n2. 剩余重量：7.3 - 1 = 6.3公斤\n\n3. 因为不足1公斤按1公斤计，6.3公斤要按7公斤计算\n\n4. 超出首重部分的费用：7公斤 × 3元/公斤 = 21元\n\n5. 总费用：8元 + 21元 = 29元\n\n让我再验证一下：\n- 首重1公斤：8元\n- 超出部分：7.3 - 1 = 6.3公斤，按7公斤计算\n- 超出部分费用：7 × 3 = 21元\n- 总费用：8 + 21 = 29元\n\n答案是29元。", "verified": true, "usage": {"completion_tokens": 592, "prompt_tokens": 162, "total_tokens": 754, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 203, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.009366, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.009366, "upstream_inference_prompt_cost": 0.000486, "upstream_inference_completions_cost": 0.00888}}, "error": null}
+{"id": "m22", "question": "一个圆的半径扩大到原来的 2 倍，它的面积是 314 平方厘米（π 取 3.14）。原来的半径是多少厘米？", "gold_answer": 5, "model": "anthropic/claude-sonnet-4.5", "content": "让我一步步推理这道题。\n\n**设定变量：**\n设原来圆的半径为 r 厘米。\n\n**理解题意：**\n- 半径扩大到原来的 2 倍后，新半径 = 2r 厘米\n- 扩大后圆的面积 = 314 平方厘米\n\n**根据圆面积公式求解：**\n\n扩大后圆的面积公式：\n$$S = \\pi \\times (2r)^2$$\n\n$$S = \\pi \\times 4r^2$$\n\n$$S = 4\\pi r^2$$\n\n**代入已知条件：**\n\n$$4\\pi r^2 = 314$$\n\n$$4 \\times 3.14 \\times r^2 = 314$$\n\n$$12.56 \\times r^2 = 314$$\n\n**求解 r²：**\n\n$$r^2 = \\frac{314}{12.56}$$\n\n$$r^2 = 25$$\n\n**求解 r：**\n\n$$r = 5$$\n\n**验证：**\n- 原半径：5 厘米\n- 扩大后半径：10 厘米\n- 扩大后面积：3.14 × 10² = 3.14 × 100 = 314 平方厘米 ✓\n\nFinal Answer: 5", "reasoning": "设原来的半径为 r 厘米。\n\n圆的半径扩大到原来的 2 倍后，新半径为 2r 厘米。\n\n新圆的面积是 π(2r)² = π × 4r² = 4πr² 平方厘米。\n\n已知新圆的面积是 314 平方厘米，π 取 3.14，所以：\n4πr² = 314\n4 × 3.14 × r² = 314\n12.56 × r² = 314\nr² = 314 ÷ 12.56\nr² = 25\nr = 5\n\n所以原来的半径是 5 厘米。\n\n让我验证一下：\n原来的半径是 5 厘米\n扩大到原来的 2 倍后，新半径是 10 厘米\n新圆的面积 = π × 10² = 3.14 × 100 = 314 平方厘米 ✓\n\n答案正确。", "verified": true, "usage": {"completion_tokens": 644, "prompt_tokens": 136, "total_tokens": 780, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 173, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.010068, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.010068, "upstream_inference_prompt_cost": 0.000408, "upstream_inference_completions_cost": 0.00966}}, "error": null}
+{"id": "m23", "question": "连续三个偶数的和是 78，其中最大的偶数是多少？", "gold_answer": 28, "model": "anthropic/claude-sonnet-4.5", "content": "让我来一步步解决这个问题。\n\n**设定未知数：**\n设三个连续的偶数分别为 $n$、$n+2$、$n+4$（其中 $n$ 是偶数）\n\n**建立方程：**\n根据题意，三个连续偶数的和是 78：\n$$n + (n+2) + (n+4) = 78$$\n\n**求解方程：**\n$$3n + 6 = 78$$\n\n$$3n = 78 - 6$$\n\n$$3n = 72$$\n\n$$n = 24$$\n\n**找出三个偶数：**\n- 第一个偶数：$n = 24$\n- 第二个偶数：$n + 2 = 26$\n- 第三个偶数：$n + 4 = 28$\n\n**验证：**\n$24 + 26 + 28 = 78$ ✓\n\n**结论：**\n最大的偶数是 28\n\nFinal Answer: 28", "reasoning": "设三个连续的偶数分别为 $n$, $n+2$, $n+4$，其中 $n$ 是偶数。\n\n根据题意，这三个偶数的和是 78：\n$n + (n+2) + (n+4) = 78$\n\n化简：\n$3n + 6 = 78$\n$3n = 72$\n$n = 24$\n\n所以这三个连续的偶数是：24, 26, 28\n\n其中最大的偶数是 28。\n\n让我验证一下：24 + 26 + 28 = 78 ✓", "verified": true, "usage": {"completion_tokens": 450, "prompt_tokens": 106, "total_tokens": 556, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 99, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.007068, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.007068, "upstream_inference_prompt_cost": 0.000318, "upstream_inference_completions_cost": 0.00675}}, "error": null}
+{"id": "m24", "question": "某班一次数学测验的平均分是 84 分。去掉一个 60 分的成绩后，其余 24 人的平均分是 85 分。这个班原来有多少人？", "gold_answer": 25, "model": "anthropic/claude-sonnet-4.5", "content": "我来一步步解决这个问题。\n\n**设定未知数：**\n设这个班原来有 $n$ 人。\n\n**根据题意列出等量关系：**\n\n1. 原来全班的总分 = 平均分 × 人数 = $84n$ 分\n\n2. 去掉一个 60 分后：\n   - 剩余人数：$n - 1 = 24$ 人\n   - 剩余总分：$84n - 60$ 分\n   - 剩余平均分：85 分\n\n**从剩余人数可以直接得出：**\n$n - 1 = 24$\n$n = 25$\n\n**验证答案：**\n- 原来 25 人，总分 = $84 \\times 25 = 2100$ 分\n- 去掉 60 分后，剩余总分 = $2100 - 60 = 2040$ 分\n- 剩余 24 人，平均分 = $2040 \\div 24 = 85$ 分 ✓\n\n验证正确！\n\nFinal Answer: 25", "reasoning": "设这个班原来有 $n$ 人。\n\n根据题意：\n- 全班平均分是 84 分\n- 全班总分是 $84n$ 分\n- 去掉一个 60 分后，剩余 24 人\n- 所以原来有 $n$ 人，去掉 1 人后剩余 24 人，即 $n - 1 = 24$，所以 $n = 25$\n\n让我验证一下：\n- 原来 25 人，总分 $= 84 \\times 25 = 2100$ 分\n- 去掉 60 分后，剩余总分 $= 2100 - 60 = 2040$ 分\n- 剩余 24 人，平均分 $= 2040 \\div 24 = 85$ 分\n\n验证正确！\n\n所以这个班原来有 25 人。", "verified": true, "usage": {"completion_tokens": 560, "prompt_tokens": 140, "total_tokens": 700, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": 0, "reasoning_tokens": 151, "rejected_prediction_tokens": null, "image_tokens": 0}, "prompt_tokens_details": {"audio_tokens": 0, "cached_tokens": 0, "cache_write_tokens": 0, "video_tokens": 0}, "cost": 0.00882, "is_byok": false, "cost_details": {"upstream_inference_cost": 0.00882, "upstream_inference_prompt_cost": 0.00042, "upstream_inference_completions_cost": 0.0084}}, "error": null}
@@ -0,0 +1,24 @@
+{"id": "m01", "question": "一个书架有 5 层，每层放 28 本书。已经借出了 37 本，书架上还剩多少本书？", "gold_answer": 103, "model": "kimi-k3", "content": "书架总书数：5 层 × 每层 28 本 = 140 本。  \n借出 37 本后剩余：140 − 37 = 103 本。  \n\nFinal Answer: 103", "reasoning": "The user wants step-by-step reasoning in Chinese, with final answer line format. Compute total books 5*28=140. Remaining after lending 37: 140-37=103. Need final last line exactly Final Answer: 103. Ensure no unit in final answer. Provide concise steps.", "verified": true, "usage": {"completion_tokens": 125, "prompt_tokens": 142, "total_tokens": 267, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 63, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m02", "question": "小明买了 3 支单价 4.5 元的笔和 2 本单价 12 元的笔记本，付了 50 元，应找回多少元？", "gold_answer": 12.5, "model": "kimi-k3", "content": "3支笔的总价：3 × 4.5 = 13.5 元  \n2本笔记本的总价：2 × 12 = 24 元  \n总花费：13.5 + 24 = 37.5 元  \n应找回：50 − 37.5 = 12.5 元  \n\nFinal Answer: 12.5", "reasoning": "小明买了3支笔，每支4.5元，共3×4.5=13.5元。\n买了2本笔记本，每本12元，共2×12=24元。\n总共花费13.5+24=37.5元。\n付了50元，应找回50-37.5=12.5元。\n最终答案：12.5。", "verified": true, "usage": {"completion_tokens": 173, "prompt_tokens": 150, "total_tokens": 323, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 81, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m03", "question": "一辆汽车以每小时 60 公里的速度行驶了 2.5 小时，又以每小时 80 公里的速度行驶了 1.5 小时，一共行驶了多少公里？", "gold_answer": 270, "model": "kimi-k3", "content": "第一段路程：60 × 2.5 = 150  \n第二段路程：80 × 1.5 = 120  \n总路程：150 + 120 = 270  \n\nFinal Answer: 270", "reasoning": "The user wants a step-by-step reasoning in Chinese, with final line format \"Final Answer: 数值\" only numeric no unit.\n\nCompute: 60 km/h * 2.5 h = 150 km. 80 km/h * 1.5 h = 120 km. Total = 270 km.\nNeed final answer only numeric. Ensure last line exactly Final Answer: 270. Could include steps before. No unit in final. Use Chinese. Need not over-explain too much but step-by-step.\n", "verified": true, "usage": {"completion_tokens": 165, "prompt_tokens": 152, "total_tokens": 317, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 107, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m04", "question": "某班有 48 名学生，其中 3/8 参加了数学兴趣小组，参加数学兴趣小组的有多少人？", "gold_answer": 18, "model": "kimi-k3", "content": "全班有 48 名学生，参加数学兴趣小组的占 \\( \\frac{3}{8} \\)。\n\n计算人数：  \n\\[\n48 \\times \\frac{3}{8} = 6 \\times 3 = 18\n\\]\n\nFinal Answer: 18", "reasoning": "题目：48名学生，3/8参加数学兴趣小组。计算48*3/8 = 6*3 =18。需要一步步推理，最后一行严格格式 Final Answer: 数值，只写数值不带单位。用中文简洁说明。确保最后一行只有 Final Answer: 18。", "verified": true, "usage": {"completion_tokens": 135, "prompt_tokens": 139, "total_tokens": 274, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 63, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m05", "question": "一个长方形的长是宽的 3 倍，周长是 64 厘米，它的面积是多少平方厘米？", "gold_answer": 192, "model": "kimi-k3", "content": "设长方形的宽为 \\(w\\) 厘米，则长为 \\(3w\\) 厘米。\n\n周长公式：  \n\\[\n2(\\text{长}+\\text{宽})=64\n\\]\n\n代入得：  \n\\[\n2(3w+w)=64\n\\]\n\\[\n2\\cdot 4w=64\n\\]\n\\[\n8w=64\n\\]\n\\[\nw=8\n\\]\n\n所以宽是 8 厘米，长是：  \n\\[\n3w=3\\times 8=24\n\\]\n\n面积为：  \n\\[\n24\\times 8=192\n\\]\n\nFinal Answer: 192", "reasoning": "用户问了一个长方形面积问题。需要一步步推理，最后格式 Final Answer: 数值（不带单位）。\n\n题目：长是宽的3倍，周长64厘米。设宽为 w，长为 3w。周长 2(长+宽)=2(3w+w)=8w=64，所以 w=8 厘米，长=24 厘米。面积=24*8=192 平方厘米。\n\n需要输出一步步推理，最后一行精确格式。确保最后一行只写 Final Answer: 192。前面可以带单位解释。最终答案行不带单位。用中文。", "verified": true, "usage": {"completion_tokens": 274, "prompt_tokens": 137, "total_tokens": 411, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 127, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m06", "question": "商店把一件商品先提价 20%，再降价 20%，现价是 96 元。这件商品的原价是多少元？", "gold_answer": 100, "model": "kimi-k3", "content": "设原价为 \\(x\\) 元。\n\n先提价 20% 后，价格变为：\n\\[\nx \\times (1+20\\%)=1.2x\n\\]\n\n再降价 20%，是在提价后的价格基础上降价，所以现价为：\n\\[\n1.2x \\times (1-20\\%)=1.2x \\times 0.8=0.96x\n\\]\n\n已知现价是 96 元：\n\\[\n0.96x=96\n\\]\n\n解得：\n\\[\nx=\\frac{96}{0.96}=100\n\\]\n\nFinal Answer: 100", "reasoning": "The user wants step-by-step reasoning and final answer format. Need compute: original price P. Increase 20% => 1.2P. Then decrease 20% => 1.2P*0.8 = 0.96P. Current price 96, so 0.96P=96 => P=100. Need answer in Chinese probably. Ensure last line exactly Final Answer: 100. Need maybe concise step-by-step. Let's craft with equations. No unit in final. Final answer only numeric after colon. They said 只写数值，不带单位 for final answer; last line format. Good.", "verified": true, "usage": {"completion_tokens": 276, "prompt_tokens": 141, "total_tokens": 417, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 130, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m07", "question": "5 台同样的机器 8 小时可以生产 600 个零件。照这样计算，8 台机器 10 小时可以生产多少个零件？", "gold_answer": 1200, "model": "kimi-k3", "content": "第一步：求 1 台机器 1 小时生产多少个零件。\n\n5 台机器 8 小时生产 600 个，  \n所以 1 台机器 8 小时生产：600 ÷ 5 = 120 个，  \n1 台机器 1 小时生产：120 ÷ 8 = 15 个。\n\n第二步：计算 8 台机器 10 小时的产量。\n\n8 台机器 1 小时生产：15 × 8 = 120 个，  \n8 台机器 10 小时生产：120 × 10 = 1200 个。\n\nFinal Answer: 1200", "reasoning": "用户要求解决一个数学问题，并一步步推理，最后以\"Final Answer: 数值\"格式给出答案。\n\n问题：5台同样的机器8小时可以生产600个零件。照这样计算，8台机器10小时可以生产多少个零件？\n\n步骤：\n1. 先求1台机器1小时生产多少个零件。\n   5台机器8小时生产600个。\n   1台机器8小时生产：600 ÷ 5 = 120个。\n   1台机器1小时生产：120 ÷ 8 = 15个。\n\n2. 8台机器10小时生产：\n   1台1小时15个。\n   8台1小时：15 × 8 = 120个。\n   8台10小时：120 × 10 = 1200个。\n\n或者用比例：\n零件数与机器台数成正比，与时间成正比。\n600 × (8/5) × (10/8) = 600 × 8/5 × 10/8\n= 600 × (8×10)/(5×8) = 600 × 80/40 = 600 × 2 = 1200。\n\n最终答案：1200。\n\n需要确保最后一行格式正确：Final Answer: 1200\n不带单位。\n", "verified": true, "usage": {"completion_tokens": 419, "prompt_tokens": 147, "total_tokens": 566, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 266, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m08", "question": "某数加上它的 1/4 等于 35，这个数是多少？", "gold_answer": 28, "model": "kimi-k3", "content": "设这个数为 \\(x\\)。\n\n根据题意：\n\\[\nx+\\frac14 x=35\n\\]\n\n合并左边：\n\\[\n\\frac54 x=35\n\\]\n\n两边同乘 \\(\\frac45\\)：\n\\[\nx=35\\times \\frac45=28\n\\]\n\n检验：\\(28+28\\times\\frac14=28+7=35\\)，正确。\n\nFinal Answer: 28", "reasoning": "The problem in Chinese: \"某数加上它的 1/4 等于 35，这个数是多少？\" Translation: A number plus its 1/4 equals 35. What is the number?\n\nLet the number be x.\nx + (1/4)x = 35\n(5/4)x = 35\nx = 35 * (4/5) = 28.\n\nCheck: 28 + 28/4 = 28 + 7 = 35. Correct.\n\nNeed to respond with step-by-step reasoning in Chinese, and final line format \"Final Answer: 数值\" only value no unit. So last line: Final Answer: 28. Ensure no extra after? The user requested last line with format. We can have reasoning then final line.\n\nNeed final answer only. Keep concise but step-by-step.", "verified": true, "usage": {"completion_tokens": 273, "prompt_tokens": 130, "total_tokens": 403, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 174, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m09", "question": "甲、乙两人从相距 240 公里的两地同时出发相向而行，甲每小时走 14 公里，乙每小时走 10 公里，几小时后两人相遇？", "gold_answer": 10, "model": "kimi-k3", "content": "两人相向而行，接近速度为两人速度之和：\n\n甲速度：14 公里/小时  \n乙速度：10 公里/小时  \n相对速度：14 + 10 = 24 公里/小时\n\n两地相距 240 公里，所以相遇时间为：\n\n240 ÷ 24 = 10\n\nFinal Answer: 10", "reasoning": "甲和乙相向而行，相距240公里。甲速度14公里/小时，乙速度10公里/小时。相向而行，相对速度为14+10=24公里/小时。相遇时间=路程/相对速度=240/24=10小时。需要最后一行格式 Final Answer: 数值，只写数值不带单位。可以一步步推理。最终行必须是 Final Answer: 10。确保不带单位。", "verified": true, "usage": {"completion_tokens": 172, "prompt_tokens": 150, "total_tokens": 322, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 88, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m10", "question": "一个等差数列的第 3 项是 11，第 7 项是 27，它的第 10 项是多少？", "gold_answer": 39, "model": "kimi-k3", "content": "设等差数列的公差为 \\(d\\)。\n\n1. 第 3 项与第 7 项相差 \\(7-3=4\\) 个公差：\n\\[\n4d=27-11=16\n\\]\n\n2. 求得公差：\n\\[\nd=\\frac{16}{4}=4\n\\]\n\n3. 第 10 项比第 7 项多 \\(3\\) 个公差：\n\\[\na_{10}=27+3\\times4=39\n\\]\n\nFinal Answer: 39", "reasoning": "We need answer Chinese, step-by-step, final exact format only numeric. Need solve arithmetic progression a3=11 a7=27. Difference 4d=16 d=4. a10=a7+3d=27+12=39. Need likely concise. Last line Final Answer: 39. Ensure last line exactly. Chinese.", "verified": true, "usage": {"completion_tokens": 206, "prompt_tokens": 144, "total_tokens": 350, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 72, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m11", "question": "农场里鸡和兔共有 35 个头、94 只脚，兔有多少只？", "gold_answer": 12, "model": "kimi-k3", "content": "设鸡有 \\(x\\) 只，兔有 \\(y\\) 只。\n\n根据头的数量：\n\\[x + y = 35\\]\n\n根据脚的数量（鸡 2 只脚，兔 4 只脚）：\n\\[2x + 4y = 94\\]\n\n由第一式得 \\(x = 35 - y\\)，代入第二式：\n\\[2(35 - y) + 4y = 94\\]\n\\[70 - 2y + 4y = 94\\]\n\\[70 + 2y = 94\\]\n\\[2y = 24\\]\n\\[y = 12\\]\n\n验证：兔 12 只（48 只脚），鸡 23 只（46 只脚），共 35 个头、94 只脚，符合题意。\n\nFinal Answer: 12", "reasoning": "The user wants me to solve a classic chicken and rabbit problem. Let me work through it.\n\nChickens have 1 head and 2 feet. Rabbits have 1 head and 4 feet.\n\nLet chickens = c, rabbits = r.\nc + r = 35 (heads)\n2c + 4r = 94 (feet)\n\nFrom first equation: c = 35 - r\nSubstitute: 2(35 - r) + 4r = 94\n70 - 2r + 4r = 94\n70 + 2r = 94\n2r = 24\nr = 12\n\nSo there are 12 rabbits.\n\nLet me verify: 12 rabbits = 48 feet, 23 chickens = 46 feet. Total = 94 feet. Heads = 12 + 23 = 35. Correct.\n\nThe user wants step-by-step reasoning in Chinese, and the last line should be \"Final Answer: 数值\" with only the number, no units.", "verified": true, "usage": {"completion_tokens": 404, "prompt_tokens": 133, "total_tokens": 537, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 214, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m12", "question": "一项工程，甲单独做 12 天完成，乙单独做 18 天完成。两人合作 4 天后，剩下的工程由乙单独完成，还需要多少天？", "gold_answer": 8, "model": "kimi-k3", "content": "甲每天完成工程的 \\( \\frac{1}{12} \\)，乙每天完成工程的 \\( \\frac{1}{18} \\)。\n\n两人合作一天完成：\n\\[\n\\frac{1}{12}+\\frac{1}{18}=\\frac{3}{36}+\\frac{2}{36}=\\frac{5}{36}\n\\]\n\n合作 4 天完成：\n\\[\n4 \\times \\frac{5}{36}=\\frac{20}{36}=\\frac{5}{9}\n\\]\n\n剩余工程：\n\\[\n1-\\frac{5}{9}=\\frac{4}{9}\n\\]\n\n由乙单独完成，需要：\n\\[\n\\frac{4}{9}\\div \\frac{1}{18}=\\frac{4}{9}\\times 18=8\n\\]\n\nFinal Answer: 8", "reasoning": "甲单独做12天完成，每天完成 1/12。\n乙单独做18天完成，每天完成 1/18。\n两人合作4天，完成 (1/12+1/18)*4。\n1/12+1/18 = 3/36+2/36=5/36。\n合作4天完成 20/36=5/9。\n剩余 1-5/9=4/9。\n乙单独完成需要 (4/9)/(1/18)= (4/9)*18=8 天。\n最终答案 8。", "verified": true, "usage": {"completion_tokens": 299, "prompt_tokens": 152, "total_tokens": 451, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 121, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m13", "question": "某商品进价 80 元，按标价卖出可赚 25%。若按标价的 9 折出售，每件可赚多少元？", "gold_answer": 10, "model": "kimi-k3", "content": "进价是 80 元。  \n按标价卖出可赚 25%，即标价比进价多 25%：\n\n标价 = 80 × (1 + 25%) = 80 × 1.25 = 100 元\n\n按标价的 9 折出售：\n\n售价 = 100 × 90% = 90 元\n\n每件可赚：\n\n90 - 80 = 10 元\n\nFinal Answer: 10", "reasoning": "We need answer in Chinese, step-by-step, final line exact format \"Final Answer: 数值\" only number no unit. Need solve: cost 80, marked price sold earns 25% => profit 25% of cost? Chinese \"按标价卖出可赚25%\" usually means profit margin on cost? In elementary math, 可赚25% means earns 25% of cost (利润率 relative to 进价), so marked price = 80*(1+25%)=100. Sell at 90% of marked price =90. Profit per item=90-80=10. Need maybe note if \"赚25%\" could be ambiguous but conventional. Provide concise step-by-step. Ensure final line only Final Answer: 10. No extra after. Need include units? final only number no unit. Good.", "verified": true, "usage": {"completion_tokens": 279, "prompt_tokens": 146, "total_tokens": 425, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 170, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m14", "question": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？", "gold_answer": 70, "model": "kimi-k3", "content": "总份数：4 + 5 = 9\n\n每份数量：126 ÷ 9 = 14\n\n乙班占 5 份：14 × 5 = 70\n\nFinal Answer: 70", "reasoning": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？