DMXAPI

Appearance

gpt 输出思考内容 API 文档 ​

GPT 模型的输出推理内容（Reasoning/CoT）是模型在生成最终答案前，通过内部“打草稿”进行逻辑拆解、自我验证与路径规划的显性化思维过程。

⭐接口地址 ​

https://www.dmxapi.cn/v1/responses

注意：

需要升级谷歌sdk为最新版

🚀支持的模型（部分） ​

  • gpt-5.2
  • gpt-5.1
  • gpt-5

🧙‍♀️输出思考内容 示例代码 ​

python
# ============================================================
#  GPT-5 深度思考模式调用示例（SDK 版）
#  功能：通过 OpenAI Python SDK 调用 Responses API
# ============================================================

from openai import OpenAI

# ---------- 初始化客户端 ----------
client = OpenAI(
    api_key="sk-*****************************",
    base_url="https://www.dmxapi.cn/v1"
)

# ---------- 发送请求 ----------
response = client.responses.create(
    model="gpt-5",                        # 使用的模型
    input="一加到十最复杂的方法",  # 用户输入的问题
    reasoning={
        "effort": "low",                  # 思考等级
        "summary": "auto"                 # 思考摘要：auto 自动返回思考过程
    }
)

# ---------- 输出结果 ----------
# 遍历 output 列表，分别提取「思考内容」和「响应结果」
for item in response.output:
    if item.type == "reasoning":
        print("【思考内容】")
        for content in item.summary:
            print(content.text)
        print()
    elif item.type == "message":
        print("【响应结果】")
        for content in item.content:
            print(content.text)
python
# ============================================================
#  GPT-5 深度思考模式调用示例
#  功能：调用 OpenAI Responses API，获取模型的思考过程与最终回答
# ============================================================

import requests
import json
import sys
import io

# ---------- 终端编码设置 ----------
# 解决 Windows 终端输出中文乱码问题
sys.stdout = io.TextIOWrapper(sys.stdout.buffer, encoding='utf-8')

# ---------- API 配置 ----------
url = "https://www.dmxapi.cn/v1/responses"
api_key = "sk-*****************************"

headers = {
    "Content-Type": "application/json",
    "Authorization": f"Bearer {api_key}"
}

# ---------- 请求参数 ----------
data = {
    "model": "gpt-5",                    # 使用的模型
    "input": "一加到十最复杂的方法",        # 用户输入的问题
    "reasoning": {
        "effort": "low",                  # 思考等级
        "summary": "auto"                 # 思考摘要：auto 自动返回思考过程
    }
}

# ---------- 发送请求 ----------·
response = requests.post(url, headers=headers, json=data)
result = response.json()

# ---------- 解析并输出结果 ----------
# 遍历 output 列表，分别提取「思考内容」和「响应结果」
for item in result.get("output", []):
    if item.get("type") == "reasoning":
        print("【思考内容】")
        for content in item.get("summary", []):
            print(content.get("text", ""))
        print()
    elif item.get("type") == "message":
        print("【响应结果】")
        for content in item.get("content", []):
            print(content.get("text", ""))
shell

curl https://www.dmxapi.cn/v1/responses \
  -H "Content-Type: application/json" \
  -H "Authorization: Bearer sk-*****************************" \
  -d '{
    "model": "gpt-5",
    "input": "What is the capital of France?",
    "reasoning": {
        "effort": "low",
        "summary": "auto"
    }
  }'

🛸返回示例 ​

以下是运行示例代码后的控制台输出，展示了 AI 的推理过程和推理结果。

json
【思考内容】
**Exploring sum methods**

I’m considering how to sum the numbers from 1 to 10 through various mathematical and computational approaches. I could use things like geometric series at x=1, inclusion-exclusion principles, or even linear algebra. Maybe applications like FFT could come into play, and I can even think about dynamic programming, SQL recursive CTEs, or even Bash scripting. I want to keep it fun and concise! One method involves using Newton's identities, though that might be a bit complex.
**Considering complex sum methods**

