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

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        "question": {
            "questionId": "3411",
            "questionFrontendId": "3145",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2772487,
            "title": "Find Products of Elements of Big Array",
            "titleSlug": "find-products-of-elements-of-big-array",
            "content": "<p>A <strong>powerful array</strong> for an integer <code>x</code> is the shortest sorted array of powers of two that sum up to <code>x</code>. For example, the powerful array for 11 is <code>[1, 2, 8]</code>.</p>\n\n<p>The array <code>big_nums</code> is created by concatenating the <strong>powerful</strong> arrays for every positive integer <code>i</code> in ascending order: 1, 2, 3, and so forth. Thus, <code>big_nums</code> starts 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 remainder of 4 under 7 is 4.</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 remainder of 8 under 3 is 2.</p>\n\n<p>Second query: <code>big_nums[7] = 2</code>. The remainder of 2 under 4 is 2.</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>一个整数 <code>x</code>&nbsp;的 <strong>强数组</strong>&nbsp;指的是满足和为 <code>x</code> 的二的幂的最短有序数组。比方说，11 的强数组为&nbsp;<code>[1, 2, 8]</code>&nbsp;。</p>\n\n<p>我们将每一个正整数 <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 class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>queries = [[1,3,7]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>[4]</span></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 ，4 对 7 取余数得到 4 。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>queries = [[2,5,3],[7,7,4]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>[2,2]</span></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 ，8 对 3 取余数得到 2 。</p>\n\n<p>第二个查询：<code>big_nums[7] = 2</code>&nbsp;，2 对 4 取余数得到 2 。</p>\n</div>\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",
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                    "code": "class Solution {\npublic:\n    vector<int> findProductsOfElements(vector<vector<long long>>& queries) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def findProductsOfElements(self, queries):\n        \"\"\"\n        :type queries: List[List[int]]\n        :rtype: List[int]\n        \"\"\"",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* findProductsOfElements(long long** queries, int queriesSize, int* queriesColSize, int* returnSize) {\n    \n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[][] $queries\n     * @return Integer[]\n     */\n    function findProductsOfElements($queries) {\n\n    }\n}",
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                    "code": "class Solution {\n    func findProductsOfElements(_ queries: [[Int]]) -> [Int] {\n\n    }\n}",
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                    "code": "class Solution {\n    fun findProductsOfElements(queries: Array<LongArray>): IntArray {\n\n    }\n}",
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                    "code": "func findProductsOfElements(queries [][]int64) []int {\n\n}",
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                    "code": "# @param {Integer[][]} queries\n# @return {Integer[]}\ndef find_products_of_elements(queries)\n\nend",
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            "hints": [
                "Find a way to calculate <code>f(n, i)</code> which is the total number of numbers in <code>[1, n]</code> when the <code>i<sup>th</sup></code> bit is set in <code>O(log(n))</code> time.",
                "Use binary search to find the last number for each query (and there might be one “incomplete” number for the query).",
                "Use a similar way to find the product (we only need to save the sum of exponents of power of <code>2</code>)."
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