\n\n总份数 = 4 + 5 = 9\n每份 = 126 ÷ 9 = 14\n乙班占 5 份 = 14 × 5 = 70\n\nFinal Answer: 70", "verified": true, "usage": {"completion_tokens": 134, "prompt_tokens": 137, "total_tokens": 271, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 73, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m15", "question": "一个水池，单开进水管 6 小时注满，单开出水管 10 小时放空。两管同时打开，多少小时可以注满空水池？", "gold_answer": 15, "model": "kimi-k3", "content": "设水池容量为 1。\n\n1. 进水管 6 小时注满，所以进水速度为每小时注入水池的：  \n   \\[\n   \\frac{1}{6}\n   \\]\n\n2. 出水管 10 小时放空，所以出水速度为每小时放掉水池的：  \n   \\[\n   \\frac{1}{10}\n   \\]\n\n3. 两管同时打开时，实际净注水速度为：  \n   \\[\n   \\frac{1}{6}-\\frac{1}{10}\n   =\\frac{5}{30}-\\frac{3}{30}\n   =\\frac{2}{30}\n   =\\frac{1}{15}\n   \\]\n\n4. 因此注满空水池所需时间为：  \n   \\[\n   1\\div \\frac{1}{15}=15\n   \\]\n\nFinal Answer: 15", "reasoning": "这是一个基础的水池进出水问题。\n\n进水管单独6小时注满，所以进水速度是 1/6 池/小时。\n出水管单独10小时放空，所以出水速度是 1/10 池/小时。\n\n两管同时打开，净速度 = 1/6 - 1/10。\n\n计算：1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15 池/小时。\n\n注满时间 = 1 ÷ (1/15) = 15 小时。\n\n需要按要求一步步推理，最后以 \"Final Answer: 数值\" 格式输出，只写数值不带单位。", "verified": true, "usage": {"completion_tokens": 346, "prompt_tokens": 149, "total_tokens": 495, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 146, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m16", "question": "小华今年的年龄是小丽的 3 倍，8 年后小华的年龄是小丽的 2 倍。小华今年多少岁？", "gold_answer": 24, "model": "kimi-k3", "content": "设小丽今年 \\(x\\) 岁，则小华今年 \\(3x\\) 岁。\n\n8 年后：\n- 小丽：\\(x+8\\) 岁\n- 小华：\\(3x+8\\) 岁\n\n根据“8 年后小华的年龄是小丽的 2 倍”列方程：\n\n\\[\n3x+8=2(x+8)\n\\]\n\n展开：\n\n\\[\n3x+8=2x+16\n\\]\n\n移项得：\n\n\\[\nx=8\n\\]\n\n所以小华今年：\n\n\\[\n3x=3\\times 8=24\n\\]\n\n验证：8 年后小华 32 岁，小丽 16 岁，32 正好是 16 的 2 倍，正确。\n\nFinal Answer: 24", "reasoning": "用户要求解一道年龄问题，需要一步步推理，最后按指定格式输出。\n\n设小丽今年 x 岁，小华今年 3x 岁。\n8 年后：小丽 x+8，小华 3x+8。\n条件：3x+8 = 2(x+8) = 2x+16。\n解得 x = 8。\n小华今年 3x = 24 岁。\n验证：8年后小华32，小丽16，32=2*16，正确。\n最终只写数值 24。\n需要包含一步步推理，最后行 Final Answer: 24。注意用户说最后一行只写数值不带单位，格式为「Final Answer: 数值」。所以最后一行应是 Final Answer: 24。推理过程可以带单位。确保最后一行 exact。", "verified": true, "usage": {"completion_tokens": 352, "prompt_tokens": 143, "total_tokens": 495, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 170, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m17", "question": "某工厂第一季度生产零件 2400 个，第二季度比第一季度增产 15%，两个季度一共生产了多少个零件？", "gold_answer": 5160, "model": "kimi-k3", "content": "第一季度生产：2400 个\n\n第二季度比第一季度增产 15%，所以第二季度生产：\n2400 × (1 + 15%) = 2400 × 1.15 = 2760 个\n\n两个季度一共生产：\n2400 + 2760 = 5160 个\n\nFinal Answer: 5160", "reasoning": "The user is asking a math word problem in Chinese. Let me solve it step by step.\n\n第一季度生产零件 2400 个\n第二季度比第一季度增产 15%\n第二季度产量 = 2400 × (1 + 15%) = 2400 × 1.15 = 2760\n两个季度一共 = 2400 + 2760 = 5160\n\nLet me verify: 2400 × 0.15 = 360, so second quarter = 2400 + 360 = 2760. Total = 2400 + 2760 = 5160.\n\nFinal Answer: 5160", "verified": true, "usage": {"completion_tokens": 213, "prompt_tokens": 139, "total_tokens": 352, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 130, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m18", "question": "一个两位数的个位数字比十位数字大 3，这个两位数等于它的两个数字之和的 4 倍，这个两位数是多少？", "gold_answer": 36, "model": "kimi-k3", "content": "设十位数字为 \\(x\\)，个位数字为 \\(y\\)。\n\n1. 个位数字比十位数字大 3：  \n   \\[\n   y=x+3\n   \\]\n\n2. 这个两位数等于两个数字之和的 4 倍：  \n   \\[\n   10x+y=4(x+y)\n   \\]\n\n3. 代入 \\(y=x+3\\)：  \n   \\[\n   10x+(x+3)=4(x+x+3)\n   \\]\n   \\[\n   11x+3=4(2x+3)=8x+12\n   \\]\n\n4. 解方程：  \n   \\[\n   11x+3=8x+12\n   \\]\n   \\[\n   3x=9\n   \\]\n   \\[\n   x=3\n   \\]\n\n5. 个位数字：  \n   \\[\n   y=3+3=6\n   \\]\n\n所以这个两位数是 36。检验：\\(3+6=9\\)，\\(9\\times4=36\\)，符合条件。\n\nFinal Answer: 36", "reasoning": "用户要求解一个数学题，并一步步推理，最后一行用指定格式给答案。\n\n题目：一个两位数的个位数字比十位数字大3，这个两位数等于它的两个数字之和的4倍。\n\n设十位数字为 t，个位数字为 u。\n条件1：u = t + 3。\n两位数数值 = 10t + u。\n数字和 = t + u。\n条件2：10t + u = 4(t+u)。\n代入 u=t+3：10t + t+3 = 4(t + t+3) = 4(2t+3)=8t+12。\n左边 11t+3 = 8t+12 => 3t=9 => t=3。u=6。两位数=36。检验：数字和9，4倍=36。正确。\n\n需要最终 answer 最后一行 \"Final Answer: 36\"。只写数值不带单位，指数值部分只写36；格式中包含 Final Answer: 。可以。\n\n需要用中文一步步推理。确保最后一行精确格式。", "verified": true, "usage": {"completion_tokens": 471, "prompt_tokens": 143, "total_tokens": 614, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 220, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m19", "question": "从 1 到 100 的所有整数中，能被 3 整除但不能被 5 整除的数有多少个？", "gold_answer": 27, "model": "kimi-k3", "content": "能被 3 整除的数有：⌊100/3⌋ = 33 个。  \n其中同时能被 5 整除，即能被 15 整除的数有：⌊100/15⌋ = 6 个。  \n所以能被 3 整除但不能被 5 整除的数有：33 − 6 = 27 个。\n\nFinal Answer: 27", "reasoning": "用户要求中文回答，并一步步推理，最后格式 Final Answer: 数值。需要计算1到100整数中能被3整除但不能被5整除的个数。\n\n计算：能被3整除：floor(100/3)=33（3到99）。能被3且能被5即能被15整除：floor(100/15)=6（15,30,45,60,75,90）。所求=33-6=27。\n\n需要简洁逐步推理。最后行精确格式。确保最后一行只有 Final Answer: 27。前面可以有步骤。不要在最终答案后添加内容。", "verified": true, "usage": {"completion_tokens": 233, "prompt_tokens": 143, "total_tokens": 376, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 127, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m20", "question": "甲容器中有浓度 20% 的盐水 300 克，乙容器中有浓度 10% 的盐水 200 克，混合后盐水的浓度是多少？（用百分数表示）", "gold_answer": 16, "model": "kimi-k3", "content": "甲容器中盐的质量：\\(300 \\times 20\\% = 60\\) 克。  \n乙容器中盐的质量：\\(200 \\times 10\\% = 20\\) 克。  \n混合后盐的总质量：\\(60 + 20 = 80\\) 克。  \n混合后盐水总质量：\\(300 + 200 = 500\\) 克。  \n浓度：\\(\\frac{80}{500} \\times 100\\% = 16\\%\\)。  \n\nFinal Answer: 16%", "reasoning": "We need answer in Chinese likely. Need solve stepwise. Compute: salt in A =20%*300=60 g. Salt in B=10%*200=20 g. Total salt=80 g. Total solution=500 g. Concentration=80/500=0.16=16%. Need final last line exactly \"Final Answer: 16%\"? User says 用百分数表示 and final line format 「Final Answer: 数值」 only write value no unit. But says 用百分数表示. They specifically: 最后一行用「Final Answer: 数值」的格式给出最终答案（只写数值，不带单位）. Since unit is percent? They say no unit, but concentration expressed as percentage. Ambiguous: Should final answer be 16% or 16? \"只写数值，不带单位\" means write numeric value no unit; but they asked use percent. The format placeholder 数值 maybe include %? They say no unit; percent sign is unit. However \"用百分数表示\" suggests final answer numeric with percent sign. Last line format \"Final Answer: 数值\" and \"只写数值，不带单位\" probably from benchmark wants e.g. Final Answer: 16%. Yet \"不带单位\" maybe no 克. For percent, sign is required? Let's parse Chinese: \"混合后盐水的浓度是多少？（用百分数表示）\" Then: \"请一步步推理，并在最后一行用「Final Answer: 数值」的格式给出最终答案（只写数值，不带单位）。\" If final answer only value no unit, since asked in percent, the numerical value is 16 (percent). They explicitly say no unit; percent sign could be considered unit. But many tasks expect Final Answer: 16% because they ask percent. The phrase 「Final Answer: 数值」 with 数值 could be 16%? They add \"只写数值，不带单位\" meaning if answer is 16%, don't write \"16%克\" or \"16 percent\". In prior similar tasks, they demand final answer only numerical no units. Need choose. It says \"用百分数表示\" so final should be a percent; \"只写数值\" maybe no Chinese unit like 克. The last line: Final Answer: 16%. This includes % not a unit? It is a percent sign. They said 不带单位; percent is the requested representation. I'd output steps and final line Final Answer: 16%. Ensure last line only that. Could mention concentration=16%. Final answer last line exactly. If strict no unit maybe should be 16. But problem says use percentage; I think 16% is okay and matches. The final answer field \"数值\" can include percent sign? Hmm. The phrase \"只写数值，不带单位\" strongly indicates not write unit; but then a percentage without % is just number 16, which is the numerical value of the percentage. Many math word problems ask \"浓度是多少（用百分数表示）\" answer \"16%\"; if they say no unit they still often expect \"16%\". The final answer format examples in these prompts usually accept \"Final Answer: 16%\". I'll use 16% because requested percent. Need step-by-step concise.\n\nNeed ensure no extra after final line. Use Chinese. Steps:\n1. 甲盐质量 300×20%=60克\n2. 乙盐质量 200×10%=20克\n3. 总盐 80克，总盐水 500克\n4. 浓度=80/500×100%=16%\nFinal Answer: 16%\n\nNeed last line exact. Good.", "verified": true, "usage": {"completion_tokens": 879, "prompt_tokens": 156, "total_tokens": 1035, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 755, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m21", "question": "某快递公司收费标准：首重 1 公斤收 8 元，之后每增加 1 公斤加收 3 元（不足 1 公斤按 1 公斤计）。寄一件 7.3 公斤的包裹要多少元？", "gold_answer": 29, "model": "kimi-k3", "content": "1. 包裹实际重量为 7.3 公斤。\n2. 因为“不足 1 公斤按 1 公斤计”，所以计费重量按 8 公斤算。\n3. 首重 1 公斤收费 8 元。\n4. 超出首重的部分为：8 - 1 = 7 公斤。\n5. 超出部分费用：7 × 3 = 21 元。\n6. 总费用：8 + 21 = 29 元。\n\nFinal Answer: 29", "reasoning": "用户要求解一个快递收费问题，并明确要求一步步推理，最后以特定格式输出。需要准确计算。\n\n题目：首重1公斤收8元，之后每增加1公斤加收3元（不足1公斤按1公斤计）。包裹7.3公斤。\n\n计算：不足1公斤按1公斤计，所以计费重量为 ceil(7.3)=8 公斤。首重1公斤8元，其余增加公斤数 = 8-1=7 公斤，每公斤3元，加收21元。总费用=8+21=29元。\n\n需要注意：是否“之后每增加1公斤”对于7.3公斤，另一种理解：首重1kg，续重部分6.3kg，不足1kg按1kg计 -> 续重按7kg计，费用8+7*3=29。结果相同。\n\n需要输出中文一步步推理，最后最后一行严格是「Final Answer: 29」。确保最后一行只写数值不带单位，按照格式包含 Final Answer:。不要多余内容在最后一行。可以前面有步骤。最终答案最后一行格式正确。", "verified": true, "usage": {"completion_tokens": 361, "prompt_tokens": 167, "total_tokens": 528, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 233, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m22", "question": "一个圆的半径扩大到原来的 2 倍，它的面积是 314 平方厘米（π 取 3.14）。原来的半径是多少厘米？", "gold_answer": 5, "model": "kimi-k3", "content": "设原来的半径为 \\(r\\) 厘米。\n\n半径扩大到原来的 2 倍后，新半径为 \\(2r\\) 厘米。\n\n扩大后的面积为：\n\\[\n\\pi (2r)^2 = 314\n\\]\n\n代入 \\(\\pi = 3.14\\)：\n\\[\n3.14 \\times 4r^2 = 314\n\\]\n\n\\[\n12.56r^2 = 314\n\\]\n\n\\[\nr^2 = 314 \\div 12.56 = 25\n\\]\n\n\\[\nr = 5\n\\]\n\nFinal Answer: 5", "reasoning": "The problem: A circle's radius is expanded to 2 times original. Its area is 314 cm² (π = 3.14). Find original radius.\n\nLet original radius = r. New radius = 2r. New area = π(2r)² = 4πr² = 314.\nWith π = 3.14: 4 × 3.14 × r² = 314.\n12.56 r² = 314.\nr² = 314 / 12.56 = 25.\nr = 5.\n\nCheck: original radius 5, new radius 10, area = 3.14 × 100 = 314. Correct.\n\nNeed to provide step-by-step reasoning in Chinese, last line format \"Final Answer: 数值\" only number no unit. So final line: Final Answer: 5.\n\nNeed to ensure last line exactly. Use Chinese explanation then final answer. No extra after final answer.\n", "verified": true, "usage": {"completion_tokens": 337, "prompt_tokens": 145, "total_tokens": 482, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 199, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m23", "question": "连续三个偶数的和是 78，其中最大的偶数是多少？", "gold_answer": 28, "model": "kimi-k3", "content": "设这三个连续偶数中最小的为 \\(x\\)，则另外两个分别为 \\(x+2\\)、\\(x+4\\)。\n\n根据题意：\n\\[\nx+(x+2)+(x+4)=78\n\\]\n\\[\n3x+6=78\n\\]\n\\[\n3x=72\n\\]\n\\[\nx=24\n\\]\n\n所以三个连续偶数是 24、26、28，其中最大的偶数是 28。\n\nFinal Answer: 28", "reasoning": "The user is asking in Chinese: The sum of three consecutive even numbers is 78, what is the largest even number?\n\nLet me solve step by step.\n\nLet the three consecutive even numbers be n, n+2, n+4.\nSum = n + (n+2) + (n+4) = 3n + 6 = 78\n3n = 72\nn = 24\nSo the numbers are 24, 26, 28.\nThe largest is 28.\n\nAlternatively, middle number = 78/3 = 26, largest = 28.\n\nFinal Answer: 28\n", "verified": true, "usage": {"completion_tokens": 241, "prompt_tokens": 128, "total_tokens": 369, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 130, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
+{"id": "m24", "question": "某班一次数学测验的平均分是 84 分。去掉一个 60 分的成绩后，其余 24 人的平均分是 85 分。这个班原来有多少人？", "gold_answer": 25, "model": "kimi-k3", "content": "去掉 1 个成绩后，其余人数是 24 人，所以原来人数为：\n\n24 + 1 = 25\n\n验证：原来总分 = 25 × 84 = 2100；去掉 60 分后总分 = 2040；其余 24 人平均分 = 2040 ÷ 24 = 85，符合题意。\n\nFinal Answer: 25", "reasoning": "The problem: A class math test average is 84. After removing one score of 60, the remaining 24 people have average 85. How many people originally?\n\nIf after removing one person there are 24 people, originally there were 25 people. Let's verify: total original = 25*84 = 2100. Remove 60 -> 2040. Remaining 24 average = 2040/24 = 85. Yes consistent.\n\nNeed provide step-by-step reasoning in Chinese and final line exactly \"Final Answer: 25\". Ensure last line format. Could mention original人数 = 24+1=25; also verify with average. But maybe trick: The statement says average is 84; removing 60 leaves 24 people average 85. If there are 24 remaining, original =25. The averages are consistent. Need answer only numeric in final.\n\nLet's craft concise stepwise. Last line must be Final Answer: 数值. No extra after. Ensure no unit. Use Chinese. Done.", "verified": true, "usage": {"completion_tokens": 309, "prompt_tokens": 152, "total_tokens": 461, "completion_tokens_details": {"accepted_prediction_tokens": null, "audio_tokens": null, "reasoning_tokens": 211, "rejected_prediction_tokens": null}, "prompt_tokens_details": null}, "error": null}
@@ -0,0 +1,24 @@
+{"messages": [{"role": "user", "content": "一个书架有 5 层，每层放 28 本书。已经借出了 37 本，书架上还剩多少本书？"}, {"role": "assistant", "content": "<think>\n让我一步步解决这个问题。\n\n1. 首先计算书架原本有多少本书：\n   - 书架有5层\n   - 每层放28本书\n   - 总共有：5 × 28 = 140本书\n\n2. 已经借出了37本书\n\n3. 书架上还剩多少本书：\n   - 140 - 37 = 103本书\n\n所以答案是103。\n</think>\n\n让我一步步计算：\n\n**第一步：计算书架原本有多少本书**\n- 书架有5层\n- 每层放28本书\n- 总共的书：5 × 28 = 140本\n\n**第二步：计算借出后还剩多少本**\n- 原本有140本书\n- 借出了37本\n- 还剩：140 - 37 = 103本\n\nFinal Answer: 103"}]}
+{"messages": [{"role": "user", "content": "小明买了 3 支单价 4.5 元的笔和 2 本单价 12 元的笔记本，付了 50 元，应找回多少元？"}, {"role": "assistant", "content": "<think>\n让我一步步计算：\n\n1. 首先计算3支笔的总价：\n   3 × 4.5 = 13.5元\n\n2. 然后计算2本笔记本的总价：\n   2 × 12 = 24元\n\n3. 计算总共花费：\n   13.5 + 24 = 37.5元\n\n4. 计算应找回的钱：\n   50 - 37.5 = 12.5元\n\n所以应该找回12.5元。\n</think>\n\n让我一步步计算小明应找回多少钱：\n\n**第一步：计算3支笔的总价**\n- 单价：4.5元/支\n- 数量：3支\n- 总价：3 × 4.5 = 13.5元\n\n**第二步：计算2本笔记本的总价**\n- 单价：12元/本\n- 数量：2本\n- 总价：2 × 12 = 24元\n\n**第三步：计算所有商品的总价**\n- 笔的总价：13.5元\n- 笔记本的总价：24元\n- 合计：13.5 + 24 = 37.5元\n\n**第四步：计算应找回的钱**\n- 付款：50元\n- 实际花费：37.5元\n- 应找回：50 - 37.5 = 12.5元\n\nFinal Answer: 12.5"}]}
+{"messages": [{"role": "user", "content": "一辆汽车以每小时 60 公里的速度行驶了 2.5 小时，又以每小时 80 公里的速度行驶了 1.5 小时，一共行驶了多少公里？"}, {"role": "assistant", "content": "<think>\n这道题需要计算两段行驶的总路程。\n\n第一段：\n速度：60 公里/小时\n时间：2.5 小时\n路程 = 速度 × 时间 = 60 × 2.5 = 150 公里\n\n第二段：\n速度：80 公里/小时\n时间：1.5 小时\n路程 = 速度 × 时间 = 80 × 1.5 = 120 公里\n\n总路程 = 第一段路程 + 第二段路程 = 150 + 120 = 270 公里\n</think>\n\n好的，让我一步步计算这道题。\n\n**第一段行驶：**\n- 速度：60 公里/小时\n- 时间：2.5 小时\n- 路程 = 速度 × 时间 = 60 × 2.5 = 150 公里\n\n**第二段行驶：**\n- 速度：80 公里/小时\n- 时间：1.5 小时\n- 路程 = 速度 × 时间 = 80 × 1.5 = 120 公里\n\n**总路程：**\n总路程 = 第一段路程 + 第二段路程 = 150 + 120 = 270 公里\n\nFinal Answer: 270"}]}
+{"messages": [{"role": "user", "content": "某班有 48 名学生，其中 3/8 参加了数学兴趣小组，参加数学兴趣小组的有多少人？"}, {"role": "assistant", "content": "<think>\n这道题是一个简单的分数乘法问题。\n\n总人数：48名学生\n参加数学兴趣小组的比例：3/8\n\n参加数学兴趣小组的人数 = 总人数 × 比例\n= 48 × 3/8\n= (48 × 3) / 8\n= 144 / 8\n= 18\n\n所以参加数学兴趣小组的有18人。\n</think>\n\n让我来解决这个问题：\n\n**已知条件：**\n- 全班学生总数：48名\n- 参加数学兴趣小组的比例：3/8\n\n**计算过程：**\n参加数学兴趣小组的人数 = 总人数 × 参加比例\n\n参加人数 = 48 × 3/8\n\n参加人数 = (48 × 3) ÷ 8\n\n参加人数 = 144 ÷ 8\n\n参加人数 = 18\n\n**验证：**\n18 ÷ 48 = 3/8 ✓\n\nFinal Answer: 18"}]}
+{"messages": [{"role": "user", "content": "一个长方形的长是宽的 3 倍，周长是 64 厘米，它的面积是多少平方厘米？"}, {"role": "assistant", "content": "<think>\n设长方形的宽为 $x$ 厘米，那么长为 $3x$ 厘米。\n\n根据周长公式：\n周长 = 2(长 + 宽)\n\n所以：\n$2(3x + x) = 64$\n$2(4x) = 64$\n$8x = 64$\n$x = 8$\n\n所以宽为 8 厘米，长为 $3 \\times 8 = 24$ 厘米。\n\n面积 = 长 × 宽 = $24 \\times 8 = 192$ 平方厘米。\n</think>\n\n设长方形的宽为 $x$ 厘米，则长为 $3x$ 厘米。\n\n根据周长公式求宽：\n- 周长 = 2(长 + 宽)\n- $64 = 2(3x + x)$\n- $64 = 2 \\times 4x$\n- $64 = 8x$\n- $x = 8$\n\n所以：\n- 宽 = 8 厘米\n- 长 = 3 × 8 = 24 厘米\n\n计算面积：\n- 面积 = 长 × 宽\n- 面积 = 24 × 8 = 192 平方厘米\n\nFinal Answer: 192"}]}
+{"messages": [{"role": "user", "content": "商店把一件商品先提价 20%，再降价 20%，现价是 96 元。这件商品的原价是多少元？"}, {"role": "assistant", "content": "<think>\n设这件商品的原价是 $x$ 元。\n\n第一步：先提价 20%\n提价后的价格 = $x \\times (1 + 20\\%) = x \\times 1.2$\n\n第二步：再降价 20%\n注意这里的降价 20% 是基于提价后的价格，不是原价。\n降价后的价格 = $x \\times 1.2 \\times (1 - 20\\%) = x \\times 1.2 \\times 0.8$\n\n第三步：列方程\n根据题意，降价后的现价是 96 元，所以：\n$x \\times 1.2 \\times 0.8 = 96$\n\n第四步：解方程\n$x \\times 0.96 = 96$\n$x = 96 \\div 0.96$\n$x = 100$\n\n验证：\n原价 100 元\n提价 20%：100 × 1.2 = 120 元\n降价 20%：120 × 0.8 = 96 元 ✓\n\n所以原价是 100 元。\n</think>\n\n让我一步步推理这道题。\n\n设这件商品的原价为 $x$ 元。\n\n**第一步：提价 20%**\n- 提价后的价格 = $x \\times (1 + 20\\%)$\n- 提价后的价格 = $x \\times 1.2$\n\n**第二步：再降价 20%**\n- 注意：这里的降价 20% 是基于提价后的价格\n- 降价后的价格 = $x \\times 1.2 \\times (1 - 20\\%)$\n- 降价后的价格 = $x \\times 1.2 \\times 0.8$\n- 降价后的价格 = $x \\times 0.96$\n\n**第三步：列方程**\n- 根据题意，现价是 96 元，所以：\n$$x \\times 0.96 = 96$$\n\n**第四步：解方程**\n$$x = \\frac{96}{0.96} = \\frac{96}{\\frac{96}{100}} = \\frac{96 \\times 100}{96} = 100$$\n\n**验证：**\n- 原价：100 元\n- 提价 20%：100 × 1.2 = 120 元\n- 降价 20%：120 × 0.8 = 96 元 ✓\n\nFinal Answer: 100"}]}
