chore: update lc problems (doocs#2051)

Author: acbin

solution/2700-2799/2756.Query Batching/README_EN.md

@@ -9,7 +9,7 @@
 <p>The constructor should accept two parameters:</p>
 
 <ul>
-	<li>An asyncronous function&nbsp;<code>queryMultiple</code>&nbsp;which accepts an array of&nbsp;string keys <code>input</code>. It will resolve with an array of values that is the same length as the input array. Each index corresponds to the value associated with&nbsp;<code>input[i]</code>.&nbsp;You can assume the promise will never reject.</li>
+	<li>An asynchronous function&nbsp;<code>queryMultiple</code>&nbsp;which accepts an array of&nbsp;string keys <code>input</code>. It will resolve with an array of values that is the same length as the input array. Each index corresponds to the value associated with&nbsp;<code>input[i]</code>.&nbsp;You can assume the promise will never reject.</li>
 	<li>A throttle time in milliseconds&nbsp;<code>t</code>.</li>
 </ul>
 

solution/2900-2999/2950.Number of Divisible Substrings/README.md

@@ -1,23 +1,24 @@
-# [2950. Number of Divisible Substrings](https://leetcode.cn/problems/number-of-divisible-substrings)
+# [2950. 可整除子串的数量](https://leetcode.cn/problems/number-of-divisible-substrings)
 
 [English Version](/solution/2900-2999/2950.Number%20of%20Divisible%20Substrings/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>Each character of the English alphabet has been mapped to a digit as shown below.</p>
+<p>每个英文字母都被映射到一个数字，如下所示。</p>
 
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2950.Number%20of%20Divisible%20Substrings/images/old_phone_digits.png" style="padding: 10px; width: 200px; height: 200px;" /></p>
 
-<p>A string is <strong>divisible</strong> if the sum of the mapped values of its characters is divisible by its length.</p>
+<p>如果字符串的字符的映射值的总和可以被字符串的长度整除，则该字符串是 <strong>可整除</strong> 的。</p>
 
-<p>Given a string <code>s</code>, return <em>the number of <strong>divisible substrings</strong> of</em> <code>s</code>.</p>
+<p>给定一个字符串 <code>s</code>，请返回 <code>s</code> 的<em> <strong>可整除子串</strong> 的数量</em>。</p>
 
-<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within a string.</p>
+<p><strong>子串</strong> 是字符串内的一个连续的非空字符序列。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong class="example">示例 1：</strong></p>
 
 <table border="1" cellspacing="3" style="border-collapse: separate; text-align: center;">
 	<tbody>
@@ -102,35 +103,36 @@
 </table>
 
 <pre>
-<strong>Input:</strong> word = &quot;asdf&quot;
-<strong>Output:</strong> 6
-<strong>Explanation:</strong> The table above contains the details about every substring of word, and we can see that 6 of them are divisible.
+<b>输入：</b>word = "asdf"
+<b>输出：</b>6
+<b>解释：</b>上表包含了有关 word 中每个子串的详细信息，我们可以看到其中有 6 个是可整除的。
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><b>示例 2:</b></p>
 
 <pre>
-<strong>Input:</strong> word = &quot;bdh&quot;
-<strong>Output:</strong> 4
-<strong>Explanation:</strong> The 4 divisible substrings are: &quot;b&quot;, &quot;d&quot;, &quot;h&quot;, &quot;bdh&quot;.
-It can be shown that there are no other substrings of word that are divisible.
+<b>输入：</b>word = "bdh"
+<b>输出：</b>4
+<b>解释：</b>4 个可整除的子串是："b"，"d"，"h"，"bdh"。
+可以证明 word 中没有其他可整除的子串。
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><b>示例 3:</b></p>
 
 <pre>
-<strong>Input:</strong> word = &quot;abcd&quot;
-<strong>Output:</strong> 6
-<strong>Explanation:</strong> The 6 divisible substrings are: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;, &quot;ab&quot;, &quot;cd&quot;.
-It can be shown that there are no other substrings of word that are divisible.
+<b>输入：</b>word = "abcd"
+<b>输出：</b>6
+<b>解释：</b>6 个可整除的子串是："a"，"b"，"c"，"d"，"ab"，"cd"。
+可以证明 word 中没有其他可整除的子串。
 </pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><b>提示：</b></p>
 
 <ul>
 	<li><code>1 &lt;= word.length &lt;= 2000</code></li>
-	<li><code>word</code> consists only of lowercase English letters.</li>
+	<li><code>word</code> 仅包含小写英文字母。</li>
 </ul>
 
 ## 解法