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存量题库数据更新

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leetcode-cn/problem (Chinese)/更新数组后处理求和查询 [handling-sum-queries-after-update].html
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<p>给你两个下标从 <strong>0</strong>&nbsp;开始的数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;，和一个二维数组&nbsp;<code>queries</code>&nbsp;表示一些操作。总共有 3 种类型的操作：</p>
<ol>
<li>操作类型 1 为&nbsp;<code>queries[i]&nbsp;= [1, l, r]</code>&nbsp;。你需要将 <code>nums1</code>&nbsp;从下标&nbsp;<code>l</code>&nbsp;到下标 <code>r</code>&nbsp;的所有 <code>0</code>&nbsp;反转成 <code>1</code>&nbsp;或将 <code>1</code>&nbsp;反转成 <code>0</code>&nbsp;。<code>l</code>&nbsp;和 <code>r</code>&nbsp;下标都从 <strong>0</strong>&nbsp;开始。</li>
<li>操作类型 1 为&nbsp;<code>queries[i]&nbsp;= [1, l, r]</code>&nbsp;。你需要将 <code>nums1</code>&nbsp;从下标&nbsp;<code>l</code>&nbsp;到下标 <code>r</code>&nbsp;的所有 <code>0</code>&nbsp;反转成 <code>1</code> 并且所有&nbsp;<code>1</code>&nbsp;反转成 <code>0</code>&nbsp;。<code>l</code>&nbsp;和 <code>r</code>&nbsp;下标都从 <strong>0</strong>&nbsp;开始。</li>
<li>操作类型 2 为&nbsp;<code>queries[i]&nbsp;= [2, p, 0]</code>&nbsp;。对于&nbsp;<code>0 &lt;= i &lt; n</code>&nbsp;中的所有下标，令&nbsp;<code>nums2[i] =&nbsp;nums2[i]&nbsp;+ nums1[i]&nbsp;* p</code>&nbsp;。</li>
<li>操作类型 3 为&nbsp;<code>queries[i]&nbsp;= [3, 0, 0]</code>&nbsp;。求&nbsp;<code>nums2</code>&nbsp;中所有元素的和。</li>
</ol>
<p>请你返回一个数组，包含所有第三种操作类型的答案。</p>
<p>请你返回一个&nbsp;<em>数组</em>，包含&nbsp;<em>所有第三种操作类型&nbsp;</em>的答案。</p>
<p>&nbsp;</p>