+{"messages": [{"role": "user", "content": "5 台同样的机器 8 小时可以生产 600 个零件。照这样计算，8 台机器 10 小时可以生产多少个零件？"}, {"role": "assistant", "content": "<think>\n让我一步步分析这道题：\n\n已知：5台机器8小时生产600个零件\n求：8台机器10小时生产多少个零件\n\n第一步：计算1台机器1小时生产多少个零件（即单位效率）\n5台机器8小时生产600个零件\n1台机器8小时生产：600 ÷ 5 = 120个零件\n1台机器1小时生产：120 ÷ 8 = 15个零件\n\n第二步：计算8台机器10小时生产多少个零件\n1台机器1小时生产15个零件\n8台机器1小时生产：15 × 8 = 120个零件\n8台机器10小时生产：120 × 10 = 1200个零件\n\n让我验证一下：\n- 5台机器8小时 = 5 × 8 = 40 台·小时，生产600个零件\n- 单位效率 = 600 ÷ 40 = 15 个零件/(台·小时)\n- 8台机器10小时 = 8 × 10 = 80 台·小时\n- 生产零件数 = 80 × 15 = 1200个零件\n\n答案是1200个零件。\n</think>\n\n让我一步步推理这道工程问题：\n\n**第一步：求单台机器每小时的生产效率**\n\n5台机器8小时生产600个零件\n\n1台机器8小时生产：600 ÷ 5 = 120（个）\n\n1台机器1小时生产：120 ÷ 8 = 15（个）\n\n**第二步：计算8台机器10小时的产量**\n\n8台机器1小时生产：15 × 8 = 120（个）\n\n8台机器10小时生产：120 × 10 = 1200（个）\n\n**验证：**\n- 5台机器8小时 = 40个\"机器·小时\"，生产600个零件\n- 单位效率 = 600 ÷ 40 = 15个/（机器·小时）\n- 8台机器10小时 = 80个\"机器·小时\"\n- 产量 = 80 × 15 = 1200个 ✓\n\nFinal Answer: 1200"}]}
+{"messages": [{"role": "user", "content": "某数加上它的 1/4 等于 35，这个数是多少？"}, {"role": "assistant", "content": "<think>\n设这个数为 x。\n\n根据题意：某数加上它的 1/4 等于 35\n\n用数学表达式写出来就是：\nx + (1/4)x = 35\n\n化简左边：\nx + x/4 = 35\n(4x + x)/4 = 35\n5x/4 = 35\n\n两边同时乘以 4：\n5x = 140\n\n两边同时除以 5：\nx = 28\n\n验证：28 + 28×(1/4) = 28 + 7 = 35 ✓\n\n所以这个数是 28。\n</think>\n\n我来一步步解决这个问题。\n\n**设这个数为 x**\n\n根据题意可以列出方程：\n- 某数加上它的 1/4 等于 35\n- 即：x + (1/4)x = 35\n\n**化简方程：**\n- x + (1/4)x = 35\n- 可以写成：x + x/4 = 35\n- 通分：(4x + x)/4 = 35\n- 合并：5x/4 = 35\n\n**求解 x：**\n- 两边同时乘以 4：5x = 140\n- 两边同时除以 5：x = 28\n\n**验证：**\n- 28 + 28×(1/4) = 28 + 7 = 35 ✓\n\nFinal Answer: 28"}]}
+{"messages": [{"role": "user", "content": "甲、乙两人从相距 240 公里的两地同时出发相向而行，甲每小时走 14 公里，乙每小时走 10 公里，几小时后两人相遇？"}, {"role": "assistant", "content": "<think>\n这是一个相遇问题。\n\n甲和乙从相距240公里的两地同时出发相向而行。\n- 甲的速度：14公里/小时\n- 乙的速度：10公里/小时\n- 两地距离：240公里\n\n相向而行时，两人的相对速度是两人速度之和。\n\n相对速度 = 甲的速度 + 乙的速度 = 14 + 10 = 24公里/小时\n\n相遇时间 = 总距离 ÷ 相对速度 = 240 ÷ 24 = 10小时\n\n验算：\n- 甲走的距离：14 × 10 = 140公里\n- 乙走的距离：10 × 10 = 100公里\n- 总距离：140 + 100 = 240公里 ✓\n\n所以两人将在10小时后相遇。\n</think>\n\n让我来一步步分析这个相遇问题。\n\n**已知条件：**\n- 两地距离：240公里\n- 甲的速度：14公里/小时\n- 乙的速度：10公里/小时\n- 两人同时出发，相向而行\n\n**解题步骤：**\n\n第一步：理解相向而行的概念\n相向而行意味着两人朝对方方向前进，他们之间的距离会越来越近。\n\n第二步：计算相对速度\n由于两人相向而行，他们的相对速度等于两人速度之和：\n相对速度 = 14 + 10 = 24公里/小时\n\n第三步：计算相遇时间\n相遇时间 = 总距离 ÷ 相对速度\n相遇时间 = 240 ÷ 24 = 10小时\n\n**验证：**\n- 甲走的距离：14 × 10 = 140公里\n- 乙走的距离：10 × 10 = 100公里\n- 两人走的总距离：140 + 100 = 240公里 ✓\n\nFinal Answer: 10"}]}
+{"messages": [{"role": "user", "content": "一个等差数列的第 3 项是 11，第 7 项是 27，它的第 10 项是多少？"}, {"role": "assistant", "content": "<think>\n等差数列的通项公式是：$a_n = a_1 + (n-1)d$\n\n其中 $a_1$ 是首项，$d$ 是公差。\n\n已知：\n- 第3项：$a_3 = 11$\n- 第7项：$a_7 = 27$\n\n根据通项公式：\n- $a_3 = a_1 + 2d = 11$ ... (1)\n- $a_7 = a_1 + 6d = 27$ ... (2)\n\n从方程(2)减去方程(1)：\n$(a_1 + 6d) - (a_1 + 2d) = 27 - 11$\n$4d = 16$\n$d = 4$\n\n将 $d = 4$ 代入方程(1)：\n$a_1 + 2(4) = 11$\n$a_1 + 8 = 11$\n$a_1 = 3$\n\n现在求第10项：\n$a_{10} = a_1 + 9d = 3 + 9(4) = 3 + 36 = 39$\n\n让我验证一下：\n- $a_3 = 3 + 2(4) = 3 + 8 = 11$ ✓\n- $a_7 = 3 + 6(4) = 3 + 24 = 27$ ✓\n- $a_{10} = 3 + 9(4) = 3 + 36 = 39$\n\n答案是39。\n</think>\n\n我需要求等差数列的第10项，已知第3项和第7项。\n\n设等差数列的首项为 $a_1$，公差为 $d$。\n\n**根据等差数列通项公式：**\n$$a_n = a_1 + (n-1)d$$\n\n**列出已知条件的方程：**\n\n第3项：$a_3 = a_1 + 2d = 11$ ... (1)\n\n第7项：$a_7 = a_1 + 6d = 27$ ... (2)\n\n**求公差 $d$：**\n\n方程(2) - 方程(1)：\n$$(a_1 + 6d) - (a_1 + 2d) = 27 - 11$$\n$$4d = 16$$\n$$d = 4$$\n\n**求首项 $a_1$：**\n\n将 $d = 4$ 代入方程(1)：\n$$a_1 + 2 \\times 4 = 11$$\n$$a_1 + 8 = 11$$\n$$a_1 = 3$$\n\n**求第10项：**\n$$a_{10} = a_1 + 9d = 3 + 9 \\times 4 = 3 + 36 = 39$$\n\n**验证：**\n- $a_3 = 3 + 2 \\times 4 = 11$ ✓\n- $a_7 = 3 + 6 \\times 4 = 27$ ✓\n\nFinal Answer: 39"}]}
+{"messages": [{"role": "user", "content": "农场里鸡和兔共有 35 个头、94 只脚，兔有多少只？"}, {"role": "assistant", "content": "<think>\n设鸡有 $x$ 只，兔有 $y$ 只。\n\n根据题意：\n- 鸡和兔共有 35 个头：$x + y = 35$\n- 鸡和兔共有 94 只脚：$2x + 4y = 94$（鸡有2只脚，兔有4只脚）\n\n从第一个方程得：$x = 35 - y$\n\n代入第二个方程：\n$2(35 - y) + 4y = 94$\n$70 - 2y + 4y = 94$\n$70 + 2y = 94$\n$2y = 24$\n$y = 12$\n\n验证：\n鸡有 $35 - 12 = 23$ 只\n脚数：$23 \\times 2 + 12 \\times 4 = 46 + 48 = 94$ ✓\n头数：$23 + 12 = 35$ ✓\n\n所以兔有 12 只。\n</think>\n\n我来用方程法解决这个鸡兔同笼问题。\n\n设鸡有 $x$ 只，兔有 $y$ 只。\n\n**建立方程组：**\n\n根据头的数量：\n$$x + y = 35 \\quad \\text{(方程1)}$$\n\n根据脚的数量（鸡有2只脚，兔有4只脚）：\n$$2x + 4y = 94 \\quad \\text{(方程2)}$$\n\n**求解方程组：**\n\n从方程1得到：\n$$x = 35 - y$$\n\n将其代入方程2：\n$$2(35 - y) + 4y = 94$$\n\n展开：\n$$70 - 2y + 4y = 94$$\n\n合并同类项：\n$$70 + 2y = 94$$\n\n解得：\n$$2y = 24$$\n$$y = 12$$\n\n**验证：**\n- 兔有 12 只，鸡有 $35 - 12 = 23$ 只\n- 头数：$23 + 12 = 35$ ✓\n- 脚数：$23 \\times 2 + 12 \\times 4 = 46 + 48 = 94$ ✓\n\nFinal Answer: 12"}]}
+{"messages": [{"role": "user", "content": "一项工程，甲单独做 12 天完成，乙单独做 18 天完成。两人合作 4 天后，剩下的工程由乙单独完成，还需要多少天？"}, {"role": "assistant", "content": "<think>\n让我一步步解决这个问题。\n\n设这项工程的总工作量为1。\n\n甲单独做12天完成，所以甲每天完成 1/12 的工作量。\n乙单独做18天完成，所以乙每天完成 1/18 的工作量。\n\n两人合作4天，完成的工作量为：\n4 × (1/12 + 1/18)\n\n先计算 1/12 + 1/18：\n1/12 + 1/18 = 3/36 + 2/36 = 5/36\n\n所以两人合作4天完成：\n4 × 5/36 = 20/36 = 5/9\n\n剩下的工程量为：\n1 - 5/9 = 4/9\n\n剩下的工程由乙单独完成，乙每天完成 1/18，所以需要的天数为：\n(4/9) ÷ (1/18) = (4/9) × 18 = 4 × 18/9 = 4 × 2 = 8\n\n所以还需要8天。\n</think>\n\n让我逐步分析这道工程问题。\n\n**第一步：确定工作效率**\n- 甲单独做12天完成，甲的工作效率 = 1/12\n- 乙单独做18天完成，乙的工作效率 = 1/18\n\n**第二步：计算两人合作的工作效率**\n两人合作的效率 = 1/12 + 1/18\n\n找公分母：\n- 1/12 = 3/36\n- 1/18 = 2/36\n- 合作效率 = 3/36 + 2/36 = 5/36\n\n**第三步：计算合作4天完成的工作量**\n合作4天完成 = 4 × 5/36 = 20/36 = 5/9\n\n**第四步：计算剩余工作量**\n剩余工作量 = 1 - 5/9 = 4/9\n\n**第五步：计算乙单独完成剩余工作需要的天数**\n乙单独完成剩余工作需要的天数 = (4/9) ÷ (1/18)\n= 4/9 × 18\n= 72/9\n= 8天\n\nFinal Answer: 8"}]}
+{"messages": [{"role": "user", "content": "某商品进价 80 元，按标价卖出可赚 25%。若按标价的 9 折出售，每件可赚多少元？"}, {"role": "assistant", "content": "<think>\n让我来一步步解决这个问题。\n\n已知信息：\n- 进价：80元\n- 按标价卖出可赚25%\n- 问题：按标价的9折出售，每件可赚多少元？\n\n第一步：计算标价\n按标价卖出可赚25%，意思是利润是进价的25%。\n利润 = 80 × 25% = 80 × 0.25 = 20元\n标价 = 进价 + 利润 = 80 + 20 = 100元\n\n第二步：计算9折售价\n9折售价 = 标价 × 0.9 = 100 × 0.9 = 90元\n\n第三步：计算9折出售时的利润\n利润 = 售价 - 进价 = 90 - 80 = 10元\n\n所以按标价的9折出售，每件可赚10元。\n</think>\n\n让我一步步来解决这个问题。\n\n**已知条件：**\n- 进价：80元\n- 按标价卖出可赚25%\n- 求：按标价的9折出售，每件可赚多少元\n\n**第一步：计算标价**\n\n按标价卖出可赚25%，意思是利润率为25%。\n\n利润 = 进价 × 25% = 80 × 0.25 = 20元\n\n标价 = 进价 + 利润 = 80 + 20 = 100元\n\n**第二步：计算9折售价**\n\n9折售价 = 标价 × 0.9 = 100 × 0.9 = 90元\n\n**第三步：计算9折出售时的利润**\n\n利润 = 售价 - 进价 = 90 - 80 = 10元\n\nFinal Answer: 10"}]}
+{"messages": [{"role": "user", "content": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？"}, {"role": "assistant", "content": "<think>\n这道题是关于比例分配的问题。\n\n已知：\n- 总共有126本书\n- 按4:5的比例分给甲、乙两个班\n- 需要求乙班分到多少本\n\n解题步骤：\n1. 首先理解比例4:5的含义：甲班得4份，乙班得5份\n2. 总份数 = 4 + 5 = 9份\n3. 每份对应的书本数 = 126 ÷ 9 = 14本\n4. 乙班分到的书本数 = 5 × 14 = 70本\n\n让我验证一下：\n- 甲班：4 × 14 = 56本\n- 乙班：5 × 14 = 70本\n- 总计：56 + 70 = 126本 ✓\n\n答案是70本。\n</think>\n\n我来一步步解决这个比例分配问题。\n\n**已知条件：**\n- 总共有126本书\n- 按4:5的比例分给甲、乙两个班\n- 其中甲班占4份，乙班占5份\n\n**解题步骤：**\n\n1. 计算总份数\n   - 总份数 = 4 + 5 = 9份\n\n2. 计算每份对应的书本数\n   - 每份 = 126 ÷ 9 = 14本\n\n3. 计算乙班分到的书本数\n   - 乙班占5份\n   - 乙班分到 = 5 × 14 = 70本\n\n**验证：**\n- 甲班：4 × 14 = 56本\n- 乙班：5 × 14 = 70本\n- 合计：56 + 70 = 126本 ✓\n\nFinal Answer: 70"}]}
+{"messages": [{"role": "user", "content": "一个水池，单开进水管 6 小时注满，单开出水管 10 小时放空。两管同时打开，多少小时可以注满空水池？"}, {"role": "assistant", "content": "<think>\n让我来分析这道题目。\n\n设水池的容量为1（单位体积）。\n\n单开进水管6小时注满，那么进水管每小时的进水速度是：1/6\n\n单开出水管10小时放空，那么出水管每小时的出水速度是：1/10\n\n两管同时打开时，每小时的净进水速度是：\n进水速度 - 出水速度 = 1/6 - 1/10\n\n计算 1/6 - 1/10：\n找最小公倍数，6和10的最小公倍数是30\n1/6 = 5/30\n1/10 = 3/30\n1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15\n\n所以每小时净进水是 1/15 的水池容量。\n\n要注满整个水池（容量为1），需要的时间是：\n1 ÷ (1/15) = 15小时\n\n答案是15小时。\n</think>\n\n让我逐步分析这道工程问题。\n\n**设定单位：**\n将水池的总容量设为单位\"1\"\n\n**计算各管道的工作效率：**\n- 进水管单独工作6小时注满水池\n  进水管每小时的工作效率 = 1/6\n\n- 出水管单独工作10小时放空水池\n  出水管每小时的工作效率 = 1/10\n\n**计算两管同时打开的净效率：**\n两管同时打开时，每小时的净进水量 = 进水量 - 出水量\n\n净效率 = 1/6 - 1/10\n\n找最小公倍数进行计算：\n- 6和10的最小公倍数是30\n- 1/6 = 5/30\n- 1/10 = 3/30\n- 净效率 = 5/30 - 3/30 = 2/30 = 1/15\n\n**计算注满时间：**\n注满水池所需时间 = 水池总容量 ÷ 净效率\n= 1 ÷ (1/15)\n= 15小时\n\nFinal Answer: 15"}]}
+{"messages": [{"role": "user", "content": "小华今年的年龄是小丽的 3 倍，8 年后小华的年龄是小丽的 2 倍。小华今年多少岁？"}, {"role": "assistant", "content": "<think>\n设小华今年的年龄为 $x$ 岁，小丽今年的年龄为 $y$ 岁。\n\n根据题意：\n1. 小华今年的年龄是小丽的 3 倍：$x = 3y$\n2. 8 年后小华的年龄是小丽的 2 倍：$x + 8 = 2(y + 8)$\n\n从第二个方程：\n$x + 8 = 2y + 16$\n$x = 2y + 8$\n\n现在我们有两个方程：\n- $x = 3y$\n- $x = 2y + 8$\n\n因此：\n$3y = 2y + 8$\n$y = 8$\n\n所以：\n$x = 3y = 3 \\times 8 = 24$\n\n验证：\n- 今年小华 24 岁，小丽 8 岁，24 = 3 × 8 ✓\n- 8 年后小华 32 岁，小丽 16 岁，32 = 2 × 16 ✓\n\n所以小华今年 24 岁。\n</think>\n\n我来逐步分析这道年龄问题。\n\n**设定未知数：**\n- 设小华今年的年龄为 x 岁\n- 设小丽今年的年龄为 y 岁\n\n**根据题意列方程：**\n\n条件1：小华今年的年龄是小丽的 3 倍\n$$x = 3y$$\n\n条件2：8 年后小华的年龄是小丽的 2 倍\n$$x + 8 = 2(y + 8)$$\n\n**求解方程组：**\n\n将第二个方程展开：\n$$x + 8 = 2y + 16$$\n$$x = 2y + 8$$\n\n现在有两个关于 x 的表达式：\n- $x = 3y$\n- $x = 2y + 8$\n\n令两式相等：\n$$3y = 2y + 8$$\n$$y = 8$$\n\n代入第一个方程求 x：\n$$x = 3y = 3 \\times 8 = 24$$\n\n**验证：**\n- 今年：小华 24 岁，小丽 8 岁，24 = 3 × 8 ✓\n- 8 年后：小华 32 岁，小丽 16 岁，32 = 2 × 16 ✓\n\nFinal Answer: 24"}]}
+{"messages": [{"role": "user", "content": "某工厂第一季度生产零件 2400 个，第二季度比第一季度增产 15%，两个季度一共生产了多少个零件？"}, {"role": "assistant", "content": "<think>\n让我一步步分析这道题：\n\n第一季度生产零件：2400个\n\n第二季度比第一季度增产15%，意思是第二季度生产的数量是第一季度的(1 + 15%) = 1.15倍\n\n第二季度生产零件数 = 2400 × (1 + 15%)\n= 2400 × 1.15\n= 2760个\n\n两个季度一共生产：\n= 2400 + 2760\n= 5160个\n\n所以答案是5160\n</think>\n\n让我一步步解决这道题：\n\n**第一步：确定第一季度产量**\n- 第一季度生产零件：2400个\n\n**第二步：计算第二季度产量**\n- 第二季度比第一季度增产15%\n- 第二季度产量 = 第一季度产量 × (1 + 15%)\n- 第二季度产量 = 2400 × (1 + 0.15)\n- 第二季度产量 = 2400 × 1.15\n- 第二季度产量 = 2760个\n\n**第三步：计算两个季度总产量**\n- 总产量 = 第一季度产量 + 第二季度产量\n- 总产量 = 2400 + 2760\n- 总产量 = 5160个\n\nFinal Answer: 5160"}]}
+{"messages": [{"role": "user", "content": "一个两位数的个位数字比十位数字大 3，这个两位数等于它的两个数字之和的 4 倍，这个两位数是多少？"}, {"role": "assistant", "content": "<think>\n设这个两位数的十位数字为 $a$，个位数字为 $b$。\n\n根据题意：\n1. 个位数字比十位数字大 3：$b = a + 3$\n2. 这个两位数等于它的两个数字之和的 4 倍：$10a + b = 4(a + b)$\n\n从第二个条件：\n$10a + b = 4a + 4b$\n$10a - 4a = 4b - b$\n$6a = 3b$\n$2a = b$\n\n现在我们有两个关系式：\n- $b = a + 3$\n- $2a = b$\n\n将第一个式子代入第二个：\n$2a = a + 3$\n$a = 3$\n\n因此：\n$b = a + 3 = 3 + 3 = 6$\n\n这个两位数是：$10a + b = 10 \\times 3 + 6 = 36$\n\n验证：\n1. 个位数字 6 比十位数字 3 大 3 ✓\n2. 两个数字之和：$3 + 6 = 9$，$9 \\times 4 = 36$ ✓\n\n答案是 36。\n</think>\n\n让我设这个两位数的十位数字为 $a$，个位数字为 $b$。\n\n**根据题意列方程：**\n\n条件1：个位数字比十位数字大 3\n$$b = a + 3$$\n\n条件2：这个两位数等于它的两个数字之和的 4 倍\n$$10a + b = 4(a + b)$$\n\n**解方程：**\n\n将条件2展开：\n$$10a + b = 4a + 4b$$\n\n移项整理：\n$$10a - 4a = 4b - b$$\n$$6a = 3b$$\n$$2a = b$$\n\n**联立两个关系式：**\n\n从 $b = a + 3$ 和 $2a = b$，得：\n$$2a = a + 3$$\n$$a = 3$$\n\n因此：\n$$b = 2 \\times 3 = 6$$\n\n**求两位数：**\n$$10a + b = 10 \\times 3 + 6 = 36$$\n\n**验证：**\n- 个位数字 6 比十位数字 3 大 3 ✓\n- 数字之和：$3 + 6 = 9$，$36 = 4 \\times 9 = 36$ ✓\n\nFinal Answer: 36"}]}
+{"messages": [{"role": "user", "content": "从 1 到 100 的所有整数中，能被 3 整除但不能被 5 整除的数有多少个？"}, {"role": "assistant", "content": "<think>\n我需要找出从1到100中能被3整除但不能被5整除的数有多少个。\n\n首先，让我找出能被3整除的数有多少个。\n从1到100中，能被3整除的数是：3, 6, 9, 12, ..., 99\n这些数可以表示为3k，其中k = 1, 2, 3, ..., 33\n所以能被3整除的数有：⌊100/3⌋ = 33个\n\n接下来，我需要从这33个数中排除那些也能被5整除的数。\n如果一个数既能被3整除又能被5整除，那么它能被15整除（因为3和5互质）。\n\n从1到100中，能被15整除的数是：15, 30, 45, 60, 75, 90\n这些数可以表示为15k，其中k = 1, 2, 3, 4, 5, 6\n所以能被15整除的数有：⌊100/15⌋ = 6个\n\n因此，能被3整除但不能被5整除的数有：33 - 6 = 27个\n\n让我验证一下：\n能被3整除的数：3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99\n共33个\n\n其中能被5整除的（即能被15整除的）：15, 30, 45, 60, 75, 90\n共6个\n\n排除这些后，剩下的是：3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, 99，共27个。答案确实是27。\n</think>\n\n好的，让我一步步分析这个问题。\n\n**第一步：找出1到100中能被3整除的数**\n\n能被3整除的数是：3, 6, 9, 12, ..., 99\n\n这些数的个数为：⌊100 ÷ 3⌋ = ⌊33.33...⌋ = 33个\n\n**第二步：找出既能被3整除又能被5整除的数**\n\n如果一个数既能被3整除又能被5整除，那么它必须能被15整除（因为3和5互质，所以最小公倍数是3×5=15）\n\n能被15整除的数是：15, 30, 45, 60, 75, 90\n\n这些数的个数为：⌊100 ÷ 15⌋ = ⌊6.66...⌋ = 6个\n\n**第三步：计算能被3整除但不能被5整除的数**\n\n使用排除法：\n- 能被3整除的数：33个\n- 减去既能被3整除又能被5整除的数：6个\n- 结果：33 - 6 = 27个\n\n**验证：**\n能被3整除但不能被5整除的数包括：\n3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, 99\n共27个 ✓\n\nFinal Answer: 27"}]}
+{"messages": [{"role": "user", "content": "甲容器中有浓度 20% 的盐水 300 克，乙容器中有浓度 10% 的盐水 200 克，混合后盐水的浓度是多少？（用百分数表示）"}, {"role": "assistant", "content": "<think>\n让我一步步计算混合后盐水的浓度。\n\n甲容器：\n- 盐水总量：300 克\n- 浓度：20%\n- 盐的质量：300 × 20% = 300 × 0.2 = 60 克\n\n乙容器：\n- 盐水总量：200 克\n- 浓度：10%\n- 盐的质量：200 × 10% = 200 × 0.1 = 20 克\n\n混合后：\n- 盐的总质量：60 + 20 = 80 克\n- 盐水的总质量：300 + 200 = 500 克\n- 浓度：80 / 500 = 0.16 = 16%\n\n所以混合后盐水的浓度是 16%。\n</think>\n\n让我一步步计算混合后盐水的浓度。\n\n**第一步：计算甲容器中盐的质量**\n- 甲容器盐水：300 克，浓度 20%\n- 甲容器中盐的质量 = 300 × 20% = 300 × 0.2 = 60 克\n\n**第二步：计算乙容器中盐的质量**\n- 乙容器盐水：200 克，浓度 10%\n- 乙容器中盐的质量 = 200 × 10% = 200 × 0.1 = 20 克\n\n**第三步：计算混合后盐的总质量**\n- 混合后盐的总质量 = 60 + 20 = 80 克\n\n**第四步：计算混合后盐水的总质量**\n- 混合后盐水的总质量 = 300 + 200 = 500 克\n\n**第五步：计算混合后的浓度**\n- 混合后的浓度 = 盐的总质量 / 盐水的总质量 × 100%\n- 混合后的浓度 = 80 / 500 × 100% = 0.16 × 100% = 16%\n\nFinal Answer: 16%"}]}
+{"messages": [{"role": "user", "content": "某快递公司收费标准：首重 1 公斤收 8 元，之后每增加 1 公斤加收 3 元（不足 1 公斤按 1 公斤计）。寄一件 7.3 公斤的包裹要多少元？"}, {"role": "assistant", "content": "<think>\n让我逐步分析这个快递收费问题。\n\n收费标准：\n- 首重1公斤收8元\n- 之后每增加1公斤加收3元\n- 不足1公斤按1公斤计\n\n包裹重量：7.3公斤\n\n步骤：\n1. 首先，首重1公斤收费：8元\n\n2. 剩余重量：7.3 - 1 = 6.3公斤\n\n3. 因为不足1公斤按1公斤计，6.3公斤要按7公斤计算\n\n4. 超出首重部分的费用：7公斤 × 3元/公斤 = 21元\n\n5. 总费用：8元 + 21元 = 29元\n\n让我再验证一下：\n- 首重1公斤：8元\n- 超出部分：7.3 - 1 = 6.3公斤，按7公斤计算\n- 超出部分费用：7 × 3 = 21元\n- 总费用：8 + 21 = 29元\n\n答案是29元。\n</think>\n\n让我逐步计算这件7.3公斤包裹的快递费用。\n\n**第一步：确定首重费用**\n- 首重1公斤收费：8元\n\n**第二步：计算超出首重的部分**\n- 包裹总重：7.3公斤\n- 超出首重部分：7.3 - 1 = 6.3公斤\n\n**第三步：按照\"不足1公斤按1公斤计\"的规则处理**\n- 6.3公斤不足7公斤，按7公斤计算\n\n**第四步：计算超重部分费用**\n- 超重部分：7公斤\n- 每公斤加收3元\n- 超重费用：7 × 3 = 21元\n\n**第五步：计算总费用**\n- 总费用 = 首重费用 + 超重费用\n- 总费用 = 8 + 21 = 29元\n\nFinal Answer: 29"}]}
+{"messages": [{"role": "user", "content": "一个圆的半径扩大到原来的 2 倍，它的面积是 314 平方厘米（π 取 3.14）。原来的半径是多少厘米？"}, {"role": "assistant", "content": "<think>\n设原来的半径为 r 厘米。\n\n圆的半径扩大到原来的 2 倍后，新半径为 2r 厘米。\n\n新圆的面积是 π(2r)² = π × 4r² = 4πr² 平方厘米。\n\n已知新圆的面积是 314 平方厘米，π 取 3.14，所以：\n4πr² = 314\n4 × 3.14 × r² = 314\n12.56 × r² = 314\nr² = 314 ÷ 12.56\nr² = 25\nr = 5\n\n所以原来的半径是 5 厘米。\n\n让我验证一下：\n原来的半径是 5 厘米\n扩大到原来的 2 倍后，新半径是 10 厘米\n新圆的面积 = π × 10² = 3.14 × 100 = 314 平方厘米 ✓\n\n答案正确。\n</think>\n\n让我一步步推理这道题。\n\n**设定变量：**\n设原来圆的半径为 r 厘米。\n\n**理解题意：**\n- 半径扩大到原来的 2 倍后，新半径 = 2r 厘米\n- 扩大后圆的面积 = 314 平方厘米\n\n**根据圆面积公式求解：**\n\n扩大后圆的面积公式：\n$$S = \\pi \\times (2r)^2$$\n\n$$S = \\pi \\times 4r^2$$\n\n$$S = 4\\pi r^2$$\n\n**代入已知条件：**\n\n$$4\\pi r^2 = 314$$\n\n$$4 \\times 3.14 \\times r^2 = 314$$\n\n$$12.56 \\times r^2 = 314$$\n\n**求解 r²：**\n\n$$r^2 = \\frac{314}{12.56}$$\n\n$$r^2 = 25$$\n\n**求解 r：**\n\n$$r = 5$$\n\n**验证：**\n- 原半径：5 厘米\n- 扩大后半径：10 厘米\n- 扩大后面积：3.14 × 10² = 3.14 × 100 = 314 平方厘米 ✓\n\nFinal Answer: 5"}]}
+{"messages": [{"role": "user", "content": "连续三个偶数的和是 78，其中最大的偶数是多少？"}, {"role": "assistant", "content": "<think>\n设三个连续的偶数分别为 $n$, $n+2$, $n+4$，其中 $n$ 是偶数。\n\n根据题意，这三个偶数的和是 78：\n$n + (n+2) + (n+4) = 78$\n\n化简：\n$3n + 6 = 78$\n$3n = 72$\n$n = 24$\n\n所以这三个连续的偶数是：24, 26, 28\n\n其中最大的偶数是 28。\n\n让我验证一下：24 + 26 + 28 = 78 ✓\n</think>\n\n让我来一步步解决这个问题。\n\n**设定未知数：**\n设三个连续的偶数分别为 $n$、$n+2$、$n+4$（其中 $n$ 是偶数）\n\n**建立方程：**\n根据题意，三个连续偶数的和是 78：\n$$n + (n+2) + (n+4) = 78$$\n\n**求解方程：**\n$$3n + 6 = 78$$\n\n$$3n = 78 - 6$$\n\n$$3n = 72$$\n\n$$n = 24$$\n\n**找出三个偶数：**\n- 第一个偶数：$n = 24$\n- 第二个偶数：$n + 2 = 26$\n- 第三个偶数：$n + 4 = 28$\n\n**验证：**\n$24 + 26 + 28 = 78$ ✓\n\n**结论：**\n最大的偶数是 28\n\nFinal Answer: 28"}]}
+{"messages": [{"role": "user", "content": "某班一次数学测验的平均分是 84 分。去掉一个 60 分的成绩后，其余 24 人的平均分是 85 分。这个班原来有多少人？"}, {"role": "assistant", "content": "<think>\n设这个班原来有 $n$ 人。\n\n根据题意：\n- 全班平均分是 84 分\n- 全班总分是 $84n$ 分\n- 去掉一个 60 分后，剩余 24 人\n- 所以原来有 $n$ 人，去掉 1 人后剩余 24 人，即 $n - 1 = 24$，所以 $n = 25$\n\n让我验证一下：\n- 原来 25 人，总分 $= 84 \\times 25 = 2100$ 分\n- 去掉 60 分后，剩余总分 $= 2100 - 60 = 2040$ 分\n- 剩余 24 人，平均分 $= 2040 \\div 24 = 85$ 分\n\n验证正确！\n\n所以这个班原来有 25 人。\n</think>\n\n我来一步步解决这个问题。\n\n**设定未知数：**\n设这个班原来有 $n$ 人。\n\n**根据题意列出等量关系：**\n\n1. 原来全班的总分 = 平均分 × 人数 = $84n$ 分\n\n2. 去掉一个 60 分后：\n   - 剩余人数：$n - 1 = 24$ 人\n   - 剩余总分：$84n - 60$ 分\n   - 剩余平均分：85 分\n\n**从剩余人数可以直接得出：**\n$n - 1 = 24$\n$n = 25$\n\n**验证答案：**\n- 原来 25 人，总分 = $84 \\times 25 = 2100$ 分\n- 去掉 60 分后，剩余总分 = $2100 - 60 = 2040$ 分\n- 剩余 24 人，平均分 = $2040 \\div 24 = 85$ 分 ✓\n\n验证正确！\n\nFinal Answer: 25"}]}
@@ -0,0 +1,24 @@
