leetcode-problemset/leetcode-cn/originData/handling-sum-queries-after-update.json

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        "question": {
            "questionId": "2703",
            "questionFrontendId": "2569",
            "categoryTitle": "Algorithms",
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            "title": "Handling Sum Queries After Update",
            "titleSlug": "handling-sum-queries-after-update",
            "content": "<p>You are given two <strong>0-indexed</strong> arrays <code>nums1</code> and <code>nums2</code> and a 2D array <code>queries</code> of queries. There are three types of queries:</p>\n\n<ol>\n\t<li>For a query of type 1, <code>queries[i]&nbsp;= [1, l, r]</code>. Flip the values from <code>0</code> to <code>1</code> and from <code>1</code> to <code>0</code> in <code>nums1 </code>from index <code>l</code> to index <code>r</code>. Both <code>l</code> and <code>r</code> are <strong>0-indexed</strong>.</li>\n\t<li>For a query of type 2, <code>queries[i]&nbsp;= [2, p, 0]</code>. For every index <code>0 &lt;= i &lt; n</code>, set&nbsp;<code>nums2[i] =&nbsp;nums2[i]&nbsp;+ nums1[i]&nbsp;* p</code>.</li>\n\t<li>For a query of type 3, <code>queries[i]&nbsp;= [3, 0, 0]</code>. Find the sum of the elements in <code>nums2</code>.</li>\n</ol>\n\n<p>Return <em>an array containing all the answers to the third type&nbsp;queries.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]\n<strong>Output:</strong> [3]\n<strong>Explanation:</strong> After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]\n<strong>Output:</strong> [5]\n<strong>Explanation:</strong> After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length,nums2.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>nums1.length = nums2.length</code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code><font face=\"monospace\">queries[i].length = 3</font></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= l &lt;= r &lt;= nums1.length - 1</font></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= p &lt;= 10<sup>6</sup></font></code></li>\n\t<li><code>0 &lt;= nums1[i] &lt;= 1</code></li>\n\t<li><code>0 &lt;= nums2[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "更新数组后处理求和查询",
            "translatedContent": "<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>\n\n<ol>\n\t<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>\n\t<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>\n\t<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>\n</ol>\n\n<p>请你返回一个数组，包含所有第三种操作类型的答案。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]\n<b>输出：</b>[3]\n<strong>解释：</strong>第一个操作后 nums1 变为 [1,1,1] 。第二个操作后，nums2 变成 [1,1,1] ，所以第三个操作的答案为 3 。所以返回 [3] 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]\n<b>输出：</b>[5]\n<b>解释：</b>第一个操作后，nums2 保持不变为 [5] ，所以第二个操作的答案是 5 。所以返回 [5] 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length,nums2.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>nums1.length = nums2.length</code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length = 3</code></li>\n\t<li><code>0 &lt;= l &lt;= r &lt;= nums1.length - 1</code></li>\n\t<li><code>0 &lt;= p &lt;= 10<sup>6</sup></code></li>\n\t<li><code>0 &lt;= nums1[i] &lt;= 1</code></li>\n\t<li><code>0 &lt;= nums2[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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