I'm exploring different methods to calculate the sum of numbers from 1 to 10. For one, I think about using a contour integral with generating functions, and maybe apply residue theory. Then, there's the idea of using a geometric series derivative and L'Hospital's rule, but that feels complex. I could also compute using FFT convolution, though that might be clunky. An easier approach could be dynamic programming with memoization, or even a Monte Carlo method. Lastly, I'm thinking about applying a SQL recursive CTE for the task!
**Exploring even more sum methods**

I’m thinking of more ways to sum from 1 to 10. For instance, I could use Church numerals to symbolize the numbers, leading to 55, but that feels like overkill. Maybe I could try matrix products, dot products with a vector, though it seems complex. Newton-Cotes integration sounds nice, but it might not fit well for a simple sum. In another alternative, I could implement methods in Python using SymPy for residues or limits. I want to keep the methods concise and provide some neat code snippets!
**Summarizing sum methods**

I need to keep things brief while reviewing ways to sum from 1 to 10. One method is using generating functions and taking the limit as \( x \) approaches 1, resulting in 55. Another approach involves linear algebra with a companion matrix, but that seems messy. The trapezoidal rule gives an approximation of 55.5, while the Euler-Maclaurin formula yields 55, which is quite effective! I also think about Monte Carlo methods and SQL recursive CTEs, plus some category theory for folding over lists. But, wow, I’ll stick to the first couple options for simplicity!

【响应结果】
下面给你几种“极其复杂但能得到 1+…+10=55”的方法，图一乐：

- 欧拉–麦克劳林公式
  Sum_{k=1}^{10} k = ∫_1^{10} x dx + (f(1)+f(10))/2，因 f'(x)=1 常数，后续校正项为 0。
  结果：(100−1)/2 + (1+10)/2 = 49.5 + 5.5 = 55

- 维埃塔定理（构造一个没有必要的大多项式）
  令 P(x)=∏_{k=1}^{10}(x−k)。根之和为系数 −a_9/a_10 = 1+2+…+10 = 55

- 生成函数极限法
  对前 n 项和 S(n)：S(n,x)=x(1−(n+1)x^n+nx^{n+1})/(1−x)^2
  则 1+…+10 = lim_{x→1} S(10,x) = 55

- 复分析（留数/系数提取）
  1+…+10 = ∑_{k=1}^{10} k = [z^≥1≤10] z/(1−z)^2
  用 Cauchy 系数公式提取 (1−z^{11})/(1−z)·z/(1−z)^2 的 z^0 系数，可由在 z=1 的孤立奇点留数算得 55

- 随机化蒙特卡洛（完全没必要）
  从 {1,…,10} 均匀独立采样 m 次，取样本均值 μ̂，再乘以 10，m→∞ 时以概率 1 收敛到 55

- SQL 递归 CTE
  WITH RECURSIVE t(n,s) AS (
    SELECT 1,1
    UNION ALL
    SELECT n+1, s+(n+1) FROM t WHERE n<10
  ) SELECT s FROM t ORDER BY n DESC LIMIT 1;  → 55

- Haskell 单子与幺半群折叠（装腔版）
  sum = getSum . mconcat . map Sum
  main = print $ sum [1..10]  -- 输出 55

- Lambda 演算（丘奇数）
  逐个把 1..10 的丘奇数用加法组合，β-约简得到丘奇数 55（然后再解码成自然数）

- 线性代数（伴随矩阵反复相乘）
  把状态 (n, s) 用矩阵 T 递推：s_{k}=s_{k-1}+k。构造齐次扩展矩阵把 “+k” 变成线性项，算 T 的 10 次幂作用到初始向量，读出 s_10=55

想要我把其中任意一种写成可运行的代码或把某个推导细节完整展开吗？
json
【思考内容】
**Explaining a complex sum**

The user is asking, in Chinese, for "the most complicated way to add one to ten," which likely refers to calculating the sum from 1 to 10, resulting in 55. They've requested a humorous and overly complicated method to do this. I can suggest several increasingly complex techniques, like using integrals, generating functions, or even advanced concepts like lambda calculus and neural networks. I'll keep it concise while ensuring I provide a few different escalating methods that all arrive at the same answer.
**Exploring complex summation methods**

I want to explore various complicated methods to calculate the sum from 1 to 10, resulting in 55. Here are a few ideas:

- Use the arithmetic progression formula but derive it through calculus.
- An integral of the floor function could help, considering the bounds carefully.
- A generating function might provide a creative approach.
- Matrix eigen decomposition is an option, though it feels excessive.
- I could even suggest using SQL for summation.
- Lastly, a whimsical Rube Goldberg-style pipeline could humorously compute this sum using Monte Carlo methods to approximate 55.

