chore(docs): sync doc metadata [skip ci]

Author: longsizhuo

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

@@ -1,21 +1,23 @@
 ---
 title: 2894. 分类求和并作差
-date: '2025.05.27 23:52'
+date: "2025.05.27 23:52"
 tags:
   - - Python
   - - answer
   - - typescript
 abbrlink: 66adcc9e
+docId: y0ntwlksnvj7ymuapqvkvmwr
 ---
-[2894. 分类求和并作差](https://leetcode.cn/problems/divisible-and-non-divisible-sums-difference/description/)
 
+[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：  
    $S_{\text{total}} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$
 
@@ -31,6 +33,7 @@ abbrlink: 66adcc9e
    $= \frac{n(n+1)}{2} - m \cdot \left\lfloor \frac{n}{m} \right\rfloor(\left\lfloor \frac{n}{m} \right\rfloor+1)$
 
 # Code
+
 ```python
 class Solution:
     def differenceOfSums(self, n: int, m: int) -> int:
@@ -40,5 +43,6 @@ class Solution:
 ```
 
 ```typescript
-const differenceOfSums = (n: number, m: number): number => -m * ((1 + Math.floor(n / m)) * Math.floor(n / m)) + (n * (n + 1) >> 1);
-```
+const differenceOfSums = (n: number, m: number): number =>
+  -m * ((1 + Math.floor(n / m)) * Math.floor(n / m)) + ((n * (n + 1)) >> 1);
+```

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