leetcode-problemset/leetcode-cn/originData/range-product-queries-of-powers.json

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        "question": {
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            "title": "Range Product Queries of Powers",
            "titleSlug": "range-product-queries-of-powers",
            "content": "<p>Given a positive integer <code>n</code>, there exists a <strong>0-indexed</strong> array called <code>powers</code>, composed of the <strong>minimum</strong> number of powers of <code>2</code> that sum to <code>n</code>. The array is sorted in <strong>non-decreasing</strong> order, and there is <strong>only one</strong> way to form the array.</p>\n\n<p>You are also given a <strong>0-indexed</strong> 2D integer array <code>queries</code>, where <code>queries[i] = [left<sub>i</sub>, right<sub>i</sub>]</code>. Each <code>queries[i]</code> represents a query where you have to find the product of all <code>powers[j]</code> with <code>left<sub>i</sub> &lt;= j &lt;= right<sub>i</sub></code>.</p>\n\n<p>Return<em> an array </em><code>answers</code><em>, equal in length to </em><code>queries</code><em>, where </em><code>answers[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query</em>. Since the answer to the <code>i<sup>th</sup></code> query may be too large, each <code>answers[i]</code> should be returned <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 15, queries = [[0,1],[2,2],[0,3]]\n<strong>Output:</strong> [2,4,64]\n<strong>Explanation:</strong>\nFor n = 15, powers = [1,2,4,8]. It can be shown that powers cannot be a smaller size.\nAnswer to 1st query: powers[0] * powers[1] = 1 * 2 = 2.\nAnswer to 2nd query: powers[2] = 4.\nAnswer to 3rd query: powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64.\nEach answer modulo 10<sup>9</sup> + 7 yields the same answer, so [2,4,64] is returned.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2, queries = [[0,0]]\n<strong>Output:</strong> [2]\n<strong>Explanation:</strong>\nFor n = 2, powers = [2].\nThe answer to the only query is powers[0] = 2. The answer modulo 10<sup>9</sup> + 7 is the same, so [2] is returned.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt; powers.length</code></li>\n</ul>\n",
            "translatedTitle": "二的幂数组中查询范围内的乘积",
            "translatedContent": "<p>给你一个正整数&nbsp;<code>n</code>&nbsp;，你需要找到一个下标从&nbsp;<strong>0</strong>&nbsp;开始的数组&nbsp;<code>powers</code>&nbsp;，它包含 <strong>最少</strong>&nbsp;数目的 <code>2</code>&nbsp;的幂，且它们的和为&nbsp;<code>n</code>&nbsp;。<code>powers</code>&nbsp;数组是&nbsp;<strong>非递减</strong>&nbsp;顺序的。根据前面描述，构造&nbsp;<code>powers</code>&nbsp;数组的方法是唯一的。</p>\n\n<p>同时给你一个下标从 <strong>0</strong>&nbsp;开始的二维整数数组&nbsp;<code>queries</code>&nbsp;，其中&nbsp;<code>queries[i] = [left<sub>i</sub>, right<sub>i</sub>]</code>&nbsp;，其中&nbsp;<code>queries[i]</code>&nbsp;表示请你求出满足&nbsp;<code>left<sub>i</sub> &lt;= j &lt;= right<sub>i</sub></code>&nbsp;的所有&nbsp;<code>powers[j]</code>&nbsp;的乘积。</p>\n\n<p>请你返回一个数组<em>&nbsp;</em><code>answers</code>&nbsp;，长度与<em>&nbsp;</em><code>queries</code>&nbsp;的长度相同，其中<em>&nbsp;</em><code>answers[i]</code>是第<em>&nbsp;</em><code>i</code>&nbsp;个查询的答案。由于查询的结果可能非常大，请你将每个&nbsp;<code>answers[i]</code>&nbsp;都对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>n = 15, queries = [[0,1],[2,2],[0,3]]\n<b>输出：</b>[2,4,64]\n<strong>解释：</strong>\n对于 n = 15 ，得到 powers = [1,2,4,8] 。没法得到元素数目更少的数组。\n第 1 个查询的答案：powers[0] * powers[1] = 1 * 2 = 2 。\n第 2 个查询的答案：powers[2] = 4 。\n第 3 个查询的答案：powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64 。\n每个答案对 10<sup>9</sup> + 7 得到的结果都相同，所以返回 [2,4,64] 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>n = 2, queries = [[0,0]]\n<b>输出：</b>[2]\n<strong>解释：</strong>\n对于 n = 2, powers = [2] 。\n唯一一个查询的答案是 powers[0] = 2 。答案对 10<sup>9</sup> + 7 取余后结果相同，所以返回 [2] 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt; powers.length</code></li>\n</ul>\n",
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