+{"messages": [{"role": "user", "content": "一个书架有 5 层，每层放 28 本书。已经借出了 37 本，书架上还剩多少本书？"}, {"role": "assistant", "content": "<think>\nThe user wants step-by-step reasoning in Chinese, with final answer line format. Compute total books 5*28=140. Remaining after lending 37: 140-37=103. Need final last line exactly Final Answer: 103. Ensure no unit in final answer. Provide concise steps.\n</think>\n\n书架总书数：5 层 × 每层 28 本 = 140 本。  \n借出 37 本后剩余：140 − 37 = 103 本。  \n\nFinal Answer: 103"}]}
+{"messages": [{"role": "user", "content": "小明买了 3 支单价 4.5 元的笔和 2 本单价 12 元的笔记本，付了 50 元，应找回多少元？"}, {"role": "assistant", "content": "<think>\n小明买了3支笔，每支4.5元，共3×4.5=13.5元。\n买了2本笔记本，每本12元，共2×12=24元。\n总共花费13.5+24=37.5元。\n付了50元，应找回50-37.5=12.5元。\n最终答案：12.5。\n</think>\n\n3支笔的总价：3 × 4.5 = 13.5 元  \n2本笔记本的总价：2 × 12 = 24 元  \n总花费：13.5 + 24 = 37.5 元  \n应找回：50 − 37.5 = 12.5 元  \n\nFinal Answer: 12.5"}]}
+{"messages": [{"role": "user", "content": "一辆汽车以每小时 60 公里的速度行驶了 2.5 小时，又以每小时 80 公里的速度行驶了 1.5 小时，一共行驶了多少公里？"}, {"role": "assistant", "content": "<think>\nThe user wants a step-by-step reasoning in Chinese, with final line format \"Final Answer: 数值\" only numeric no unit.\n\nCompute: 60 km/h * 2.5 h = 150 km. 80 km/h * 1.5 h = 120 km. Total = 270 km.\nNeed final answer only numeric. Ensure last line exactly Final Answer: 270. Could include steps before. No unit in final. Use Chinese. Need not over-explain too much but step-by-step.\n</think>\n\n第一段路程：60 × 2.5 = 150  \n第二段路程：80 × 1.5 = 120  \n总路程：150 + 120 = 270  \n\nFinal Answer: 270"}]}
+{"messages": [{"role": "user", "content": "某班有 48 名学生，其中 3/8 参加了数学兴趣小组，参加数学兴趣小组的有多少人？"}, {"role": "assistant", "content": "<think>\n题目：48名学生，3/8参加数学兴趣小组。计算48*3/8 = 6*3 =18。需要一步步推理，最后一行严格格式 Final Answer: 数值，只写数值不带单位。用中文简洁说明。确保最后一行只有 Final Answer: 18。\n</think>\n\n全班有 48 名学生，参加数学兴趣小组的占 \\( \\frac{3}{8} \\)。\n\n计算人数：  \n\\[\n48 \\times \\frac{3}{8} = 6 \\times 3 = 18\n\\]\n\nFinal Answer: 18"}]}
+{"messages": [{"role": "user", "content": "一个长方形的长是宽的 3 倍，周长是 64 厘米，它的面积是多少平方厘米？"}, {"role": "assistant", "content": "<think>\n用户问了一个长方形面积问题。需要一步步推理，最后格式 Final Answer: 数值（不带单位）。\n\n题目：长是宽的3倍，周长64厘米。设宽为 w，长为 3w。周长 2(长+宽)=2(3w+w)=8w=64，所以 w=8 厘米，长=24 厘米。面积=24*8=192 平方厘米。\n\n需要输出一步步推理，最后一行精确格式。确保最后一行只写 Final Answer: 192。前面可以带单位解释。最终答案行不带单位。用中文。\n</think>\n\n设长方形的宽为 \\(w\\) 厘米，则长为 \\(3w\\) 厘米。\n\n周长公式：  \n\\[\n2(\\text{长}+\\text{宽})=64\n\\]\n\n代入得：  \n\\[\n2(3w+w)=64\n\\]\n\\[\n2\\cdot 4w=64\n\\]\n\\[\n8w=64\n\\]\n\\[\nw=8\n\\]\n\n所以宽是 8 厘米，长是：  \n\\[\n3w=3\\times 8=24\n\\]\n\n面积为：  \n\\[\n24\\times 8=192\n\\]\n\nFinal Answer: 192"}]}
+{"messages": [{"role": "user", "content": "商店把一件商品先提价 20%，再降价 20%，现价是 96 元。这件商品的原价是多少元？"}, {"role": "assistant", "content": "<think>\nThe user wants step-by-step reasoning and final answer format. Need compute: original price P. Increase 20% => 1.2P. Then decrease 20% => 1.2P*0.8 = 0.96P. Current price 96, so 0.96P=96 => P=100. Need answer in Chinese probably. Ensure last line exactly Final Answer: 100. Need maybe concise step-by-step. Let's craft with equations. No unit in final. Final answer only numeric after colon. They said 只写数值，不带单位 for final answer; last line format. Good.\n</think>\n\n设原价为 \\(x\\) 元。\n\n先提价 20% 后，价格变为：\n\\[\nx \\times (1+20\\%)=1.2x\n\\]\n\n再降价 20%，是在提价后的价格基础上降价，所以现价为：\n\\[\n1.2x \\times (1-20\\%)=1.2x \\times 0.8=0.96x\n\\]\n\n已知现价是 96 元：\n\\[\n0.96x=96\n\\]\n\n解得：\n\\[\nx=\\frac{96}{0.96}=100\n\\]\n\nFinal Answer: 100"}]}
+{"messages": [{"role": "user", "content": "5 台同样的机器 8 小时可以生产 600 个零件。照这样计算，8 台机器 10 小时可以生产多少个零件？"}, {"role": "assistant", "content": "<think>\n用户要求解决一个数学问题，并一步步推理，最后以\"Final Answer: 数值\"格式给出答案。\n\n问题：5台同样的机器8小时可以生产600个零件。照这样计算，8台机器10小时可以生产多少个零件？\n\n步骤：\n1. 先求1台机器1小时生产多少个零件。\n   5台机器8小时生产600个。\n   1台机器8小时生产：600 ÷ 5 = 120个。\n   1台机器1小时生产：120 ÷ 8 = 15个。\n\n2. 8台机器10小时生产：\n   1台1小时15个。\n   8台1小时：15 × 8 = 120个。\n   8台10小时：120 × 10 = 1200个。\n\n或者用比例：\n零件数与机器台数成正比，与时间成正比。\n600 × (8/5) × (10/8) = 600 × 8/5 × 10/8\n= 600 × (8×10)/(5×8) = 600 × 80/40 = 600 × 2 = 1200。\n\n最终答案：1200。\n\n需要确保最后一行格式正确：Final Answer: 1200\n不带单位。\n</think>\n\n第一步：求 1 台机器 1 小时生产多少个零件。\n\n5 台机器 8 小时生产 600 个，  \n所以 1 台机器 8 小时生产：600 ÷ 5 = 120 个，  \n1 台机器 1 小时生产：120 ÷ 8 = 15 个。\n\n第二步：计算 8 台机器 10 小时的产量。\n\n8 台机器 1 小时生产：15 × 8 = 120 个，  \n8 台机器 10 小时生产：120 × 10 = 1200 个。\n\nFinal Answer: 1200"}]}
+{"messages": [{"role": "user", "content": "某数加上它的 1/4 等于 35，这个数是多少？"}, {"role": "assistant", "content": "<think>\nThe problem in Chinese: \"某数加上它的 1/4 等于 35，这个数是多少？\" Translation: A number plus its 1/4 equals 35. What is the number?\n\nLet the number be x.\nx + (1/4)x = 35\n(5/4)x = 35\nx = 35 * (4/5) = 28.\n\nCheck: 28 + 28/4 = 28 + 7 = 35. Correct.\n\nNeed to respond with step-by-step reasoning in Chinese, and final line format \"Final Answer: 数值\" only value no unit. So last line: Final Answer: 28. Ensure no extra after? The user requested last line with format. We can have reasoning then final line.\n\nNeed final answer only. Keep concise but step-by-step.\n</think>\n\n设这个数为 \\(x\\)。\n\n根据题意：\n\\[\nx+\\frac14 x=35\n\\]\n\n合并左边：\n\\[\n\\frac54 x=35\n\\]\n\n两边同乘 \\(\\frac45\\)：\n\\[\nx=35\\times \\frac45=28\n\\]\n\n检验：\\(28+28\\times\\frac14=28+7=35\\)，正确。\n\nFinal Answer: 28"}]}
+{"messages": [{"role": "user", "content": "甲、乙两人从相距 240 公里的两地同时出发相向而行，甲每小时走 14 公里，乙每小时走 10 公里，几小时后两人相遇？"}, {"role": "assistant", "content": "<think>\n甲和乙相向而行，相距240公里。甲速度14公里/小时，乙速度10公里/小时。相向而行，相对速度为14+10=24公里/小时。相遇时间=路程/相对速度=240/24=10小时。需要最后一行格式 Final Answer: 数值，只写数值不带单位。可以一步步推理。最终行必须是 Final Answer: 10。确保不带单位。\n</think>\n\n两人相向而行，接近速度为两人速度之和：\n\n甲速度：14 公里/小时  \n乙速度：10 公里/小时  \n相对速度：14 + 10 = 24 公里/小时\n\n两地相距 240 公里，所以相遇时间为：\n\n240 ÷ 24 = 10\n\nFinal Answer: 10"}]}
+{"messages": [{"role": "user", "content": "一个等差数列的第 3 项是 11，第 7 项是 27，它的第 10 项是多少？"}, {"role": "assistant", "content": "<think>\nWe need answer Chinese, step-by-step, final exact format only numeric. Need solve arithmetic progression a3=11 a7=27. Difference 4d=16 d=4. a10=a7+3d=27+12=39. Need likely concise. Last line Final Answer: 39. Ensure last line exactly. Chinese.\n</think>\n\n设等差数列的公差为 \\(d\\)。\n\n1. 第 3 项与第 7 项相差 \\(7-3=4\\) 个公差：\n\\[\n4d=27-11=16\n\\]\n\n2. 求得公差：\n\\[\nd=\\frac{16}{4}=4\n\\]\n\n3. 第 10 项比第 7 项多 \\(3\\) 个公差：\n\\[\na_{10}=27+3\\times4=39\n\\]\n\nFinal Answer: 39"}]}
+{"messages": [{"role": "user", "content": "农场里鸡和兔共有 35 个头、94 只脚，兔有多少只？"}, {"role": "assistant", "content": "<think>\nThe user wants me to solve a classic chicken and rabbit problem. Let me work through it.\n\nChickens have 1 head and 2 feet. Rabbits have 1 head and 4 feet.\n\nLet chickens = c, rabbits = r.\nc + r = 35 (heads)\n2c + 4r = 94 (feet)\n\nFrom first equation: c = 35 - r\nSubstitute: 2(35 - r) + 4r = 94\n70 - 2r + 4r = 94\n70 + 2r = 94\n2r = 24\nr = 12\n\nSo there are 12 rabbits.\n\nLet me verify: 12 rabbits = 48 feet, 23 chickens = 46 feet. Total = 94 feet. Heads = 12 + 23 = 35. Correct.\n\nThe user wants step-by-step reasoning in Chinese, and the last line should be \"Final Answer: 数值\" with only the number, no units.\n</think>\n\n设鸡有 \\(x\\) 只，兔有 \\(y\\) 只。\n\n根据头的数量：\n\\[x + y = 35\\]\n\n根据脚的数量（鸡 2 只脚，兔 4 只脚）：\n\\[2x + 4y = 94\\]\n\n由第一式得 \\(x = 35 - y\\)，代入第二式：\n\\[2(35 - y) + 4y = 94\\]\n\\[70 - 2y + 4y = 94\\]\n\\[70 + 2y = 94\\]\n\\[2y = 24\\]\n\\[y = 12\\]\n\n验证：兔 12 只（48 只脚），鸡 23 只（46 只脚），共 35 个头、94 只脚，符合题意。\n\nFinal Answer: 12"}]}
+{"messages": [{"role": "user", "content": "一项工程，甲单独做 12 天完成，乙单独做 18 天完成。两人合作 4 天后，剩下的工程由乙单独完成，还需要多少天？"}, {"role": "assistant", "content": "<think>\n甲单独做12天完成，每天完成 1/12。\n乙单独做18天完成，每天完成 1/18。\n两人合作4天，完成 (1/12+1/18)*4。\n1/12+1/18 = 3/36+2/36=5/36。\n合作4天完成 20/36=5/9。\n剩余 1-5/9=4/9。\n乙单独完成需要 (4/9)/(1/18)= (4/9)*18=8 天。\n最终答案 8。\n</think>\n\n甲每天完成工程的 \\( \\frac{1}{12} \\)，乙每天完成工程的 \\( \\frac{1}{18} \\)。\n\n两人合作一天完成：\n\\[\n\\frac{1}{12}+\\frac{1}{18}=\\frac{3}{36}+\\frac{2}{36}=\\frac{5}{36}\n\\]\n\n合作 4 天完成：\n\\[\n4 \\times \\frac{5}{36}=\\frac{20}{36}=\\frac{5}{9}\n\\]\n\n剩余工程：\n\\[\n1-\\frac{5}{9}=\\frac{4}{9}\n\\]\n\n由乙单独完成，需要：\n\\[\n\\frac{4}{9}\\div \\frac{1}{18}=\\frac{4}{9}\\times 18=8\n\\]\n\nFinal Answer: 8"}]}
+{"messages": [{"role": "user", "content": "某商品进价 80 元，按标价卖出可赚 25%。若按标价的 9 折出售，每件可赚多少元？"}, {"role": "assistant", "content": "<think>\nWe need answer in Chinese, step-by-step, final line exact format \"Final Answer: 数值\" only number no unit. Need solve: cost 80, marked price sold earns 25% => profit 25% of cost? Chinese \"按标价卖出可赚25%\" usually means profit margin on cost? In elementary math, 可赚25% means earns 25% of cost (利润率 relative to 进价), so marked price = 80*(1+25%)=100. Sell at 90% of marked price =90. Profit per item=90-80=10. Need maybe note if \"赚25%\" could be ambiguous but conventional. Provide concise step-by-step. Ensure final line only Final Answer: 10. No extra after. Need include units? final only number no unit. Good.\n</think>\n\n进价是 80 元。  \n按标价卖出可赚 25%，即标价比进价多 25%：\n\n标价 = 80 × (1 + 25%) = 80 × 1.25 = 100 元\n\n按标价的 9 折出售：\n\n售价 = 100 × 90% = 90 元\n\n每件可赚：\n\n90 - 80 = 10 元\n\nFinal Answer: 10"}]}
+{"messages": [{"role": "user", "content": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？"}, {"role": "assistant", "content": "<think>\n把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？\n\n总份数 = 4 + 5 = 9\n每份 = 126 ÷ 9 = 14\n乙班占 5 份 = 14 × 5 = 70\n\nFinal Answer: 70\n</think>\n\n总份数：4 + 5 = 9\n\n每份数量：126 ÷ 9 = 14\n\n乙班占 5 份：14 × 5 = 70\n\nFinal Answer: 70"}]}
+{"messages": [{"role": "user", "content": "一个水池，单开进水管 6 小时注满，单开出水管 10 小时放空。两管同时打开，多少小时可以注满空水池？"}, {"role": "assistant", "content": "<think>\n这是一个基础的水池进出水问题。\n\n进水管单独6小时注满，所以进水速度是 1/6 池/小时。\n出水管单独10小时放空，所以出水速度是 1/10 池/小时。\n\n两管同时打开，净速度 = 1/6 - 1/10。\n\n计算：1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15 池/小时。\n\n注满时间 = 1 ÷ (1/15) = 15 小时。\n\n需要按要求一步步推理，最后以 \"Final Answer: 数值\" 格式输出，只写数值不带单位。\n</think>\n\n设水池容量为 1。\n\n1. 进水管 6 小时注满，所以进水速度为每小时注入水池的：  \n   \\[\n   \\frac{1}{6}\n   \\]\n\n2. 出水管 10 小时放空，所以出水速度为每小时放掉水池的：  \n   \\[\n   \\frac{1}{10}\n   \\]\n\n3. 两管同时打开时，实际净注水速度为：  \n   \\[\n   \\frac{1}{6}-\\frac{1}{10}\n   =\\frac{5}{30}-\\frac{3}{30}\n   =\\frac{2}{30}\n   =\\frac{1}{15}\n   \\]\n\n4. 因此注满空水池所需时间为：  \n   \\[\n   1\\div \\frac{1}{15}=15\n   \\]\n\nFinal Answer: 15"}]}
+{"messages": [{"role": "user", "content": "小华今年的年龄是小丽的 3 倍，8 年后小华的年龄是小丽的 2 倍。小华今年多少岁？"}, {"role": "assistant", "content": "<think>\n用户要求解一道年龄问题，需要一步步推理，最后按指定格式输出。\n\n设小丽今年 x 岁，小华今年 3x 岁。\n8 年后：小丽 x+8，小华 3x+8。\n条件：3x+8 = 2(x+8) = 2x+16。\n解得 x = 8。\n小华今年 3x = 24 岁。\n验证：8年后小华32，小丽16，32=2*16，正确。\n最终只写数值 24。\n需要包含一步步推理，最后行 Final Answer: 24。注意用户说最后一行只写数值不带单位，格式为「Final Answer: 数值」。所以最后一行应是 Final Answer: 24。推理过程可以带单位。确保最后一行 exact。\n</think>\n\n设小丽今年 \\(x\\) 岁，则小华今年 \\(3x\\) 岁。\n\n8 年后：\n- 小丽：\\(x+8\\) 岁\n- 小华：\\(3x+8\\) 岁\n\n根据“8 年后小华的年龄是小丽的 2 倍”列方程：\n\n\\[\n3x+8=2(x+8)\n\\]\n\n展开：\n\n\\[\n3x+8=2x+16\n\\]\n\n移项得：\n\n\\[\nx=8\n\\]\n\n所以小华今年：\n\n\\[\n3x=3\\times 8=24\n\\]\n\n验证：8 年后小华 32 岁，小丽 16 岁，32 正好是 16 的 2 倍，正确。\n\nFinal Answer: 24"}]}
+{"messages": [{"role": "user", "content": "某工厂第一季度生产零件 2400 个，第二季度比第一季度增产 15%，两个季度一共生产了多少个零件？"}, {"role": "assistant", "content": "<think>\nThe user is asking a math word problem in Chinese. Let me solve it step by step.\n\n第一季度生产零件 2400 个\n第二季度比第一季度增产 15%\n第二季度产量 = 2400 × (1 + 15%) = 2400 × 1.15 = 2760\n两个季度一共 = 2400 + 2760 = 5160\n\nLet me verify: 2400 × 0.15 = 360, so second quarter = 2400 + 360 = 2760. Total = 2400 + 2760 = 5160.\n\nFinal Answer: 5160\n</think>\n\n第一季度生产：2400 个\n\n第二季度比第一季度增产 15%，所以第二季度生产：\n2400 × (1 + 15%) = 2400 × 1.15 = 2760 个\n\n两个季度一共生产：\n2400 + 2760 = 5160 个\n\nFinal Answer: 5160"}]}
+{"messages": [{"role": "user", "content": "一个两位数的个位数字比十位数字大 3，这个两位数等于它的两个数字之和的 4 倍，这个两位数是多少？"}, {"role": "assistant", "content": "<think>\n用户要求解一个数学题，并一步步推理，最后一行用指定格式给答案。\n\n题目：一个两位数的个位数字比十位数字大3，这个两位数等于它的两个数字之和的4倍。\n\n设十位数字为 t，个位数字为 u。\n条件1：u = t + 3。\n两位数数值 = 10t + u。\n数字和 = t + u。\n条件2：10t + u = 4(t+u)。\n代入 u=t+3：10t + t+3 = 4(t + t+3) = 4(2t+3)=8t+12。\n左边 11t+3 = 8t+12 => 3t=9 => t=3。u=6。两位数=36。检验：数字和9，4倍=36。正确。\n\n需要最终 answer 最后一行 \"Final Answer: 36\"。只写数值不带单位，指数值部分只写36；格式中包含 Final Answer: 。可以。\n\n需要用中文一步步推理。确保最后一行精确格式。\n</think>\n\n设十位数字为 \\(x\\)，个位数字为 \\(y\\)。\n\n1. 个位数字比十位数字大 3：  \n   \\[\n   y=x+3\n   \\]\n\n2. 这个两位数等于两个数字之和的 4 倍：  \n   \\[\n   10x+y=4(x+y)\n   \\]\n\n3. 代入 \\(y=x+3\\)：  \n   \\[\n   10x+(x+3)=4(x+x+3)\n   \\]\n   \\[\n   11x+3=4(2x+3)=8x+12\n   \\]\n\n4. 解方程：  \n   \\[\n   11x+3=8x+12\n   \\]\n   \\[\n   3x=9\n   \\]\n   \\[\n   x=3\n   \\]\n\n5. 个位数字：  \n   \\[\n   y=3+3=6\n   \\]\n\n所以这个两位数是 36。检验：\\(3+6=9\\)，\\(9\\times4=36\\)，符合条件。\n\nFinal Answer: 36"}]}
+{"messages": [{"role": "user", "content": "从 1 到 100 的所有整数中，能被 3 整除但不能被 5 整除的数有多少个？"}, {"role": "assistant", "content": "<think>\n用户要求中文回答，并一步步推理，最后格式 Final Answer: 数值。需要计算1到100整数中能被3整除但不能被5整除的个数。\n\n计算：能被3整除：floor(100/3)=33（3到99）。能被3且能被5即能被15整除：floor(100/15)=6（15,30,45,60,75,90）。所求=33-6=27。\n\n需要简洁逐步推理。最后行精确格式。确保最后一行只有 Final Answer: 27。前面可以有步骤。不要在最终答案后添加内容。\n</think>\n\n能被 3 整除的数有：⌊100/3⌋ = 33 个。  \n其中同时能被 5 整除，即能被 15 整除的数有：⌊100/15⌋ = 6 个。  \n所以能被 3 整除但不能被 5 整除的数有：33 − 6 = 27 个。\n\nFinal Answer: 27"}]}
+{"messages": [{"role": "user", "content": "甲容器中有浓度 20% 的盐水 300 克，乙容器中有浓度 10% 的盐水 200 克，混合后盐水的浓度是多少？（用百分数表示）"}, {"role": "assistant", "content": "<think>\nWe need answer in Chinese likely. Need solve stepwise. Compute: salt in A =20%*300=60 g. Salt in B=10%*200=20 g. Total salt=80 g. Total solution=500 g. Concentration=80/500=0.16=16%. Need final last line exactly \"Final Answer: 16%\"? User says 用百分数表示 and final line format 「Final Answer: 数值」 only write value no unit. But says 用百分数表示. They specifically: 最后一行用「Final Answer: 数值」的格式给出最终答案（只写数值，不带单位）. Since unit is percent? They say no unit, but concentration expressed as percentage. Ambiguous: Should final answer be 16% or 16? \"只写数值，不带单位\" means write numeric value no unit; but they asked use percent. The format placeholder 数值 maybe include %? They say no unit; percent sign is unit. However \"用百分数表示\" suggests final answer numeric with percent sign. Last line format \"Final Answer: 数值\" and \"只写数值，不带单位\" probably from benchmark wants e.g. Final Answer: 16%. Yet \"不带单位\" maybe no 克. For percent, sign is required? Let's parse Chinese: \"混合后盐水的浓度是多少？（用百分数表示）\" Then: \"请一步步推理，并在最后一行用「Final Answer: 数值」的格式给出最终答案（只写数值，不带单位）。\" If final answer only value no unit, since asked in percent, the numerical value is 16 (percent). They explicitly say no unit; percent sign could be considered unit. But many tasks expect Final Answer: 16% because they ask percent. The phrase 「Final Answer: 数值」 with 数值 could be 16%? They add \"只写数值，不带单位\" meaning if answer is 16%, don't write \"16%克\" or \"16 percent\". In prior similar tasks, they demand final answer only numerical no units. Need choose. It says \"用百分数表示\" so final should be a percent; \"只写数值\" maybe no Chinese unit like 克. The last line: Final Answer: 16%. This includes % not a unit? It is a percent sign. They said 不带单位; percent is the requested representation. I'd output steps and final line Final Answer: 16%. Ensure last line only that. Could mention concentration=16%. Final answer last line exactly. If strict no unit maybe should be 16. But problem says use percentage; I think 16% is okay and matches. The final answer field \"数值\" can include percent sign? Hmm. The phrase \"只写数值，不带单位\" strongly indicates not write unit; but then a percentage without % is just number 16, which is the numerical value of the percentage. Many math word problems ask \"浓度是多少（用百分数表示）\" answer \"16%\"; if they say no unit they still often expect \"16%\". The final answer format examples in these prompts usually accept \"Final Answer: 16%\". I'll use 16% because requested percent. Need step-by-step concise.\n\nNeed ensure no extra after final line. Use Chinese. Steps:\n1. 甲盐质量 300×20%=60克\n2. 乙盐质量 200×10%=20克\n3. 总盐 80克，总盐水 500克\n4. 浓度=80/500×100%=16%\nFinal Answer: 16%\n\nNeed last line exact. Good.\n</think>\n\n甲容器中盐的质量：\\(300 \\times 20\\% = 60\\) 克。  \n乙容器中盐的质量：\\(200 \\times 10\\% = 20\\) 克。  \n混合后盐的总质量：\\(60 + 20 = 80\\) 克。  \n混合后盐水总质量：\\(300 + 200 = 500\\) 克。  \n浓度：\\(\\frac{80}{500} \\times 100\\% = 16\\%\\)。  \n\nFinal Answer: 16%"}]}
