leetcode-problemset/leetcode-cn/originData/find-products-of-elements-of-big-array.json

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        "question": {
            "questionId": "3411",
            "questionFrontendId": "3145",
            "categoryTitle": "Algorithms",
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            "title": "Find Products of Elements of Big Array",
            "titleSlug": "find-products-of-elements-of-big-array",
            "content": "<p>The <strong>powerful array</strong> of a non-negative integer <code>x</code> is defined as the shortest sorted array of powers of two that sum up to <code>x</code>. The table below illustrates examples of how the <strong>powerful array</strong> is determined. It can be proven that the powerful array of <code>x</code> is unique.</p>\n\n<table border=\"1\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>num</th>\n\t\t\t<th>Binary Representation</th>\n\t\t\t<th>powerful array</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>0000<u>1</u></td>\n\t\t\t<td>[1]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>8</td>\n\t\t\t<td>0<u>1</u>000</td>\n\t\t\t<td>[8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>10</td>\n\t\t\t<td>0<u>1</u>0<u>1</u>0</td>\n\t\t\t<td>[2, 8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>13</td>\n\t\t\t<td>0<u>11</u>0<u>1</u></td>\n\t\t\t<td>[1, 4, 8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>23</td>\n\t\t\t<td><u>1</u>0<u>111</u></td>\n\t\t\t<td>[1, 2, 4, 16]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>The array <code>big_nums</code> is created by concatenating the <strong>powerful arrays</strong> for every positive integer <code>i</code> in ascending order: 1, 2, 3, and so on. Thus, <code>big_nums</code> begins as <code>[<u>1</u>, <u>2</u>, <u>1, 2</u>, <u>4</u>, <u>1, 4</u>, <u>2, 4</u>, <u>1, 2, 4</u>, <u>8</u>, ...]</code>.</p>\n\n<p>You are given a 2D integer matrix <code>queries</code>, where for <code>queries[i] = [from<sub>i</sub>, to<sub>i</sub>, mod<sub>i</sub>]</code> you should calculate <code>(big_nums[from<sub>i</sub>] * big_nums[from<sub>i</sub> + 1] * ... * big_nums[to<sub>i</sub>]) % mod<sub>i</sub></code><!-- notionvc: a71131cc-7b52-4786-9a4b-660d6d864f89 -->.</p>\n\n<p>Return an integer array <code>answer</code> such that <code>answer[i]</code> is the answer to the <code>i<sup>th</sup></code> query.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">queries = [[1,3,7]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[4]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There is one query.</p>\n\n<p><code>big_nums[1..3] = [2,1,2]</code>. The product of them is 4. The result is <code>4 % 7 = 4.</code></p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">queries = [[2,5,3],[7,7,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[2,2]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There are two queries.</p>\n\n<p>First query: <code>big_nums[2..5] = [1,2,4,1]</code>. The product of them is 8. The result is <code>8 % 3 = 2</code>.</p>\n\n<p>Second query: <code>big_nums[7] = 2</code>. The result is <code>2 % 4 = 2</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= queries.length &lt;= 500</code></li>\n\t<li><code>queries[i].length == 3</code></li>\n\t<li><code>0 &lt;= queries[i][0] &lt;= queries[i][1] &lt;= 10<sup>15</sup></code></li>\n\t<li><code>1 &lt;= queries[i][2] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "大数组元素的乘积",
            "translatedContent": "<p>一个非负整数 <code>x</code>&nbsp;的 <strong>强数组</strong>&nbsp;指的是满足元素为 2 的幂且元素总和为 <code>x</code> 的最短有序数组。下表说明了如何确定 <strong>强数组</strong> 的示例。可以证明，<code>x</code>&nbsp;对应的强数组是独一无二的。</p>\n\n<table border=\"1\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>数字</th>\n\t\t\t<th>二进制表示</th>\n\t\t\t<th>强数组</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>0000<u>1</u></td>\n\t\t\t<td>[1]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>8</td>\n\t\t\t<td>0<u>1</u>000</td>\n\t\t\t<td>[8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>10</td>\n\t\t\t<td>0<u>1</u>0<u>1</u>0</td>\n\t\t\t<td>[2, 8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>13</td>\n\t\t\t<td>0<u>11</u>0<u>1</u></td>\n\t\t\t<td>[1, 4, 8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>23</td>\n\t\t\t<td><u>1</u>0<u>111</u></td>\n\t\t\t<td>[1, 2, 4, 16]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>&nbsp;</p>\n\n<p>我们将每一个升序的正整数 <code>i</code>&nbsp;（即1，2，3等等）的 <strong>强数组</strong>&nbsp;连接得到数组&nbsp;<code>big_nums</code>&nbsp;，<code>big_nums</code>&nbsp;开始部分为&nbsp;<code>[<u>1</u>, <u>2</u>, <u>1, 2</u>, <u>4</u>, <u>1, 4</u>, <u>2, 4</u>, <u>1, 2, 4</u>, <u>8</u>, ...]</code>&nbsp;。</p>\n\n<p>给你一个二维整数数组&nbsp;<code>queries</code>&nbsp;，其中&nbsp;<code>queries[i] = [from<sub>i</sub>, to<sub>i</sub>, mod<sub>i</sub>]</code>&nbsp;，你需要计算&nbsp;<code>(big_nums[from<sub>i</sub>] * big_nums[from<sub>i</sub> + 1] * ... * big_nums[to<sub>i</sub>]) % mod<sub>i</sub></code>&nbsp;。</p>\n\n<p>请你返回一个整数数组&nbsp;<code>answer</code>&nbsp;，其中&nbsp;<code>answer[i]</code>&nbsp;是第 <code>i</code>&nbsp;个查询的答案。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><b>输入：</b>queries = [[1,3,7]]</p>\n\n<p><b>输出：</b>[4]</p>\n\n<p><strong>解释：</strong></p>\n\n<p>只有一个查询。</p>\n\n<p><code>big_nums[1..3] = [2,1,2]</code>&nbsp;。它们的乘积为 4。结果为&nbsp;<code>4 % 7 = 4</code>。</p>\n\n<p><strong>示例 2：</strong></p>\n\n<p><b>输入：</b>queries = [[2,5,3],[7,7,4]]</p>\n\n<p><b>输出：</b>[2,2]</p>\n\n<p><strong>解释：</strong></p>\n\n<p>有两个查询。</p>\n\n<p>第一个查询：<code>big_nums[2..5] = [1,2,4,1]</code>&nbsp;。它们的乘积为 8 。结果为&nbsp; <code>8 % 3 = 2</code>。</p>\n\n<p>第二个查询：<code>big_nums[7] = 2</code>&nbsp;。结果为 <code>2 % 4 = 2</code>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= queries.length &lt;= 500</code></li>\n\t<li><code>queries[i].length == 3</code></li>\n\t<li><code>0 &lt;= queries[i][0] &lt;= queries[i][1] &lt;= 10<sup>15</sup></code></li>\n\t<li><code>1 &lt;= queries[i][2] &lt;= 10<sup>5</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n",
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