chore: add daily leetcode post 2894. 分类求和并作差

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2894. 分类求和并作差.md

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+---
+title: 2894. 分类求和并作差
+date: '2025.05.27 23:52'
+tags:
+  - - Python
+  - - answer
+  - - typescript
+abbrlink: 66adcc9e
+---
+[2894. 分类求和并作差](https://leetcode.cn/problems/divisible-and-non-divisible-sums-difference/description/)
+
+
+# Thought
+
+**题意：**  
+在 $[1, n]$ 中，求所有不能被 $m$ 整除的数的总和与能被整除的数的总和之差。
+
+**思路推导：**
+1. 总和：  
+   $S_{\text{total}} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$
+
+2. 能被 $m$ 整除的数为：$m, 2m, 3m, \dots, \left\lfloor \frac{n}{m} \right\rfloor m$
+   $S_{\text{divisible}} = m \cdot \left(1 + 2 + \dots + k\right) = m \cdot \frac{k(k+1)}{2}$
+
+3. 所以不能被整除的和为：  
+   $S_{\text{not\_div}} = S_{\text{total}} - S_{\text{divisible}}$
+
+4. 所求答案为：  
+   $\text{差值} = S_{\text{not\_div}} - S_{\text{divisible}}$  
+   $= \frac{n(n+1)}{2} - m \cdot \frac{\left\lfloor \frac{n}{m} \right\rfloor(\left\lfloor \frac{n}{m} \right\rfloor+1)}{2} - m \cdot \frac{\left\lfloor \frac{n}{m} \right\rfloor(\left\lfloor \frac{n}{m} \right\rfloor+1)}{2}$  
+   $= \frac{n(n+1)}{2} - m \cdot \left\lfloor \frac{n}{m} \right\rfloor(\left\lfloor \frac{n}{m} \right\rfloor+1)$
+
+# Code
+```python
+class Solution:
+    def differenceOfSums(self, n: int, m: int) -> int:
+        # divisible = m * (1 + n // m) * (n // m) // 2
+        # undivisible = n * (n + 1) // 2 - n * ((1 + n // m) * (n // m) // 2)
+        return - m * ((1 + n // m) * (n // m)) + (n * (n + 1) >> 1)
+```
+
+```typescript
+const differenceOfSums = (n: number, m: number): number => -m * ((1 + Math.floor(n / m)) * Math.floor(n / m)) + (n * (n + 1) >> 1);
+```