+{"messages": [{"role": "user", "content": "某快递公司收费标准：首重 1 公斤收 8 元，之后每增加 1 公斤加收 3 元（不足 1 公斤按 1 公斤计）。寄一件 7.3 公斤的包裹要多少元？"}, {"role": "assistant", "content": "<think>\n用户要求解一个快递收费问题，并明确要求一步步推理，最后以特定格式输出。需要准确计算。\n\n题目：首重1公斤收8元，之后每增加1公斤加收3元（不足1公斤按1公斤计）。包裹7.3公斤。\n\n计算：不足1公斤按1公斤计，所以计费重量为 ceil(7.3)=8 公斤。首重1公斤8元，其余增加公斤数 = 8-1=7 公斤，每公斤3元，加收21元。总费用=8+21=29元。\n\n需要注意：是否“之后每增加1公斤”对于7.3公斤，另一种理解：首重1kg，续重部分6.3kg，不足1kg按1kg计 -> 续重按7kg计，费用8+7*3=29。结果相同。\n\n需要输出中文一步步推理，最后最后一行严格是「Final Answer: 29」。确保最后一行只写数值不带单位，按照格式包含 Final Answer:。不要多余内容在最后一行。可以前面有步骤。最终答案最后一行格式正确。\n</think>\n\n1. 包裹实际重量为 7.3 公斤。\n2. 因为“不足 1 公斤按 1 公斤计”，所以计费重量按 8 公斤算。\n3. 首重 1 公斤收费 8 元。\n4. 超出首重的部分为：8 - 1 = 7 公斤。\n5. 超出部分费用：7 × 3 = 21 元。\n6. 总费用：8 + 21 = 29 元。\n\nFinal Answer: 29"}]}
+{"messages": [{"role": "user", "content": "一个圆的半径扩大到原来的 2 倍，它的面积是 314 平方厘米（π 取 3.14）。原来的半径是多少厘米？"}, {"role": "assistant", "content": "<think>\nThe problem: A circle's radius is expanded to 2 times original. Its area is 314 cm² (π = 3.14). Find original radius.\n\nLet original radius = r. New radius = 2r. New area = π(2r)² = 4πr² = 314.\nWith π = 3.14: 4 × 3.14 × r² = 314.\n12.56 r² = 314.\nr² = 314 / 12.56 = 25.\nr = 5.\n\nCheck: original radius 5, new radius 10, area = 3.14 × 100 = 314. Correct.\n\nNeed to provide step-by-step reasoning in Chinese, last line format \"Final Answer: 数值\" only number no unit. So final line: Final Answer: 5.\n\nNeed to ensure last line exactly. Use Chinese explanation then final answer. No extra after final answer.\n</think>\n\n设原来的半径为 \\(r\\) 厘米。\n\n半径扩大到原来的 2 倍后，新半径为 \\(2r\\) 厘米。\n\n扩大后的面积为：\n\\[\n\\pi (2r)^2 = 314\n\\]\n\n代入 \\(\\pi = 3.14\\)：\n\\[\n3.14 \\times 4r^2 = 314\n\\]\n\n\\[\n12.56r^2 = 314\n\\]\n\n\\[\nr^2 = 314 \\div 12.56 = 25\n\\]\n\n\\[\nr = 5\n\\]\n\nFinal Answer: 5"}]}
+{"messages": [{"role": "user", "content": "连续三个偶数的和是 78，其中最大的偶数是多少？"}, {"role": "assistant", "content": "<think>\nThe user is asking in Chinese: The sum of three consecutive even numbers is 78, what is the largest even number?\n\nLet me solve step by step.\n\nLet the three consecutive even numbers be n, n+2, n+4.\nSum = n + (n+2) + (n+4) = 3n + 6 = 78\n3n = 72\nn = 24\nSo the numbers are 24, 26, 28.\nThe largest is 28.\n\nAlternatively, middle number = 78/3 = 26, largest = 28.\n\nFinal Answer: 28\n</think>\n\n设这三个连续偶数中最小的为 \\(x\\)，则另外两个分别为 \\(x+2\\)、\\(x+4\\)。\n\n根据题意：\n\\[\nx+(x+2)+(x+4)=78\n\\]\n\\[\n3x+6=78\n\\]\n\\[\n3x=72\n\\]\n\\[\nx=24\n\\]\n\n所以三个连续偶数是 24、26、28，其中最大的偶数是 28。\n\nFinal Answer: 28"}]}
+{"messages": [{"role": "user", "content": "某班一次数学测验的平均分是 84 分。去掉一个 60 分的成绩后，其余 24 人的平均分是 85 分。这个班原来有多少人？"}, {"role": "assistant", "content": "<think>\nThe problem: A class math test average is 84. After removing one score of 60, the remaining 24 people have average 85. How many people originally?\n\nIf after removing one person there are 24 people, originally there were 25 people. Let's verify: total original = 25*84 = 2100. Remove 60 -> 2040. Remaining 24 average = 2040/24 = 85. Yes consistent.\n\nNeed provide step-by-step reasoning in Chinese and final line exactly \"Final Answer: 25\". Ensure last line format. Could mention original人数 = 24+1=25; also verify with average. But maybe trick: The statement says average is 84; removing 60 leaves 24 people average 85. If there are 24 remaining, original =25. The averages are consistent. Need answer only numeric in final.\n\nLet's craft concise stepwise. Last line must be Final Answer: 数值. No extra after. Ensure no unit. Use Chinese. Done.\n</think>\n\n去掉 1 个成绩后，其余人数是 24 人，所以原来人数为：\n\n24 + 1 = 25\n\n验证：原来总分 = 25 × 84 = 2100；去掉 60 分后总分 = 2040；其余 24 人平均分 = 2040 ÷ 24 = 85，符合题意。\n\nFinal Answer: 25"}]}
@@ -0,0 +1,278 @@
+#!/usr/bin/env python3
+"""Baseline/student/teacher acceptance campaign for Experiment 8-9."""
+
+from __future__ import annotations
+
+import argparse
+import hashlib
+import json
+import math
+import re
+from datetime import datetime, timezone
+from pathlib import Path
+from typing import Any
+
+from generate_data import ANSWER_SUFFIX, extract_predicted_number, verify
+
+BEHAVIORS = {
+    "reflection": r"\b(reflect|reconsider|wait|actually|mistake|not right)\b|反思|等等|不对|重新",
+    "backtracking": r"\b(backtrack|another approach|instead|alternative)\b|回溯|换一种|另一种方法",
+    "verification": r"\b(verify|check|substitute|sanity check)\b|验算|检查|代回|核对",
+}
+
+
+def sha256(path: Path) -> str:
+    return hashlib.sha256(path.read_bytes()).hexdigest()
+
+
+def load_jsonl(path: Path) -> list[dict[str, Any]]:
+    with path.open(encoding="utf-8") as handle:
+        return [json.loads(line) for line in handle if line.strip()]
+
+
+def behavior_flags(text: str) -> dict[str, bool]:
+    return {name: bool(re.search(pattern, text, re.IGNORECASE)) for name, pattern in BEHAVIORS.items()}
+
+
+def exact_two_sided_sign_p_value(baseline_only: int, student_only: int) -> float:
+    """Exact two-sided paired sign test over discordant binary outcomes."""
+    n = baseline_only + student_only
+    if n == 0:
+        return 1.0
+    k = min(baseline_only, student_only)
+    tail = sum(math.comb(n, i) for i in range(k + 1)) / (2 ** n)
+    return min(1.0, 2 * tail)
+
+
+def compare_binary(baseline: dict[str, bool], student: dict[str, bool]) -> dict[str, Any]:
+    ids = sorted(set(baseline) & set(student))
+    both_correct = sum(baseline[i] and student[i] for i in ids)
+    baseline_only = sum(baseline[i] and not student[i] for i in ids)
+    student_only = sum(student[i] and not baseline[i] for i in ids)
+    both_wrong = len(ids) - both_correct - baseline_only - student_only
+    return {
+        "paired_cases": len(ids),
+        "both_correct": both_correct,
+        "baseline_only": baseline_only,
+        "student_only": student_only,
+        "both_wrong": both_wrong,
+        "exact_two_sided_p_value": exact_two_sided_sign_p_value(baseline_only, student_only),
+    }
+
+
+def completion_and_findings(
+    *,
+    problem_ids: set[str],
+    baseline: dict[str, Any],
+    student: dict[str, Any],
+    teacher: dict[str, Any],
+    paired: dict[str, Any],
+    student_training_complete: bool,
+    teacher_outputs_complete: bool,
+) -> tuple[dict[str, bool], dict[str, Any]]:
+    """Separate execution/evidence gates from potentially negative hypotheses."""
+    arm_ids = [
+        {str(record["id"]) for record in arm.get("records", [])}
+        for arm in (baseline, student, teacher)
+    ]
+    completion = {
+        "same_problem_ids_across_three_arms": all(ids == problem_ids for ids in arm_ids),
+        "real_student_training": student_training_complete,
+        "teacher_outputs_complete": teacher_outputs_complete,
+        "paired_quality_comparison_complete": paired.get("paired_cases") == len(problem_ids),
+        "behavior_inspection_complete": all(
+            set(arm.get("behavior_rates", {})) == set(BEHAVIORS)
+            for arm in (baseline, student, teacher)
+        ),
+    }
+    completion["complete"] = all(completion.values())
+    findings = {
+        "student_improves_over_baseline": student["accuracy"] > baseline["accuracy"],
+        "paired_improvement_significant_p_lt_0_05": paired["exact_two_sided_p_value"] < 0.05,
+        "teacher_style_reflection_backtracking_or_verification_observed": any(
+            student["behavior_rates"].values()
+        ),
+    }
+    return completion, findings
+
+
+def teacher_outputs(path: Path) -> dict[str, str]:
+    outputs: dict[str, str] = {}
+    for row in load_jsonl(path):
+        if "id" in row:
+            outputs[str(row["id"])] = "\n".join(
+                part for part in (row.get("reasoning") or "", row.get("content") or "") if part
+            )
+            continue
+        messages = row.get("messages") or []
+        if len(messages) >= 2:
+            question = str(messages[0].get("content", ""))
+            outputs[question] = str(messages[1].get("content", ""))
+    return outputs
+
+
+def generate_local(model_name: str, questions: list[str], max_new_tokens: int) -> list[str]:
+    try:
+        import torch
+        from transformers import AutoModelForCausalLM, AutoTokenizer
+    except ImportError as exc:
+        raise SystemExit("Install the full requirements.txt before local evaluation") from exc
+    tokenizer = AutoTokenizer.from_pretrained(model_name)
+    model = AutoModelForCausalLM.from_pretrained(
+        model_name,
+        torch_dtype="auto",
+        device_map="auto",
+    )
+    results = []
+    for question in questions:
+        prompt = tokenizer.apply_chat_template(
+            [{"role": "user", "content": question + ANSWER_SUFFIX}],
+            tokenize=False,
+            add_generation_prompt=True,
+        )
+        inputs = tokenizer(prompt, return_tensors="pt").to(model.device)
+        with torch.inference_mode():
+            generated = model.generate(
+                **inputs,
+                max_new_tokens=max_new_tokens,
+                do_sample=False,
+                pad_token_id=tokenizer.eos_token_id,
+            )
+        results.append(tokenizer.decode(
+            generated[0][inputs["input_ids"].shape[1]:], skip_special_tokens=True
+        ))
+    return results
+
+
+def score_arm(name: str, problems: list[dict[str, Any]], outputs: list[str]) -> dict[str, Any]:
+    records = []
+    for problem, output in zip(problems, outputs):
+        flags = behavior_flags(output)
+        records.append({
+            "id": problem["id"],
+            "gold_answer": problem["answer"],
+            "predicted_answer": extract_predicted_number(output),
+            "correct": verify(output, problem["answer"]),
+            "behaviors": flags,
+            "output": output,
+        })
+    correct = sum(record["correct"] for record in records)
+    return {
+        "name": name,
+        "cases": len(records),
+        "correct": correct,
+        "accuracy": correct / len(records) if records else 0.0,
+        "behavior_rates": {
+            behavior: sum(r["behaviors"][behavior] for r in records) / len(records)
+            if records else 0.0
+            for behavior in BEHAVIORS
+        },
+        "records": records,
+    }
+
+
+def parse_args() -> argparse.Namespace:
+    parser = argparse.ArgumentParser(
+        description="Experiment 8-9 paired baseline/student/teacher evaluation",
+        formatter_class=argparse.ArgumentDefaultsHelpFormatter,
+    )
+    parser.add_argument("--problems", type=Path, default=Path("problems.jsonl"))
+    parser.add_argument("--baseline-model", default="Qwen/Qwen2.5-1.5B-Instruct")
+    parser.add_argument("--student-model", required=True, help="Real checkpoint emitted by train_student.py")
+    parser.add_argument("--teacher-data", type=Path, default=Path("data/raw_trajectories_aime_kimi_k3.jsonl"))
+    parser.add_argument(
+        "--reuse-local-arms-from",
+        type=Path,
+        help="Reuse retained baseline/student records from a prior evaluation; teacher data is always rescored",
+    )
+    parser.add_argument("--max-new-tokens", type=int, default=4096)
+    parser.add_argument("--output", type=Path, default=Path("validation/experiment_8_9.json"))
+    return parser.parse_args()
+
+
+def main() -> None:
+    args = parse_args()
+    problems = load_jsonl(args.problems)
+    if not problems:
+        raise SystemExit("No evaluation problems")
+    student_manifest = Path(args.student_model) / "training_manifest.json"
+    if not student_manifest.is_file():
+        raise SystemExit(
+            "student-model lacks training_manifest.json; a mechanism/demo model cannot pass Experiment 8-9"
+        )
+
+    questions = [str(problem["question"]) for problem in problems]
+    reused_local_arms = None
+    if args.reuse_local_arms_from:
+        prior = json.loads(args.reuse_local_arms_from.read_text(encoding="utf-8"))
+        arms = {arm.get("name"): arm for arm in prior.get("arms", [])}
+        if set(arms) < {"baseline", "student"}:
+            raise SystemExit("reuse source lacks retained baseline and student arms")
+        baseline = arms["baseline"]
+        student = arms["student"]
+        reused_local_arms = {
+            "path": str(args.reuse_local_arms_from),
+            "sha256": sha256(args.reuse_local_arms_from),
+        }
+    else:
+        baseline = score_arm(
+            "baseline", problems, generate_local(args.baseline_model, questions, args.max_new_tokens)
+        )
+        student = score_arm(
+            "student", problems, generate_local(args.student_model, questions, args.max_new_tokens)
+        )
+    cached_teacher = teacher_outputs(args.teacher_data)
+    teacher_texts = [cached_teacher.get(str(p["id"]), cached_teacher.get(str(p["question"]), "")) for p in problems]
+    teacher = score_arm("teacher", problems, teacher_texts)
+
+    baseline_map = {r["id"]: r["correct"] for r in baseline["records"]}
+    student_map = {r["id"]: r["correct"] for r in student["records"]}
+    paired = compare_binary(baseline_map, student_map)
+    baseline_accuracy = baseline["accuracy"]
+    teacher_gap = max(0.0, teacher["accuracy"] - baseline_accuracy)
+    recovered = (
+        (student["accuracy"] - baseline_accuracy) / teacher_gap if teacher_gap > 0 else None
+    )
+    completion, findings = completion_and_findings(
+        problem_ids={str(problem["id"]) for problem in problems},
+        baseline=baseline,
+        student=student,
+        teacher=teacher,
+        paired=paired,
+        student_training_complete=json.loads(student_manifest.read_text(encoding="utf-8")).get("status") == "complete",
+        teacher_outputs_complete=all(bool(text) for text in teacher_texts),
+    )
+    payload = {
+        "schema_version": 1,
+        "experiment": "8-9",
+        "generated_at_utc": datetime.now(timezone.utc).isoformat(),
+        "status": "complete" if completion["complete"] else "incomplete",
+        "inputs": {
+            "problems": {"path": str(args.problems), "sha256": sha256(args.problems)},
+            "teacher_data": {"path": str(args.teacher_data), "sha256": sha256(args.teacher_data)},
+            "student_training_manifest": json.loads(student_manifest.read_text(encoding="utf-8")),
+            "reused_local_arms": reused_local_arms,
+        },
+        "models": {
+            "baseline": args.baseline_model,
+            "student": args.student_model,
+            "teacher": "cached real API trajectories",
+        },
+        "paired_comparison": paired,
+        "teacher_capability_recovered": recovered,
+        "completion": completion,
+        "findings": findings,
+        "arms": [baseline, student, teacher],
+    }
+    args.output.parent.mkdir(parents=True, exist_ok=True)
+    args.output.write_text(json.dumps(payload, ensure_ascii=False, indent=2) + "\n", encoding="utf-8")
+    print(json.dumps({
+        "output": str(args.output),
+        "status": payload["status"],
+        "accuracies": {arm["name"]: arm["accuracy"] for arm in payload["arms"]},
+        "paired_p": paired["exact_two_sided_p_value"],
+    }, ensure_ascii=False, indent=2))
+
+
+if __name__ == "__main__":
+    main()
@@ -0,0 +1,320 @@
+"""
+CoT 蒸馏数据采集脚本（实验 8-9 配套代码）
+
+方法（对应书中实验 8-9 的三步流程之第一步"采集轨迹"）：
+  1. 从 problems.jsonl 读取带标准答案的数学题（规则可验证的任务分布）；
+  2. 通过 OpenRouter 调用前沿教师模型（默认 Claude），开启 reasoning 获取
+     完整"思考 + 答案"轨迹（Claude 4 系列返回的是 summarized thinking——由另一个
+     模型对原始思维链做的高保真摘要，原始思维链只存在于加密的 signature 字段中）；
+  3. 用规则验证器核对最终答案，只把答对的轨迹写成 SFT 训练数据
+     （"问题 → <think>思考</think> + 最终答案" 的 messages 格式）。
+
+注意：本实验只使用各厂商官方 API 的 reasoning/thinking 能力获取思维链，
+不涉及任何绕过厂商安全机制的手段。原始轨迹（含未通过验证的）保存在
+raw_trajectories.jsonl，便于分析教师的错误模式。
+"""
+
+import argparse
+import asyncio
+import json
+import os
+import re
+from pathlib import Path
+from typing import Optional
+
+from openai import AsyncOpenAI
+
+ANSWER_SUFFIX = "\n\n请一步步推理，并在最后一行用「Final Answer: 数值」的格式给出最终答案（只写数值，不带单位）。"
+
+def load_problems(path: str) -> list[dict]:
+    problems = []
+    with open(path, "r", encoding="utf-8") as f:
+        for line in f:
+            line = line.strip()
+            if line:
+                problems.append(json.loads(line))
+    return problems
+
+
+def load_jsonl(path: str | Path) -> list[dict]:
+    path = Path(path)
+    if not path.is_file():
+        return []
+    with path.open("r", encoding="utf-8") as f:
+        return [json.loads(line) for line in f if line.strip()]
+
+
+def write_jsonl_atomic(path: str | Path, rows: list[dict]) -> None:
+    """Replace a JSONL file without exposing a partially rewritten dataset."""
