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chore: update lc problems #2051

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chore: update lc problems #2051

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2 changes: 1 addition & 1 deletion solution/2700-2799/2756.Query Batching/README_EN.md
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<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>

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44 changes: 23 additions & 21 deletions solution/2900-2999/2950.Number of Divisible Substrings/README.md
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# [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>
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</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
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