【响应结果】
你要“最复杂”的，那我给你一堆过度工程版的做法，结论都一样：结果是 55。

- 皮亚诺公理递归法
  定义 S(n)=S(n−1)+n, S(0)=0，用后继运算逐个构造到 n=10，展开十次递归得到 55。

- 数学归纳+解差分方程
  设部分和 T(n)=1+…+n，满足 T(n)−T(n−1)=n，假设 T(n) 为二次多项式 an²+bn+c，解出 a=1/2,b=1/2,c=0，故 T(10)=10×11/2=55。

- 生成函数
  S(x)=∑_{n≥1} n x^n = x/(1−x)²。部分和是系数卷积：∑_{k=1}^{10}k = [x^{≤10}] S(x)·1/(1−x)。做部分分式或系数求和得 55。

- 积分-取整表示
  ∑_{k=1}^{10} k = ∫_{0}^{10} ⌊x⌋ dx + ∫_{0}^{10} 1 dx。分段积分各区间 [m,m+1) 上值为 m，累加得 0+1+…+9 + 10 = 55。

- 群论双计数（把高斯配对搞得很绕）
  作用群 S_10 置换 {1,…,10}，数有序对 (i,σ) 满足 i≤σ(i)。用伯恩赛德引理按循环分解计数，再化简即得到配对和 11×10/2=55。

- 线性代数拟合
  用范德蒙德矩阵拟合 T(n) 为二次多项式：解
  [1 1 1; 4 2 1; 9 3 1]·[a b c]^T = [1 3 6]^T，
  得 a=1/2,b=1/2,c=0，故 T(10)=55。

- SQL 递归 CTE
  WITH RECURSIVE t(n,s) AS (SELECT 1,1 UNION ALL SELECT n+1, s+(n+1) FROM t WHERE n<10) SELECT s FROM t WHERE n=10; 返回 55。

- 蒙特卡罗无偏估计
  令 X 在 {1,…,10} 等概率，估计 μ=E[X]，则 ∑=10·μ。采样 N 次取均值 X̄，估计 10·X̄ → 55（方差随 N 增大而降）。

- 有限差分积分
  对 f(k)=k 取一次前向差 Δf=1，再对差分做离散积分 ΣΔf = f(n)−f(0)，累计二次求和得到三角数公式并算 T(10)=55。

- Church 数与λ演算
  用 Church 数表示 10，构造加法与前缀和组合子，归约到三角数组合子 Tri = λn. … ，对 10 做 β-归约，最终得到 Church(55)，再反解回 55。

不管怎么拧巴，答案都是 55。需要我把某一种方法展开成可运行代码或详细推导吗？
json
[
    {
        "id": "rs_6876cf02e0bc8192b74af0fb64b715ff06fa2fcced15a5ac",
        "type": "reasoning",
        "summary": [
            {
                "type": "summary_text",
                "text": "**Answering a simple question**\n\nI\u2019m looking at a straightforward question: the capital of France is Paris. It\u2019s a well-known fact, and I want to keep it brief and to the point. Paris is known for its history, art, and culture, so it might be nice to add just a hint of that charm. But mostly, I\u2019ll aim to focus on delivering a clear and direct answer, ensuring the user gets what they\u2019re looking for without any extra fluff."
            }
        ]
    },
    {
        "id": "msg_6876cf054f58819284ecc1058131305506fa2fcced15a5ac",
        "type": "message",
        "status": "completed",
        "content": [
            {
                "type": "output_text",
                "annotations": [],
                "logprobs": [],
                "text": "The capital of France is Paris."
            }
        ],
        "role": "assistant"
    }
]

🌻官方网址 ​

https://platform.openai.com/docs/guides/reasoning?api-mode=chat

© 2026 DMXAPI gpt 输出思考内容

一个 Key 用全球大模型

Copyright © 2025 DMXAPI 版权所有