+    path = Path(path)
+    path.parent.mkdir(parents=True, exist_ok=True)
+    temporary = path.with_name(f".{path.name}.{os.getpid()}.tmp")
+    try:
+        with temporary.open("w", encoding="utf-8") as f:
+            for row in rows:
+                f.write(json.dumps(row, ensure_ascii=False) + "\n")
+            f.flush()
+            os.fsync(f.fileno())
+        os.replace(temporary, path)
+    finally:
+        temporary.unlink(missing_ok=True)
+
+
+def records_in_problem_order(problems: list[dict], records_by_id: dict[str, dict]) -> list[dict]:
+    """Return canonical problem order while retaining any legacy extra records."""
+    known_ids = [str(problem["id"]) for problem in problems]
+    rows = [records_by_id[problem_id] for problem_id in known_ids if problem_id in records_by_id]
+    rows.extend(record for problem_id, record in records_by_id.items() if problem_id not in known_ids)
+    return rows
+
+
+def extract_predicted_number(text: str) -> Optional[float]:
+    """从模型输出中解析最终答案数值。优先匹配 Final Answer 标记，否则取最后一个数字。"""
+    m = re.findall(r"Final Answer[:：]\s*(-?[\d,]+(?:\.\d+)?)", text, re.IGNORECASE)
+    if not m:
+        m = re.findall(r"-?[\d,]+(?:\.\d+)?", text)
+    if not m:
+        return None
+    try:
+        return float(m[-1].replace(",", ""))
+    except ValueError:
+        return None
+
+
+def verify(text: str, gold: float, tol: float = 1e-6) -> bool:
+    """规则验证器：核对最终答案是否与标准答案一致。"""
+    pred = extract_predicted_number(text)
+    if pred is None:
+        return False
+    return abs(pred - float(gold)) <= tol * max(1.0, abs(float(gold)))
+
+
+def get_reasoning(message) -> str:
+    """从返回的 message 中提取思维链。
+
+    依次尝试：OpenRouter 的 reasoning / reasoning_details 字段，
+    以及 Moonshot、DeepSeek 等原生 API 的 reasoning_content 字段。
+    """
+    reasoning = getattr(message, "reasoning", None)
+    if reasoning:
+        return reasoning
+    reasoning_content = getattr(message, "reasoning_content", None)
+    if reasoning_content:
+        return reasoning_content
+    details = getattr(message, "reasoning_details", None) or []
+    parts = []
+    for d in details:
+        if isinstance(d, dict):
+            parts.append(d.get("text") or d.get("summary") or "")
+        else:
+            parts.append(getattr(d, "text", None) or getattr(d, "summary", None) or "")
+    return "\n".join(p for p in parts if p)
+
+
+def reasoning_extra_body(base_url: str, effort: str, max_tokens: int) -> dict:
+    """Build the provider-specific reasoning control without silently ignoring it."""
+    if effort:
+        if "api.moonshot.cn" in base_url:
+            # Moonshot's native OpenAI-compatible endpoint accepts the same
+            # top-level control used by the Experiment 8-8 Kimi campaign.
+            return {"reasoning_effort": effort}
+        return {"reasoning": {"effort": effort}}
+    if max_tokens:
+        return {"reasoning": {"max_tokens": max_tokens}}
+    return {}
+
+
+async def distill_one(client: AsyncOpenAI, problem: dict, args, semaphore) -> dict:
+    """对单道题调用教师模型，返回完整轨迹记录。"""
+    record = {
+        "id": problem["id"],
+        "question": problem["question"],
+        "gold_answer": problem["answer"],
+        "model": args.model,
+        "content": None,
+        "reasoning": None,
+        "verified": False,
+        "usage": None,
+        "error": None,
+        "attempts": [],
+    }
+    async with semaphore:
+        for attempt in range(args.max_retries + 1):
+            try:
+                kwargs = {}
+                reasoning_body = reasoning_extra_body(
+                    args.base_url, args.reasoning_effort, args.reasoning_max_tokens
+                )
+                if reasoning_body:
+                    kwargs["extra_body"] = reasoning_body
+                resp = await asyncio.wait_for(
+                    client.chat.completions.create(
+                        model=args.model,
+                        messages=[{"role": "user", "content": problem["question"] + args.answer_suffix}],
+                        max_tokens=args.max_tokens,
+                        # 重试时升温换取不同轨迹；Kimi K3 等锁定 temperature=1 的模型除外
+                        temperature=args.temperature + (0.2 * attempt if args.temperature < 1.0 else 0),
+                        **kwargs,
+                    ),
+                    timeout=args.request_timeout,  # 硬超时：防止半开连接挂死
+                )
+                msg = resp.choices[0].message
+                record["content"] = msg.content or ""
+                record["reasoning"] = get_reasoning(msg)
+                record["usage"] = resp.usage.model_dump() if resp.usage else None
+                record["verified"] = verify(record["content"], problem["answer"])
+                record["error"] = None
+                record["attempts"].append({
+                    "attempt": attempt,
+                    "content": record["content"],
+                    "reasoning": record["reasoning"],
+                    "usage": record["usage"],
+                    "verified": record["verified"],
+                    "error": None,
+                })
+                if record["verified"]:
+                    break
+            except Exception as e:
+                record["error"] = f"attempt {attempt}: {type(e).__name__}: {e}"
+                record["attempts"].append({
+                    "attempt": attempt,
+                    "content": None,
+                    "reasoning": None,
+                    "usage": None,
+                    "verified": False,
+                    "error": record["error"],
+                })
+        status = "OK" if record["verified"] else ("ERR" if record["error"] else "WRONG")
+        print(f"  [{status}] {record['id']}", flush=True)
+        return record
+
+
+def to_sft_sample(record: dict) -> dict:
+    """把验证通过的轨迹转成 SFT 训练样本（messages 格式，思考包在 <think> 标签里）。"""
+    if record["reasoning"]:
+        assistant = f"<think>\n{record['reasoning'].strip()}\n</think>\n\n{record['content'].strip()}"
+    else:
+        assistant = record["content"].strip()
+    return {
+        "messages": [
+            {"role": "user", "content": record["question"]},
+            {"role": "assistant", "content": assistant},
+        ]
+    }
+
+
+async def main():
+    parser = argparse.ArgumentParser(
+        description="用前沿云模型（经 OpenRouter）蒸馏 CoT 轨迹，生成 SFT 数据",
+        formatter_class=argparse.ArgumentDefaultsHelpFormatter,
+    )
+    parser.add_argument("--input", default="./problems.jsonl", help="题目文件（JSONL，含 question/answer）")
+    parser.add_argument("--sft_output", default="./data/sft_cot_distill.jsonl", help="SFT 训练数据输出路径")
+    parser.add_argument("--raw_output", default="./data/raw_trajectories.jsonl", help="原始轨迹（含失败样本）输出路径")
+    parser.add_argument("--model", default="anthropic/claude-opus-4.8", help="教师模型 ID")
+    parser.add_argument("--base_url", default="https://openrouter.ai/api/v1", help="OpenAI 兼容 API 端点")
+    parser.add_argument("--api_key_env", default="OPENROUTER_API_KEY", help="存放 API Key 的环境变量名")
+    parser.add_argument("--reasoning_effort", default="",
+                        help="OpenRouter 风格 reasoning effort（如 high/medium/low；设置后优先于 --reasoning_max_tokens，"
+                             "用于 Claude Opus 4.8 等只支持自适应思考的模型）")
+    parser.add_argument("--reasoning_max_tokens", type=int, default=4096,
+                        help="思维链最大 token 数（OpenRouter 风格 reasoning 参数；0 = 不传该参数，"
+                             "用于 Moonshot/DeepSeek 等默认返回 reasoning_content 的原生 API）")
+    parser.add_argument("--max_problems", type=int, default=0, help="最多处理多少题（0 = 全部，调试用）")
+    parser.add_argument(
+        "--problem-id",
+        action="append",
+        default=[],
+        help="只运行指定题目 ID；可重复传入。用于定点重试而不重跑整套题",
+    )
+    parser.add_argument(
+        "--resume",
+        action="store_true",
+        help="保留 raw_output 中已验证记录，只重试缺失或未验证题目，并原子更新数据集",
+    )
+    parser.add_argument("--concurrency", type=int, default=8, help="并发请求数")
+    parser.add_argument("--temperature", type=float, default=0.3, help="采样温度")
+    parser.add_argument("--max_tokens", type=int, default=8192, help="单条回复最大 token 数（须大于 reasoning tokens）")
+    parser.add_argument("--max_retries", type=int, default=1, help="失败/出错后的最大重试次数")
+    parser.add_argument("--request_timeout", type=float, default=600, help="单次请求超时（秒），超时后按失败重试")
+    parser.add_argument("--answer_suffix", default=ANSWER_SUFFIX, help="附加在题目后的作答格式要求")
+    args = parser.parse_args()
+
+    api_key = os.environ.get(args.api_key_env)
+    if not api_key:
+        raise SystemExit(f"请先设置环境变量 {args.api_key_env}")
+
+    all_problems = load_problems(args.input)
+    problem_ids = {str(problem["id"]) for problem in all_problems}
+    requested_ids = set(args.problem_id)
+    unknown_ids = sorted(requested_ids - problem_ids)
+    if unknown_ids:
+        raise SystemExit(f"未知题目 ID: {', '.join(unknown_ids)}")
+    problems = [
+        problem for problem in all_problems
+        if not requested_ids or str(problem["id"]) in requested_ids
+    ]
+    if args.max_problems:
+        problems = problems[: args.max_problems]
+
+    existing_rows = load_jsonl(args.raw_output) if args.resume else []
+    records_by_id = {
+        str(record["id"]): record for record in existing_rows if record.get("id") is not None
+    }
+    pending = [
+        problem for problem in problems
+        if not records_by_id.get(str(problem["id"]), {}).get("verified", False)
+    ]
+    print(
+        f"选中 {len(problems)} 道题，待运行 {len(pending)} 道，"
+        f"教师模型：{args.model} @ {args.base_url}"
+    )
+
+    client = AsyncOpenAI(base_url=args.base_url, api_key=api_key, timeout=args.request_timeout)
+    semaphore = asyncio.Semaphore(args.concurrency)
+
+    # 每题完成后原子替换：中断最多损失当前请求，不会破坏已有数据集。
+    run_records = []
+    tasks = [distill_one(client, p, args, semaphore) for p in pending]
+    for coro in asyncio.as_completed(tasks):
+        record = await coro
+        previous = records_by_id.get(str(record["id"]))
+        if previous and not previous.get("verified", False):
+            prior_failures = list(previous.get("prior_failures") or [])
+            prior_failures.append({
+                "model": previous.get("model"),
+                "verified": False,
+                "error": previous.get("error"),
+                "usage": previous.get("usage"),
+            })
+            record["prior_failures"] = prior_failures
+        records_by_id[str(record["id"])] = record
+        run_records.append(record)
+        write_jsonl_atomic(
+            args.raw_output,
+            records_in_problem_order(all_problems, records_by_id),
+        )
+
+    records = records_in_problem_order(all_problems, records_by_id)
+    write_jsonl_atomic(args.raw_output, records)
+    passed = [record for record in records if record.get("verified", False)]
+    write_jsonl_atomic(args.sft_output, [to_sft_sample(record) for record in passed])
+
+    total_in = sum((r["usage"] or {}).get("prompt_tokens", 0) for r in run_records)
+    total_out = sum((r["usage"] or {}).get("completion_tokens", 0) for r in run_records)
+    n_err = sum(1 for r in run_records if r["error"])
+    print(f"\n{'=' * 50}")
+    # Empty problems JSONL yields zero records; avoid ZeroDivisionError on the rate.
+    pass_rate = (len(passed) / len(records) * 100) if records else 0.0
+    print(f"数据集验证通过：{len(passed)}/{len(records)}（{pass_rate:.1f}%）")
+    print(
+        f"本次请求：{len(run_records)}  API 最终出错：{n_err}  "
+        f"无思维链返回：{sum(1 for r in run_records if not r['reasoning'])}"
+    )
+    print(f"本次 Token 消耗：输入 {total_in}，输出 {total_out}")
+    print(f"SFT 数据已写入：{args.sft_output}")
+    print(f"原始轨迹已写入：{args.raw_output}")
+
+
+if __name__ == "__main__":
+    asyncio.run(main())
@@ -0,0 +1,24 @@
+{"id": "aime-1986-1", "question": "What is the sum of the solutions to the equation $\\sqrt[4]{x} = \\frac{12}{7 - \\sqrt[4]{x}}$ ?", "answer": 337}
+{"id": "aime-1986-7", "question": "The increasing sequence $1,3,4,9,10,12,13\\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\\mbox{th}}$ term of this sequence.", "answer": 981}
+{"id": "aime-1986-8", "question": "Let $S$ be the sum of the base $10$ logarithms of all the proper divisors of $1000000$ . What is the integer nearest to $S$ ?", "answer": 141}
+{"id": "aime-1986-14", "question": "The shortest distances between an interior diagonal of a rectangular parallelepiped , $P$ , and the edges it does not meet are $2\\sqrt{5}$ , $\\frac{30}{\\sqrt{13}}$ , and $\\frac{15}{\\sqrt{10}}$ . Determine the volume of $P$ .", "answer": 750}
+{"id": "aime-1993-8", "question": "Let $S\\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\\,$ so that the union of the two subsets is $S\\,$ ?  The order of selection does not matter; for example, the pair of subsets $\\{a, c\\},\\{b, c, d, e, f\\}$ represents the same selection as the pair $\\{b, c, d, e, f\\},\\{a, c\\}.$", "answer": 365}
+{"id": "aime-1994-9", "question": "A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,\\,$ where $p\\,$ and $q\\,$ are relatively prime positive integers.  Find $p+q.\\,$", "answer": 394}
+{"id": "aime-1996-2", "question": "For each real number $x$ , let $\\lfloor x \\rfloor$ denote the greatest integer that does not exceed $x$ . For how many positive integers $n$ is it true that $n<1000$ and that $\\lfloor \\log_{2} n \\rfloor$ is a positive even integer?", "answer": 340}
+{"id": "aime-1997-2", "question": "The nine horizontal and nine vertical lines on an $8\\times8$ checkerboard form $r$ rectangles, of which $s$ are squares.  The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$", "answer": 125}
+{"id": "aime-2000-2-I", "question": "Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$ .", "answer": 21}
+{"id": "aime-2003-13-II", "question": "A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$", "answer": 683}
+{"id": "aime-2004-5-II", "question": "In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?", "answer": 766}
+{"id": "aime-2004-9-I", "question": "Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$", "answer": 35}
+{"id": "aime-2004-15-II", "question": "A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?", "answer": 593}
+{"id": "aime-2005-8-I", "question": "The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $\\frac mn$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$", "answer": 113}
+{"id": "aime-2006-1-I", "question": "In quadrilateral $ABCD , \\angle B$ is a right angle, diagonal $\\overline{AC}$ is perpendicular to $\\overline{CD},  AB=18, BC=21,$ and $CD=14.$ Find the perimeter of $ABCD.$", "answer": 84}
+{"id": "aime-2007-2-II", "question": "Find the number of ordered triples $(a,b,c)$ where $a$ , $b$ , and $c$ are positive integers , $a$ is a factor of $b$ , $a$ is a factor of $c$ , and $a+b+c=100$ .", "answer": 200}
+{"id": "aime-2008-13-I", "question": "Let $p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3$ . Suppose that $p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1) = p(1,1) = p(1, - 1) = p(2,2) = 0$ . There is a point $\\left(\\frac {a}{c},\\frac {b}{c}\\right)$ for which $p\\left(\\frac {a}{c},\\frac {b}{c}\\right) = 0$ for all such polynomials, where $a$ , $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ .  Find $a + b + c$ .", "answer": 40}
+{"id": "aime-2016-9-I", "question": "Triangle $ABC$ has $AB=40,AC=31,$ and $\\sin{A}=\\frac{1}{5}$ . This triangle is inscribed in rectangle $AQRS$ with $B$ on $\\overline{QR}$ and $C$ on $\\overline{RS}$ . Find the maximum possible area of $AQRS$ .", "answer": 744}
+{"id": "aime-2016-12-I", "question": "Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.", "answer": 132}
+{"id": "aime-2017-15-II", "question": "Tetrahedron $ABCD$ has $AD=BC=28$ , $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, define $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .", "answer": 682}
+{"id": "aime-2020-8-II", "question": "Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ .", "answer": 101}
+{"id": "aime-2022-1-II", "question": "Adults made up $\\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.", "answer": 154}
+{"id": "aime-2023-12-I", "question": "Let $\\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\\overline{BC},$ $\\overline{CA},$ and $\\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\\triangle ABC$ has the property that \\[\\angle AEP = \\angle BFP = \\angle CDP.\\] Find $\\tan^2(\\angle AEP).$", "answer": 75}
+{"id": "aime-2024-14-II", "question": "Let $b \\geq 2$ be an integer. Call a positive integer $n$ $b\\textit{-eautiful}$ if it has exactly two digits when expressed in base $b$ , and these two digits sum to $\\sqrt{n}$ . For example, $81$ is $13$ -eautiful because $81=\\underline{6}$ $\\underline{3}_{13}$ and $6+3=\\sqrt{81}$ . Find the least integer $b\\geq 2$ for which there are more than ten $b$ -eautiful integers.", "answer": 211}
@@ -0,0 +1,24 @@
+{"id": "m01", "question": "一个书架有 5 层，每层放 28 本书。已经借出了 37 本，书架上还剩多少本书？", "answer": 103}
+{"id": "m02", "question": "小明买了 3 支单价 4.5 元的笔和 2 本单价 12 元的笔记本，付了 50 元，应找回多少元？", "answer": 12.5}
+{"id": "m03", "question": "一辆汽车以每小时 60 公里的速度行驶了 2.5 小时，又以每小时 80 公里的速度行驶了 1.5 小时，一共行驶了多少公里？", "answer": 270}
+{"id": "m04", "question": "某班有 48 名学生，其中 3/8 参加了数学兴趣小组，参加数学兴趣小组的有多少人？", "answer": 18}
+{"id": "m05", "question": "一个长方形的长是宽的 3 倍，周长是 64 厘米，它的面积是多少平方厘米？", "answer": 192}
+{"id": "m06", "question": "商店把一件商品先提价 20%，再降价 20%，现价是 96 元。这件商品的原价是多少元？", "answer": 100}
+{"id": "m07", "question": "5 台同样的机器 8 小时可以生产 600 个零件。照这样计算，8 台机器 10 小时可以生产多少个零件？", "answer": 1200}
+{"id": "m08", "question": "某数加上它的 1/4 等于 35，这个数是多少？", "answer": 28}
+{"id": "m09", "question": "甲、乙两人从相距 240 公里的两地同时出发相向而行，甲每小时走 14 公里，乙每小时走 10 公里，几小时后两人相遇？", "answer": 10}
+{"id": "m10", "question": "一个等差数列的第 3 项是 11，第 7 项是 27，它的第 10 项是多少？", "answer": 39}
+{"id": "m11", "question": "农场里鸡和兔共有 35 个头、94 只脚，兔有多少只？", "answer": 12}
+{"id": "m12", "question": "一项工程，甲单独做 12 天完成，乙单独做 18 天完成。两人合作 4 天后，剩下的工程由乙单独完成，还需要多少天？", "answer": 8}
+{"id": "m13", "question": "某商品进价 80 元，按标价卖出可赚 25%。若按标价的 9 折出售，每件可赚多少元？", "answer": 10}
+{"id": "m14", "question": "把 126 本书按 4:5 分给甲、乙两个班，乙班分到多少本？", "answer": 70}
+{"id": "m15", "question": "一个水池，单开进水管 6 小时注满，单开出水管 10 小时放空。两管同时打开，多少小时可以注满空水池？", "answer": 15}
+{"id": "m16", "question": "小华今年的年龄是小丽的 3 倍，8 年后小华的年龄是小丽的 2 倍。小华今年多少岁？", "answer": 24}
+{"id": "m17", "question": "某工厂第一季度生产零件 2400 个，第二季度比第一季度增产 15%，两个季度一共生产了多少个零件？", "answer": 5160}
+{"id": "m18", "question": "一个两位数的个位数字比十位数字大 3，这个两位数等于它的两个数字之和的 4 倍，这个两位数是多少？", "answer": 36}
+{"id": "m19", "question": "从 1 到 100 的所有整数中，能被 3 整除但不能被 5 整除的数有多少个？", "answer": 27}
+{"id": "m20", "question": "甲容器中有浓度 20% 的盐水 300 克，乙容器中有浓度 10% 的盐水 200 克，混合后盐水的浓度是多少？（用百分数表示）", "answer": 16}
+{"id": "m21", "question": "某快递公司收费标准：首重 1 公斤收 8 元，之后每增加 1 公斤加收 3 元（不足 1 公斤按 1 公斤计）。寄一件 7.3 公斤的包裹要多少元？", "answer": 29}
+{"id": "m22", "question": "一个圆的半径扩大到原来的 2 倍，它的面积是 314 平方厘米（π 取 3.14）。原来的半径是多少厘米？", "answer": 5}
+{"id": "m23", "question": "连续三个偶数的和是 78，其中最大的偶数是多少？", "answer": 28}
+{"id": "m24", "question": "某班一次数学测验的平均分是 84 分。去掉一个 60 分的成绩后，其余 24 人的平均分是 85 分。这个班原来有多少人？", "answer": 25}
@@ -0,0 +1,7 @@
+openai>=1.40,<3
+torch>=2.3,<3
+# Keep the trainer trio together.  Broad lower bounds allowed pip to combine
+# transformers 5.x with an old PEFT installation, which cannot import Trainer.
+transformers==4.48.3
+accelerate==1.2.1
+peft==0.14.0
@@ -0,0 +1,665 @@
+"""
+SFT training-data quality auditor (chapter 8 CoT distillation).
+
+Every chapter 8 SFT experiment (8-8, 8-9, 8-17, 8-18, 8-19) consumes JSONL
+training data: one JSON object per line, each carrying a ``messages`` array of
+``{"role", "content"}`` pairs. ``generate_data.py`` synthesizes that data and
+``analyze_data.py`` reports coarse statistics, but neither flags the quality
+issues that corrupt a fine-tune *before* training starts. This module fills
+that gap.
+
+:class:`SFTDataQualityAuditor` walks a dataset (a file or an in-memory list of
+parsed lines) and produces an :class:`AuditReport` of :class:`QualityIssue`
+records covering six concerns:
+
+1. **Format consistency** — every line has a ``messages`` list whose entries
+   carry ``role`` and non-empty ``content`` and whose roles alternate
+   ``user``/``assistant``.
+2. **Token-length distribution** — per-example approximate token count (word
+   count) with outliers flagged below ``min_length`` or above ``max_length``.
+3. **Duplicate detection** — exact duplicate examples and near-duplicates
+   (same user message, different assistant response = potential label noise).
+4. **Label noise** — assistant responses containing placeholder markers
+   (``TODO``, ``FIXME``, ``[insert``, ``[TBD``) or self-contradictory
+   affirm/deny pairs.
+5. **Boundary coverage** — the dataset should span diverse input lengths, not
+   cluster around a single bucket.
+6. **Tokenizer compatibility** — characters that tokenize inconsistently
+   across tokenizers (curly quotes, zero-width spaces, BOM markers).
+
+The auditor is fully offline: it never loads a model or touches the network.
+"""
+from __future__ import annotations
+
+import json
+import re
+import statistics
+from dataclasses import dataclass, field
+from pathlib import Path
+from typing import Any
+
+
+# --------------------------------------------------------------------------- #
+# Data shapes
+# --------------------------------------------------------------------------- #
+@dataclass
+class QualityIssue:
+    """A single quality problem found in one example."""
+
+    line_number: int
+    issue_type: str  # format_error, length_outlier, duplicate, label_noise, boundary_gap, tokenizer_risk
+    severity: str  # warning, error
+    description: str
+    evidence: dict[str, Any] = field(default_factory=dict)
+
+
+@dataclass
+class AuditReport:
+    """Aggregate quality report for a whole dataset."""
+
+    total_examples: int = 0
+    total_issues: int = 0
+    issues_by_severity: dict[str, int] = field(default_factory=dict)
+    issues_by_type: dict[str, int] = field(default_factory=dict)
+    length_stats: dict[str, float] = field(default_factory=dict)
+    duplicate_count: int = 0
+    near_duplicate_count: int = 0
+    issues: list[QualityIssue] = field(default_factory=list)
+    overall_quality_score: float = 0.0
+
+
+# --------------------------------------------------------------------------- #
+# Auditor
+# --------------------------------------------------------------------------- #
+_PLACEHOLDER_RE = re.compile(
+    r"\b(TODO|FIXME)\b|\[insert|\[TBD", re.IGNORECASE
+)
+# Self-contradiction: an affirmative followed later by its negation (or vice
+# versa) inside the same assistant turn — "Yes ... No" / "True ... False".
+_CONTRADICTION_RE = re.compile(
+    r"\b(yes|true|correct|right)\b.*\b(no|false|wrong|incorrect)\b"
+    r"|\b(no|false|wrong|incorrect)\b.*\b(yes|true|correct|right)\b",
+    re.IGNORECASE,
+)
+# Tokenizer-hostile characters: curly quotes, zero-width spaces, BOM.
+_SPECIAL_CHARS = {
+    "\u2018": "left single curly quote",
+    "\u2019": "right single curly quote",
+    "\u201c": "left double curly quote",
+    "\u201d": "right double curly quote",
+    "\u200b": "zero-width space",
+    "\ufeff": "BOM / zero-width no-break space",
+    "\u200c": "zero-width non-joiner",
+    "\u200d": "zero-width joiner",
+}
+
+_VALID_ROLES = {"user", "assistant", "system", "tool"}
+# Number of length buckets used for boundary-coverage analysis.
+_LENGTH_BUCKETS = 5
+
+
+class SFTDataQualityAuditor:
+    """Audit SFT JSONL training data for common quality issues.
+
+    Parameters
+    ----------
+    max_length:
+        Approximate token (word) count above which an example is flagged as a
+        length outlier (may truncate at training time).
+    min_length:
+        Approximate token (word) count below which an example is flagged as a
+        length outlier (likely uninformative).
+    """
+
+    def __init__(self, max_length: int = 4096, min_length: int = 10) -> None:
+        if max_length <= 0:
+            raise ValueError("max_length must be positive")
+        if min_length < 0:
+            raise ValueError("min_length must be non-negative")
+        if min_length >= max_length:
+            raise ValueError("min_length must be less than max_length")
+        self.max_length = max_length
+        self.min_length = min_length
+
+    # ------------------------------------------------------------------ #
+    # Public entry points
+    # ------------------------------------------------------------------ #
+    def audit_file(self, path: str | Path) -> AuditReport:
+        """Read a JSONL file and audit every non-blank line."""
+        p = Path(path)
+        lines: list[dict[str, Any]] = []
+        with p.open(encoding="utf-8") as f:
+            for raw in f:
+                stripped = raw.strip()
+                if not stripped:
+                    continue
+                lines.append(json.loads(stripped))
+        return self.audit_lines(lines)
+
+    def audit_lines(self, lines: list[dict[str, Any]]) -> AuditReport:
+        """Audit an in-memory list of parsed JSONL examples."""
+        examples: list[dict[str, Any]] = list(lines)
+        total = len(examples)
+
+        issues: list[QualityIssue] = []
+        # Per-example approximate token counts (word counts). Format-invalid
+        # examples contribute 0 so they don't skew the distribution.
+        token_counts: list[int] = []
+
+        for idx, example in enumerate(examples):
+            line_number = idx + 1
+            fmt_issues = self.check_format(example)
+            for issue in fmt_issues:
+                issue.line_number = line_number
+            issues.extend(fmt_issues)
+
+            token_counts.append(self._example_token_count(example))
+
+            issues.extend(self.check_length(example, line_number))
+            issues.extend(self.check_label_noise(example, line_number))
+            issues.extend(self.check_tokenizer_compatibility(example, line_number))
+
+        issues.extend(self.find_duplicates(examples))
+        issues.extend(self._find_boundary_gaps(token_counts))
+
+        # Deduplicate duplicate/near-duplicate counts from the issue list.
+        duplicate_count = sum(
+            1 for i in issues if i.issue_type == "duplicate" and i.evidence.get("kind") == "exact"
+        )
+        near_duplicate_count = sum(
+            1 for i in issues if i.issue_type == "duplicate" and i.evidence.get("kind") == "near"
+        )
+
+        length_stats = self._length_stats(token_counts)
+        issues_by_severity = _count_by(issues, lambda i: i.severity)
+        issues_by_type = _count_by(issues, lambda i: i.issue_type)
+        score = self._quality_score(total, issues)
+
+        return AuditReport(
+            total_examples=total,
+            total_issues=len(issues),
+            issues_by_severity=issues_by_severity,
+            issues_by_type=issues_by_type,
+            length_stats=length_stats,
+            duplicate_count=duplicate_count,
+            near_duplicate_count=near_duplicate_count,
+            issues=issues,
+            overall_quality_score=score,
+        )
+
+    # ------------------------------------------------------------------ #
+    # Individual checks
+    # ------------------------------------------------------------------ #
+    def check_format(self, example: dict[str, Any]) -> list[QualityIssue]:
+        """Validate the structural shape of one example.
+
+        Issues are returned with ``line_number=0``; the caller (``audit_lines``)
+        stamps the real line number. When called directly the caller is
+        responsible for setting it.
+        """
+        issues: list[QualityIssue] = []
+
+        messages = example.get("messages")
+        if not isinstance(messages, list):
+            issues.append(
+                QualityIssue(
+                    line_number=0,
+                    issue_type="format_error",
+                    severity="error",
+                    description="'messages' is missing or not a list",
+                    evidence={"messages_type": type(messages).__name__},
+                )
+            )
+            return issues
+
+        if len(messages) == 0:
+            issues.append(
+                QualityIssue(
+                    line_number=0,
+                    issue_type="format_error",
+                    severity="error",
+                    description="'messages' is an empty list",
+                    evidence={},
+                )
+            )
+            return issues
+
+        expected_role = "user"
+        for pos, msg in enumerate(messages):
+            if not isinstance(msg, dict):
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] is not a dict",
+                        evidence={"position": pos, "type": type(msg).__name__},
+                    )
+                )
+                continue
+
+            role = msg.get("role")
+            content = msg.get("content")
+
+            if not isinstance(role, str) or not role:
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] missing or invalid 'role'",
+                        evidence={"position": pos, "role": role},
+                    )
+                )
+            elif role not in _VALID_ROLES:
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] has unknown role {role!r}",
+                        evidence={"position": pos, "role": role},
+                    )
+                )
+
+            if content is None:
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] missing 'content'",
+                        evidence={"position": pos},
+                    )
+                )
+            elif not isinstance(content, str):
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] 'content' is not a string",
+                        evidence={"position": pos, "content_type": type(content).__name__},
+                    )
+                )
+            elif not content.strip():
+                issues.append(
+                    QualityIssue(
+                        line_number=0,
+                        issue_type="format_error",
+                        severity="error",
+                        description=f"message[{pos}] has empty 'content'",
+                        evidence={"position": pos},
+                    )
+                )
+
+            # Role alternation: the first message should be 'user', then
+            # 'assistant', then 'user', etc. System/tool messages are allowed
+            # but do not reset the expected alternation.
+            if isinstance(role, str) and role in {"user", "assistant"}:
+                if role != expected_role:
+                    issues.append(
+                        QualityIssue(
+                            line_number=0,
+                            issue_type="format_error",
+                            severity="error",
+                            description=(
+                                f"message[{pos}] role {role!r} breaks expected "
+                                f"alternation (expected {expected_role!r})"
+                            ),
+                            evidence={
+                                "position": pos,
+                                "role": role,
+                                "expected": expected_role,
+                            },
+                        )
+                    )
+                expected_role = "assistant" if expected_role == "user" else "user"
+
+        return issues
+
+    def check_length(self, example: dict[str, Any], line_number: int) -> list[QualityIssue]:
+        """Flag examples whose approximate token count is an outlier."""
+        issues: list[QualityIssue] = []
+        count = self._example_token_count(example)
+        if count == 0:
+            # Format errors already cover empty/missing content.
+            return issues
+
+        if count < self.min_length:
+            issues.append(
+                QualityIssue(
+                    line_number=line_number,
+                    issue_type="length_outlier",
+                    severity="warning",
+                    description=(
+                        f"example is very short (~{count} tokens, below "
+                        f"min_length={self.min_length}); likely uninformative"
+                    ),
+                    evidence={"token_count": count, "threshold": "min", "limit": self.min_length},
+                )
+            )
+        elif count > self.max_length:
+            issues.append(
+                QualityIssue(
+                    line_number=line_number,
+                    issue_type="length_outlier",
+                    severity="warning",
+                    description=(
+                        f"example is very long (~{count} tokens, above "
+                        f"max_length={self.max_length}); may truncate at training time"
+                    ),
+                    evidence={"token_count": count, "threshold": "max", "limit": self.max_length},
+                )
+            )
+        return issues
+
+    def find_duplicates(self, examples: list[dict[str, Any]]) -> list[QualityIssue]:
+        """Detect exact duplicates and near-duplicates (same user, different assistant)."""
+        issues: list[QualityIssue] = []
+
+        # Exact duplicates: identical serialised form seen more than once.
+        seen: dict[str, list[int]] = {}
+        for idx, example in enumerate(examples):
+            key = json.dumps(example, sort_keys=True, ensure_ascii=False)
+            seen.setdefault(key, []).append(idx + 1)
+        for key, line_numbers in seen.items():
+            if len(line_numbers) > 1:
+                for ln in line_numbers[1:]:  # keep the first occurrence clean
+                    issues.append(
+                        QualityIssue(
+                            line_number=ln,
+                            issue_type="duplicate",
+                            severity="error",
+                            description=(
+                                f"exact duplicate of line {line_numbers[0]} "
+                                f"({len(line_numbers)} copies total)"
+                            ),
+                            evidence={
+                                "kind": "exact",
+                                "first_line": line_numbers[0],
+                                "copy_count": len(line_numbers),
+                            },
+                        )
+                    )
+
+        # Near-duplicates: same user message, different assistant response.
+        by_user: dict[str, list[int]] = {}
+        for idx, example in enumerate(examples):
+            user_msg = self._first_user_content(example)
+            if user_msg is None:
+                continue
+            by_user.setdefault(user_msg, []).append(idx + 1)
+        for user_msg, line_numbers in by_user.items():
+            if len(line_numbers) < 2:
+                continue
+            # Only flag as near-duplicate when the assistant responses differ.
+            assistant_responses: dict[int, str] = {}
+            for ln in line_numbers:
+                resp = self._assistant_content(examples[ln - 1])
+                if resp is not None:
+                    assistant_responses[ln] = resp
+            unique_responses = set(assistant_responses.values())
+            if len(unique_responses) > 1:
+                for ln in line_numbers[1:]:
+                    issues.append(
+                        QualityIssue(
+                            line_number=ln,
+                            issue_type="duplicate",
+                            severity="warning",
+                            description=(
+                                f"near-duplicate: same user message as line "
+                                f"{line_numbers[0]} but different assistant response "
+                                f"(potential label noise)"
+                            ),
+                            evidence={
+                                "kind": "near",
+                                "first_line": line_numbers[0],
+                                "user_preview": user_msg[:80],
+                                "distinct_responses": len(unique_responses),
+                            },
+                        )
+                    )
+
+        return issues
+
+    def check_label_noise(self, example: dict[str, Any], line_number: int) -> list[QualityIssue]:
+        """Flag placeholder text and self-contradictory assistant responses."""
+        issues: list[QualityIssue] = []
+        assistant = self._assistant_content(example)
+        if assistant is None:
+            return issues
+
+        placeholders = _PLACEHOLDER_RE.findall(assistant)
+        if placeholders:
+            matched = [m if isinstance(m, str) else m[0] for m in placeholders]
+            issues.append(
+                QualityIssue(
+                    line_number=line_number,
+                    issue_type="label_noise",
+                    severity="error",
+                    description=(
+                        "assistant response contains placeholder text "
+                        f"{matched}; likely unfinished generation"
+                    ),
+                    evidence={"markers": matched},
+                )
+            )
+
+        if _CONTRADICTION_RE.search(assistant):
+            snippet = _CONTRADICTION_RE.search(assistant)
+            issues.append(
+                QualityIssue(
+                    line_number=line_number,
+                    issue_type="label_noise",
+                    severity="warning",
+                    description=(
+                        "assistant response contains a potential self-contradiction "
+                        "(affirm/deny pair in the same turn)"
+                    ),
+                    evidence={"match": snippet.group(0)[:80] if snippet else ""},
+                )
+            )
+
+        return issues
+
+    def check_tokenizer_compatibility(
+        self, example: dict[str, Any], line_number: int
+    ) -> list[QualityIssue]:
+        """Flag characters that tokenize inconsistently across tokenizers."""
+        issues: list[QualityIssue] = []
+        messages = example.get("messages")
+        if not isinstance(messages, list):
+            return issues
+
+        found: dict[str, list[str]] = {}
+        for pos, msg in enumerate(messages):
+            if not isinstance(msg, dict):
+                continue
+            content = msg.get("content")
+            if not isinstance(content, str):
+                continue
+            for ch, label in _SPECIAL_CHARS.items():
+                if ch in content:
+                    found.setdefault(label, []).append(f"message[{pos}]")
+
+        if found:
+            issues.append(
+                QualityIssue(
+                    line_number=line_number,
+                    issue_type="tokenizer_risk",
+                    severity="warning",
+                    description=(
+                        "example contains characters that may tokenize "
+                        "differently across tokenizers: "
+                        + ", ".join(found.keys())
+                    ),
+                    evidence={"characters": found},
+                )
+            )
+        return issues
+
+    # ------------------------------------------------------------------ #
+    # Boundary coverage (internal)
+    # ------------------------------------------------------------------ #
+    def _find_boundary_gaps(self, token_counts: list[int]) -> list[QualityIssue]:
+        """Flag when examples cluster into too few length buckets."""
+        issues: list[QualityIssue] = []
+        valid = [c for c in token_counts if c > 0]
+        if len(valid) < _LENGTH_BUCKETS:
+            # Not enough examples to meaningfully demand spread across buckets.
+            return issues
+
+        lo = float(min(valid))
+        hi = float(max(valid))
+        if hi <= lo:
+            issues.append(
+                QualityIssue(
+                    line_number=0,
+                    issue_type="boundary_gap",
+                    severity="warning",
+                    description=(
+                        "all examples share the same length; no boundary diversity"
+                    ),
+                    evidence={
+                        "occupied_buckets": 1,
+                        "total_buckets": _LENGTH_BUCKETS,
+                        "min": lo,
+                        "max": hi,
+                    },
+                )
+            )
+            return issues
+        bucket_size = (hi - lo) / _LENGTH_BUCKETS
+        occupied = 0
+        for b in range(_LENGTH_BUCKETS):
+            low_edge = lo + b * bucket_size
+            high_edge = lo + (b + 1) * bucket_size
+            if b == _LENGTH_BUCKETS - 1:
+                in_bucket = any(low_edge <= c <= high_edge for c in valid)
+            else:
+                in_bucket = any(low_edge <= c < high_edge for c in valid)
+            if in_bucket:
+                occupied += 1
+
+        if occupied <= _LENGTH_BUCKETS // 2:
+            issues.append(
+                QualityIssue(
+                    line_number=0,
+                    issue_type="boundary_gap",
+                    severity="warning",
+                    description=(
+                        f"length distribution covers only {occupied}/"
+                        f"{_LENGTH_BUCKETS} buckets; input lengths are clustered"
+                    ),
+                    evidence={
+                        "occupied_buckets": occupied,
+                        "total_buckets": _LENGTH_BUCKETS,
+                        "min": lo,
+                        "max": hi,
+                    },
+                )
+            )
+        return issues
+
+    # ------------------------------------------------------------------ #
+    # Helpers
+    # ------------------------------------------------------------------ #
+    def _example_token_count(self, example: dict[str, Any]) -> int:
+        """Approximate token count as the total word count across all messages."""
+        messages = example.get("messages")
+        if not isinstance(messages, list):
+            return 0
+        total = 0
+        for msg in messages:
+            if not isinstance(msg, dict):
+                continue
+            content = msg.get("content")
+            if isinstance(content, str):
+                total += len(content.split())
+        return total
+
+    @staticmethod
+    def _first_user_content(example: dict[str, Any]) -> str | None:
+        messages = example.get("messages")
+        if not isinstance(messages, list):
+            return None
+        for msg in messages:
+            if isinstance(msg, dict) and msg.get("role") == "user":
+                content = msg.get("content")
+                if isinstance(content, str):
+                    return content
+        return None
+
+    @staticmethod
+    def _assistant_content(example: dict[str, Any]) -> str | None:
+        messages = example.get("messages")
+        if not isinstance(messages, list):
+            return None
+        for msg in messages:
+            if isinstance(msg, dict) and msg.get("role") == "assistant":
+                content = msg.get("content")
+                if isinstance(content, str):
+                    return content
+        return None
+
+    @staticmethod
+    def _length_stats(token_counts: list[int]) -> dict[str, float]:
+        valid = [c for c in token_counts if c > 0]
+        if not valid:
+            return {"min": 0.0, "max": 0.0, "mean": 0.0, "median": 0.0, "std": 0.0}
+        return {
+            "min": float(min(valid)),
+            "max": float(max(valid)),
+            "mean": statistics.fmean(valid),
+            "median": float(statistics.median(valid)),
+            "std": float(statistics.pstdev(valid)) if len(valid) > 1 else 0.0,
+        }
+
+    @staticmethod
+    def _quality_score(total: int, issues: list[QualityIssue]) -> float:
+        """Compute a 0.0–1.0 quality score.
+
+        Starts at 1.0 and is penalised per issue: errors cost more than
+        warnings. An empty dataset scores 0.0 (no data to train on).
+        """
+        if total == 0:
+            return 0.0
+        penalty = 0.0
+        for issue in issues:
+            if issue.severity == "error":
+                penalty += 0.05
+            else:
+                penalty += 0.02
+        # Normalise by dataset size so one issue in a million-example dataset
+        # does not dominate, but never let a single issue go free.
+        normalised = penalty / max(total, 1)
+        score = 1.0 - min(normalised, 1.0)
+        # A dataset with any error should not score a perfect 1.0.
+        if score == 1.0 and any(i.severity == "error" for i in issues):
+            score = max(1.0 - 1.0 / total, 0.0)
+        return round(max(score, 0.0), 4)
+
+
+# --------------------------------------------------------------------------- #
+# Internal helpers
+# --------------------------------------------------------------------------- #
+def _count_by(issues: list[QualityIssue], key) -> dict[str, int]:
+    counts: dict[str, int] = {}
+    for issue in issues:
+        k = key(issue)
+        counts[k] = counts.get(k, 0) + 1
+    return counts
+
+
+if __name__ == "__main__":  # pragma: no cover - manual smoke
+    import sys
+
+    if len(sys.argv) < 2:
+        print("usage: sft_data_auditor.py <sft.jsonl>")
+        sys.exit(1)
+    report = SFTDataQualityAuditor().audit_file(sys.argv[1])
+    print(f"examples={report.total_examples} issues={report.total_issues} "
+          f"score={report.overall_quality_score}")
+    for issue in report.issues[:20]:
+        print(f" line {issue.line_number}: [{issue.severity}] {issue.issue_type} - {issue.description}")
@@ -0,0 +1,66 @@
+"""SFT rows with null assistant content must not TypeError in analyze_data."""
+
+import json
+import sys
+
+import analyze_data as ad
+
+
+def test_null_assistant_content_skipped(tmp_path, monkeypatch, capsys):
+    sft = tmp_path / "sft.jsonl"
+    rows = [
+        {
+            "messages": [
+                {"role": "user", "content": "q"},
+                {"role": "assistant", "content": None},
+            ]
+        },
+        {
+            "messages": [
+                {"role": "user", "content": "q"},
+                {
+                    "role": "assistant",
+                    "content": "<think>\n验算一遍\n</think>\nFinal Answer: 1",
+                },
+            ]
+        },
+    ]
+    sft.write_text(
+        "\n".join(json.dumps(r, ensure_ascii=False) for r in rows) + "\n",
+        encoding="utf-8",
+    )
+    monkeypatch.setattr(
+        sys,
+        "argv",
+        ["analyze_data.py", "--sft", str(sft), "--raw", str(tmp_path / "missing.jsonl")],
+    )
+    ad.main()
+    out = capsys.readouterr().out
+    assert "SFT 样本数：2" in out
+    assert "跳过 messages 不足 2 条的样本：1" in out
+    assert "含反思/验算行为的样本：1/1" in out
+
+
+def test_missing_content_key_skipped(tmp_path, monkeypatch, capsys):
+    sft = tmp_path / "sft.jsonl"
+    sft.write_text(
+        json.dumps(
+            {
+                "messages": [
+                    {"role": "user", "content": "q"},
+                    {"role": "assistant"},
+                ]
+            },
+            ensure_ascii=False,
+        )
+        + "\n",
+        encoding="utf-8",
+    )
+    monkeypatch.setattr(
+        sys,
+        "argv",
+        ["analyze_data.py", "--sft", str(sft), "--raw", str(tmp_path / "missing.jsonl")],
+    )
+    ad.main()
+    out = capsys.readouterr().out
+    assert "跳过 messages 不足 2 条的样本：1" in out
@@ -0,0 +1,64 @@
+"""SFT rows with messages shorter than 2 must not IndexError in analyze_data."""
+
+import json
+import sys
+from pathlib import Path
+
+import analyze_data as ad
+
+
+def test_short_messages_skipped_without_index_error(tmp_path, monkeypatch, capsys):
+    sft = tmp_path / "sft.jsonl"
+    rows = [
+        {"messages": [{"role": "user", "content": "only user"}]},
+        {"messages": []},
+        {
+            "messages": [
+                {"role": "user", "content": "q"},
+                {
+                    "role": "assistant",
+                    "content": "<think>\n验算一遍\n</think>\nFinal Answer: 1",
+                },
+            ]
+        },
+    ]
+    sft.write_text(
+        "\n".join(json.dumps(r, ensure_ascii=False) for r in rows) + "\n",
+        encoding="utf-8",
+    )
+    monkeypatch.setattr(
+        sys,
+        "argv",
+        ["analyze_data.py", "--sft", str(sft), "--raw", str(tmp_path / "missing.jsonl")],
+    )
+    ad.main()
+    out = capsys.readouterr().out
+    assert "SFT 样本数：3" in out
+    assert "跳过 messages 不足 2 条的样本：2" in out
+    assert "含反思/验算行为的样本：1/1" in out
+
+
+def test_normal_two_message_row_still_scored(tmp_path, monkeypatch, capsys):
+    sft = tmp_path / "sft.jsonl"
+    sft.write_text(
+        json.dumps(
+            {
+                "messages": [
+                    {"role": "user", "content": "q"},
+                    {"role": "assistant", "content": "<think>\nok\n</think>\n1"},
+                ]
+            },
+            ensure_ascii=False,
+        )
+        + "\n",
+        encoding="utf-8",
+    )
+    monkeypatch.setattr(
+        sys,
+        "argv",
+        ["analyze_data.py", "--sft", str(sft), "--raw", str(tmp_path / "missing.jsonl")],
+    )
+    ad.main()
+    out = capsys.readouterr().out
+    assert "跳过" not in out
+    assert "含反思/验算行为的样本：0/1" in out
@@ -0,0 +1,118 @@
+"""Empty problems JSONL must not ZeroDivisionError in the pass-rate summary."""
+
+import asyncio
+import json
+import os
+from types import ModuleType
+import sys
+
+# generate_data imports openai; stub if missing so the test stays offline.
+try:
+    import openai  # noqa: F401
+except ImportError:
+    _oai = ModuleType("openai")
+
+    class _AsyncOpenAI:
+        def __init__(self, *a, **k):
+            pass
+
+    _oai.AsyncOpenAI = _AsyncOpenAI
+    sys.modules["openai"] = _oai
+
+import generate_data as gd
+
+
+def test_empty_problems_summary_does_not_divide_by_zero(tmp_path, monkeypatch):
+    empty = tmp_path / "empty.jsonl"
+    empty.write_text("", encoding="utf-8")
+    raw = tmp_path / "raw.jsonl"
+    sft = tmp_path / "sft.jsonl"
+    monkeypatch.setenv("OPENROUTER_API_KEY", "test-key-not-used")
+
+    argv = [
+        "generate_data.py",
+        "--input",
+        str(empty),
+        "--raw_output",
+        str(raw),
+        "--sft_output",
+        str(sft),
+    ]
+    monkeypatch.setattr(sys, "argv", argv)
+    asyncio.run(gd.main())
+    assert raw.exists() and sft.exists()
+    assert sft.read_text(encoding="utf-8") == ""
+
+
+def test_nonempty_pass_rate_still_computes():
+    records = [{"verified": True, "error": None, "reasoning": "x", "usage": {}}]
+    passed = [r for r in records if r["verified"]]
+    pass_rate = (len(passed) / len(records) * 100) if records else 0.0
+    assert pass_rate == 100.0
+
+
+def test_native_moonshot_reasoning_effort_uses_supported_top_level_control():
+    assert gd.reasoning_extra_body("https://api.moonshot.cn/v1", "low", 0) == {
+        "reasoning_effort": "low"
+    }
+    assert gd.reasoning_extra_body("https://openrouter.ai/api/v1", "low", 0) == {
+        "reasoning": {"effort": "low"}
+    }
+
+
+def test_targeted_resume_preserves_verified_rows_and_retries_failure(tmp_path, monkeypatch):
+    problems = tmp_path / "problems.jsonl"
+    raw = tmp_path / "raw.jsonl"
+    sft = tmp_path / "sft.jsonl"
+    problem_rows = [
+        {"id": "one", "question": "1+1?", "answer": 2},
+        {"id": "two", "question": "2+2?", "answer": 4},
+    ]
+    problems.write_text(
+        "".join(json.dumps(row) + "\n" for row in problem_rows), encoding="utf-8"
+    )
+    raw.write_text(
+        "".join(
+            json.dumps(row) + "\n"
+            for row in (
+                {
+                    "id": "one", "question": "1+1?", "gold_answer": 2,
+                    "model": "teacher", "content": "Final Answer: 2", "reasoning": "ok",
+                    "verified": True, "usage": {}, "error": None,
+                },
+                {
+                    "id": "two", "question": "2+2?", "gold_answer": 4,
+                    "model": "teacher", "content": None, "reasoning": None,
+                    "verified": False, "usage": None, "error": "timeout",
+                },
+            )
+        ),
+        encoding="utf-8",
+    )
+    calls = []
+
+    async def fake_distill(client, problem, args, semaphore):
+        calls.append(problem["id"])
+        return {
+            "id": problem["id"], "question": problem["question"],
+            "gold_answer": problem["answer"], "model": args.model,
+            "content": "Final Answer: 4", "reasoning": "worked",
+            "verified": True, "usage": {}, "error": None, "attempts": [],
+        }
+
+    monkeypatch.setattr(gd, "distill_one", fake_distill)
+    monkeypatch.setattr(gd, "AsyncOpenAI", lambda **kwargs: object())
+    monkeypatch.setenv("OPENROUTER_API_KEY", "test-key-not-used")
+    monkeypatch.setattr(sys, "argv", [
+        "generate_data.py", "--input", str(problems), "--raw_output", str(raw),
+        "--sft_output", str(sft), "--problem-id", "two", "--resume",
+    ])
+
+    asyncio.run(gd.main())
+
+    raw_rows = gd.load_jsonl(raw)
+    assert calls == ["two"]
+    assert [row["id"] for row in raw_rows] == ["one", "two"]
+    assert all(row["verified"] for row in raw_rows)
+    assert raw_rows[1]["prior_failures"][0]["error"] == "timeout"
+    assert len(gd.load_jsonl(sft)) == 2
@@ -0,0 +1,32 @@
+"""Regression: load_verified_messages must accept fullwidth colon and case-insensitive Final Answer."""
+
+import json
+from pathlib import Path
+from train_student import load_verified_messages
+
+
+def test_load_verified_messages_fullwidth_colon(tmp_path: Path):
+    sample = {
+        "messages": [
+            {"role": "user", "content": "What is 2 + 2?"},
+            {"role": "assistant", "content": "<think>\n2+2=4\n</think>\n\nFinal Answer：4"},
+        ]
+    }
+    dataset = tmp_path / "dataset.jsonl"
+    dataset.write_text(json.dumps(sample) + "\n", encoding="utf-8")
+    rows = load_verified_messages(dataset)
+    assert len(rows) == 1
+    assert rows[0][1]["content"].endswith("Final Answer：4")
+
+
+def test_load_verified_messages_lowercase_colon(tmp_path: Path):
+    sample = {
+        "messages": [
+            {"role": "user", "content": "What is 2 + 2?"},
+            {"role": "assistant", "content": "<think>\n2+2=4\n</think>\n\nfinal answer: 4"},
+        ]
+    }
+    dataset = tmp_path / "dataset.jsonl"
+    dataset.write_text(json.dumps(sample) + "\n", encoding="utf-8")
+    rows = load_verified_messages(dataset)
+    assert len(rows) == 1
@@ -0,0 +1,59 @@
+import json
+
+from evaluate_student import (
+    BEHAVIORS,
+    behavior_flags,
+    compare_binary,
+    completion_and_findings,
+    exact_two_sided_sign_p_value,
+)
+from train_student import load_verified_messages
+
+
+def test_load_verified_messages_rejects_unverified_shape(tmp_path):
+    path = tmp_path / "bad.jsonl"
+    path.write_text(json.dumps({"messages": [{"role": "user", "content": "q"}]}) + "\n")
+    try:
+        load_verified_messages(path)
+    except ValueError as exc:
+        assert "exactly two" in str(exc)
+    else:
+        raise AssertionError("invalid collection row was accepted for parameter training")
+
+
+def test_paired_sign_test_detects_one_sided_student_gain():
+    baseline = {str(i): False for i in range(8)}
+    student = {str(i): True for i in range(8)}
+    result = compare_binary(baseline, student)
+    assert result["student_only"] == 8
+    assert result["baseline_only"] == 0
+    assert result["exact_two_sided_p_value"] == exact_two_sided_sign_p_value(0, 8)
+    assert result["exact_two_sided_p_value"] < 0.05
+
+
+def test_behavior_flags_cover_acceptance_categories():
+    flags = behavior_flags("Wait, that is not right. Use another approach, then verify by substitution.")
+    assert flags == {"reflection": True, "backtracking": True, "verification": True}
+
+
+def test_negative_uplift_finding_does_not_make_executed_campaign_incomplete():
+    def arm(name, correct):
+        return {
+            "name": name,
+            "accuracy": float(correct),
+            "behavior_rates": {key: 0.0 for key in BEHAVIORS},
+            "records": [{"id": "case-1", "correct": correct}],
+        }
+
+    completion, findings = completion_and_findings(
+        problem_ids={"case-1"},
+        baseline=arm("baseline", False),
+        student=arm("student", False),
+        teacher=arm("teacher", True),
+        paired={"paired_cases": 1, "exact_two_sided_p_value": 1.0},
+        student_training_complete=True,
+        teacher_outputs_complete=True,
+    )
+    assert completion["complete"] is True
+    assert findings["student_improves_over_baseline"] is False
+    assert findings["paired_improvement_significant_p_lt_0_05"] is False
@@ -0,0 +1,341 @@
+#!/usr/bin/env python3
+"""Train the Experiment 8-9 student on verified teacher CoT trajectories.
+
+This is the parameter-update stage missing from the original collection-only
+companion.  It deliberately has no mock training mode: a successful run writes
+a real Hugging Face/PEFT checkpoint plus a provenance manifest.
+"""
+
+from __future__ import annotations
+
+import argparse
+import hashlib
+import importlib.metadata
+import importlib.util
+import json
+import os
+import re
+import platform
+import subprocess
+from dataclasses import dataclass
+from datetime import datetime, timezone
+from pathlib import Path
+from typing import Any
+
+
+def sha256(path: Path) -> str:
+    digest = hashlib.sha256()
+    with path.open("rb") as handle:
+        for chunk in iter(lambda: handle.read(1024 * 1024), b""):
+            digest.update(chunk)
+    return digest.hexdigest()
+
+
+def load_verified_messages(path: Path) -> list[list[dict[str, str]]]:
+    """Load only complete user/assistant rows with a non-empty final answer."""
+    rows: list[list[dict[str, str]]] = []
+    with path.open(encoding="utf-8") as handle:
+        for line_number, line in enumerate(handle, 1):
+            if not line.strip():
+                continue
+            item = json.loads(line)
+            messages = item.get("messages")
+            if not isinstance(messages, list) or len(messages) != 2:
+                raise ValueError(f"{path}:{line_number}: expected exactly two messages")
+            if [m.get("role") for m in messages] != ["user", "assistant"]:
+                raise ValueError(f"{path}:{line_number}: expected user then assistant")
+            if not all(isinstance(m.get("content"), str) and m["content"].strip() for m in messages):
+                raise ValueError(f"{path}:{line_number}: empty message content")
+            if not re.search(r"Final Answer[:：]", messages[1]["content"], re.IGNORECASE):
+                raise ValueError(f"{path}:{line_number}: assistant lacks verified Final Answer")
+            rows.append(messages)
+    if not rows:
+        raise ValueError(f"{path}: no training samples")
+    return rows
+
+
+@dataclass
+class EncodedExample:
+    input_ids: list[int]
+    labels: list[int]
+
+
+def _chat_template_ids(encoded: Any) -> list[int]:
+    """Normalize Transformers 4.x/5.x chat-template return values.
+
+    Transformers 4.x returned a bare list from ``apply_chat_template`` when
+    ``tokenize=True``.  Transformers 5.x returns a BatchEncoding containing
+    both ``input_ids`` and ``attention_mask``.  Calling ``len`` or slicing the
+    latter operates on mapping keys, which can make every assistant trajectory
+    appear to have only two tokens and defeats the loss-mask safety check.
+    """
+    if isinstance(encoded, dict) or hasattr(encoded, "keys"):
+        encoded = encoded["input_ids"]
+    if hasattr(encoded, "tolist"):
+        encoded = encoded.tolist()
+    if encoded and isinstance(encoded[0], list):
+        if len(encoded) != 1:
+            raise ValueError("expected one chat-template sequence")
+        encoded = encoded[0]
+    if not isinstance(encoded, list) or not all(isinstance(token, int) for token in encoded):
+        raise TypeError("chat template did not return a one-dimensional integer token sequence")
+    return encoded
+
+
+def encode_messages(tokenizer: Any, messages: list[dict[str, str]], max_length: int) -> EncodedExample:
+    """Mask user/prompt tokens and supervise only the teacher assistant trajectory."""
+    prompt_ids = _chat_template_ids(
+        tokenizer.apply_chat_template(messages[:1], tokenize=True, add_generation_prompt=True)
+    )
+    full_ids = _chat_template_ids(
+        tokenizer.apply_chat_template(messages, tokenize=True, add_generation_prompt=False)
+    )
+    if len(full_ids) > max_length:
+        full_ids = full_ids[:max_length]
+    prompt_length = min(len(prompt_ids), len(full_ids))
+    labels = [-100] * prompt_length + full_ids[prompt_length:]
+    if not any(label != -100 for label in labels):
+        raise ValueError("max_length truncates the entire assistant response")
+    return EncodedExample(input_ids=full_ids, labels=labels)
+
+
+def _git_commit(root: Path) -> str | None:
+    try:
+        return subprocess.run(
+            ["git", "rev-parse", "HEAD"], cwd=root, check=True,
+            capture_output=True, text=True,
+        ).stdout.strip()
+    except (OSError, subprocess.CalledProcessError):
+        return None
+
+
+def parse_args() -> argparse.Namespace:
+    parser = argparse.ArgumentParser(
+        description="Experiment 8-9: real student SFT on verified CoT trajectories",
+        formatter_class=argparse.ArgumentDefaultsHelpFormatter,
+    )
+    parser.add_argument("--train-data", type=Path, default=Path("data/sft_cot_distill_aime_kimi_k3.jsonl"))
+    parser.add_argument("--base-model", default="Qwen/Qwen2.5-1.5B-Instruct")
+    parser.add_argument("--output-dir", type=Path, default=Path("checkpoints/cot-student"))
+    parser.add_argument("--max-length", type=int, default=4096)
+    parser.add_argument("--epochs", type=float, default=3.0)
+    parser.add_argument("--learning-rate", type=float, default=2e-5)
+    parser.add_argument("--batch-size", type=int, default=1)
+    parser.add_argument("--gradient-accumulation", type=int, default=16)
+    parser.add_argument("--seed", type=int, default=42)
+    parser.add_argument("--lora-rank", type=int, default=32, help="0 disables LoRA and updates all weights")
+    parser.add_argument("--lora-alpha", type=int, default=64)
+    parser.add_argument("--gradient-checkpointing", action=argparse.BooleanOptionalAction, default=True)
+    parser.add_argument("--trust-remote-code", action="store_true")
+    parser.add_argument("--preflight", action="store_true", help="write dependency/GPU readiness evidence without training")
+    parser.add_argument("--preflight-output", type=Path, default=Path("validation/student_sft_preflight.json"))
+    return parser.parse_args()
+
+
+def main() -> None:
+    args = parse_args()
+    if args.max_length <= 0 or args.batch_size <= 0 or args.gradient_accumulation <= 0:
+        raise SystemExit("max-length, batch-size, and gradient-accumulation must be positive")
+    messages = load_verified_messages(args.train_data)
+
+    if args.preflight:
+        dependencies = {
+            name: importlib.util.find_spec(name) is not None
+            for name in ("torch", "transformers", "accelerate", "peft")
+        }
+        dependency_versions = {
+            name: importlib.metadata.version(name) if installed else None
+            for name, installed in dependencies.items()
+        }
+        cuda_available = False
+        gpu_names: list[str] = []
+        torch_version = None
+        trainer_stack_error = None
+        if dependencies["torch"]:
+            import torch
+            torch_version = torch.__version__
+            cuda_available = torch.cuda.is_available()
+            if cuda_available:
+                gpu_names = [torch.cuda.get_device_name(i) for i in range(torch.cuda.device_count())]
+        try:
+            from transformers import Trainer  # noqa: F401
+        except Exception as exc:  # integration errors include incompatible peft/transformers versions
+            trainer_stack_error = f"{type(exc).__name__}: {exc}"
+        trainer_stack_importable = trainer_stack_error is None
+        payload = {
+            "schema_version": 1,
+            "experiment": "8-9",
+            "stage": "student_sft_preflight",
+            "generated_at_utc": datetime.now(timezone.utc).isoformat(),
+            "ready": all(dependencies.values()) and trainer_stack_importable and cuda_available,
+            "training_data": {
+                "path": str(args.train_data.resolve()),
+                "sha256": sha256(args.train_data.resolve()),
+                "samples": len(messages),
+            },
+            "host": {
+                "platform": platform.platform(),
+                "machine": platform.machine(),
+                "torch": torch_version,
+                "cuda_available": cuda_available,
+                "gpu_names": gpu_names,
+            },
+            "dependencies": dependencies,
+            "dependency_versions": dependency_versions,
+            "trainer_stack_importable": trainer_stack_importable,
+            "trainer_stack_error": trainer_stack_error,
+            "blockers": [
+                *[f"missing Python dependency: {name}" for name, ok in dependencies.items() if not ok],
+                *([] if trainer_stack_importable else ["transformers/peft trainer stack is not importable"]),
+                *([] if cuda_available else ["no CUDA device available"]),
+            ],
+        }
+        args.preflight_output.parent.mkdir(parents=True, exist_ok=True)
+        args.preflight_output.write_text(
+            json.dumps(payload, ensure_ascii=False, indent=2) + "\n", encoding="utf-8"
+        )
+        print(json.dumps({"preflight": str(args.preflight_output), "ready": payload["ready"]}, ensure_ascii=False))
+        return
+
+    try:
+        import torch
+    except ImportError as exc:
+        raise SystemExit("PyTorch is missing. Install requirements.txt before training.") from exc
+    if not torch.cuda.is_available():
+        raise SystemExit(
+            "Experiment 8-9 student SFT requires a CUDA host; this runner has no synthetic/CPU success fallback."
+        )
+    try:
+        from torch.utils.data import Dataset
+        from transformers import (
+            AutoModelForCausalLM,
+            AutoTokenizer,
+            Trainer,
+            TrainingArguments,
+            set_seed,
+        )
+    except (ImportError, RuntimeError) as exc:
+        raise SystemExit(
+            f"The transformers/peft training stack is not importable: {type(exc).__name__}: {exc}"
+        ) from exc
+
+    set_seed(args.seed)
+    tokenizer = AutoTokenizer.from_pretrained(
+        args.base_model, trust_remote_code=args.trust_remote_code
+    )
+    if tokenizer.pad_token_id is None:
+        tokenizer.pad_token = tokenizer.eos_token
+    encoded = [encode_messages(tokenizer, item, args.max_length) for item in messages]
+
+    class CotDataset(Dataset):
+        def __len__(self) -> int:
+            return len(encoded)
+
+        def __getitem__(self, index: int) -> dict[str, list[int]]:
+            item = encoded[index]
+            return {"input_ids": item.input_ids, "labels": item.labels}
+
+    def collate(batch: list[dict[str, list[int]]]) -> dict[str, Any]:
+        width = max(len(item["input_ids"]) for item in batch)
+        ids, masks, labels = [], [], []
+        for item in batch:
+            padding = width - len(item["input_ids"])
+            ids.append(item["input_ids"] + [tokenizer.pad_token_id] * padding)
+            masks.append([1] * len(item["input_ids"]) + [0] * padding)
+            labels.append(item["labels"] + [-100] * padding)
+        return {
+            "input_ids": torch.tensor(ids, dtype=torch.long),
+            "attention_mask": torch.tensor(masks, dtype=torch.long),
+            "labels": torch.tensor(labels, dtype=torch.long),
+        }
+
+    model = AutoModelForCausalLM.from_pretrained(
+        args.base_model,
+        torch_dtype=torch.bfloat16 if torch.cuda.is_bf16_supported() else torch.float16,
+        trust_remote_code=args.trust_remote_code,
+    )
+    if args.gradient_checkpointing:
+        model.gradient_checkpointing_enable()
+        model.config.use_cache = False
+    if args.lora_rank:
+        try:
+            from peft import LoraConfig, get_peft_model
+        except ImportError as exc:
+            raise SystemExit("LoRA requested but peft is not installed") from exc
+        model = get_peft_model(model, LoraConfig(
+            r=args.lora_rank,
+            lora_alpha=args.lora_alpha,
+            lora_dropout=0.05,
+            bias="none",
+            task_type="CAUSAL_LM",
+            target_modules="all-linear",
+        ))
+
+    args.output_dir.mkdir(parents=True, exist_ok=True)
+    training_args = TrainingArguments(
+        output_dir=str(args.output_dir),
+        num_train_epochs=args.epochs,
+        learning_rate=args.learning_rate,
+        per_device_train_batch_size=args.batch_size,
+        gradient_accumulation_steps=args.gradient_accumulation,
+        logging_steps=1,
+        save_strategy="epoch",
+        seed=args.seed,
+        bf16=torch.cuda.is_bf16_supported(),
+        fp16=not torch.cuda.is_bf16_supported(),
+        report_to="none",
+        remove_unused_columns=False,
+    )
+    trainer = Trainer(
+        model=model,
+        args=training_args,
+        train_dataset=CotDataset(),
+        data_collator=collate,
+    )
+    result = trainer.train()
+    trainer.save_model(str(args.output_dir))
+    tokenizer.save_pretrained(str(args.output_dir))
+
+    root = Path(__file__).resolve().parents[2]
+    manifest = {
+        "schema_version": 1,
+        "experiment": "8-9",
+        "stage": "student_sft",
+        "status": "complete",
+        "generated_at_utc": datetime.now(timezone.utc).isoformat(),
+        "book_git_commit": _git_commit(root),
+        "training_data": {
+            "path": str(args.train_data.resolve()),
+            "sha256": sha256(args.train_data.resolve()),
+            "samples": len(messages),
+        },
+        "base_model": args.base_model,
+        "output_dir": str(args.output_dir.resolve()),
+        "host": {
+            "platform": platform.platform(),
+            "gpu_names": [torch.cuda.get_device_name(i) for i in range(torch.cuda.device_count())],
+            "torch": torch.__version__,
+        },
+        "dependency_versions": {
+            name: importlib.metadata.version(name)
+            for name in ("torch", "transformers", "accelerate", "peft")
+        },
+        "training": {
+            "epochs": args.epochs,
+            "learning_rate": args.learning_rate,
+            "max_length": args.max_length,
+            "batch_size": args.batch_size,
+            "gradient_accumulation": args.gradient_accumulation,
+            "lora_rank": args.lora_rank,
+            "seed": args.seed,
+            "metrics": result.metrics,
+        },
+    }
+    (args.output_dir / "training_manifest.json").write_text(
+        json.dumps(manifest, ensure_ascii=False, indent=2) + "\n", encoding="utf-8"
+    )
+    print(json.dumps({"checkpoint": str(args.output_dir), "samples": len(messages)}, ensure_ascii=False))
+
+
+if __name__ == "__main__":
+    main()
@@ -0,0 +1,36 @@
+{
+  "schema_version": 1,
+  "experiment": "8-9",
+  "stage": "student_sft_preflight",
+  "generated_at_utc": "2026-07-30T04:51:13.181684+00:00",
+  "ready": false,
+  "training_data": {
+    "path": "/Users/boj/book/ai-agent-book/chapter8/cot-distillation/data/sft_cot_distill_aime_kimi_k3.jsonl",
+    "sha256": "0c6cab7cb8e0bd13671eda4e2bd0dc2530f7483f09ae42e946b87600a979968e",
+    "samples": 23
+  },
+  "host": {
+    "platform": "macOS-26.3-arm64-arm-64bit",
+    "machine": "arm64",
+    "torch": "2.7.0",
+    "cuda_available": false,
+    "gpu_names": []
+  },
+  "dependencies": {
+    "torch": true,
+    "transformers": true,
+    "accelerate": true,
+    "peft": true
+  },
+  "dependency_versions": {
+    "torch": "2.7.0",
+    "transformers": "4.48.3",
+    "accelerate": "1.2.1",
+    "peft": "0.14.0"
+  },
+  "trainer_stack_importable": true,
+  "trainer_stack_error": null,
+  "blockers": [
+    "no CUDA device available"
+  ]
+}
@@ -0,0 +1,36 @@
+{
+  "schema_version": 1,
+  "experiment": "8-9",
+  "stage": "student_sft_preflight",
+  "generated_at_utc": "2026-08-01T06:28:06.142472+00:00",
+  "ready": true,
+  "training_data": {
+    "path": "/home/ubuntu/ai-agent-book/chapter8/cot-distillation/data/sft_cot_distill_aime_kimi_k3.jsonl",
+    "sha256": "0c6cab7cb8e0bd13671eda4e2bd0dc2530f7483f09ae42e946b87600a979968e",
+    "samples": 23
+  },
+  "host": {
+    "platform": "Linux-6.8.0-111-generic-x86_64-with-glibc2.35",
+    "machine": "x86_64",
+    "torch": "2.11.0+cu130",
+    "cuda_available": true,
+    "gpu_names": [
+      "NVIDIA RTX PRO 6000 Blackwell Workstation Edition"
+    ]
+  },
+  "dependencies": {
+    "torch": true,
+    "transformers": true,
+    "accelerate": true,
+    "peft": true
+  },
+  "dependency_versions": {
+    "torch": "2.11.0",
+    "transformers": "5.14.1",
+    "accelerate": "1.14.0",
+    "peft": "0.19.1"
+  },
+  "trainer_stack_importable": true,
+  "trainer_stack_error": null,
+  "blockers": []
+}