mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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182 lines
21 KiB
JSON
182 lines
21 KiB
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"title": "Number of Ways Where Square of Number Is Equal to Product of Two Numbers",
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"content": "<p>Given two arrays of integers <code>nums1</code> and <code>nums2</code>, return the number of triplets formed (type 1 and type 2) under the following rules:</p>\n\n<ul>\n\t<li>Type 1: Triplet (i, j, k) if <code>nums1[i]<sup>2</sup> == nums2[j] * nums2[k]</code> where <code>0 <= i < nums1.length</code> and <code>0 <= j < k < nums2.length</code>.</li>\n\t<li>Type 2: Triplet (i, j, k) if <code>nums2[i]<sup>2</sup> == nums1[j] * nums1[k]</code> where <code>0 <= i < nums2.length</code> and <code>0 <= j < k < nums1.length</code>.</li>\n</ul>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [7,4], nums2 = [5,2,8,9]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> Type 1: (1, 1, 2), nums1[1]<sup>2</sup> = nums2[1] * nums2[2]. (4<sup>2</sup> = 2 * 8). \n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,1], nums2 = [1,1,1]\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> All Triplets are valid, because 1<sup>2</sup> = 1 * 1.\nType 1: (0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2). nums1[i]<sup>2</sup> = nums2[j] * nums2[k].\nType 2: (0,0,1), (1,0,1), (2,0,1). nums2[i]<sup>2</sup> = nums1[j] * nums1[k].\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [7,7,8,3], nums2 = [1,2,9,7]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> There are 2 valid triplets.\nType 1: (3,0,2). nums1[3]<sup>2</sup> = nums2[0] * nums2[2].\nType 2: (3,0,1). nums2[3]<sup>2</sup> = nums1[0] * nums1[1].\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums1.length, nums2.length <= 1000</code></li>\n\t<li><code>1 <= nums1[i], nums2[i] <= 10<sup>5</sup></code></li>\n</ul>\n",
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"translatedTitle": "数的平方等于两数乘积的方法数",
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"translatedContent": "<p>给你两个整数数组 <code>nums1</code> 和 <code>nums2</code> ，请你返回根据以下规则形成的三元组的数目（类型 1 和类型 2 ）：</p>\n\n<ul>\n\t<li>类型 1：三元组 <code>(i, j, k)</code> ，如果 <code>nums1[i]<sup>2</sup> == nums2[j] * nums2[k]</code> 其中 <code>0 <= i < nums1.length</code> 且 <code>0 <= j < k < nums2.length</code></li>\n\t<li>类型 2：三元组 <code>(i, j, k)</code> ，如果 <code>nums2[i]<sup>2</sup> == nums1[j] * nums1[k]</code> 其中 <code>0 <= i < nums2.length</code> 且 <code>0 <= j < k < nums1.length</code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>nums1 = [7,4], nums2 = [5,2,8,9]\n<strong>输出：</strong>1\n<strong>解释：</strong>类型 1：(1,1,2), nums1[1]^2 = nums2[1] * nums2[2] (4^2 = 2 * 8)</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>nums1 = [1,1], nums2 = [1,1,1]\n<strong>输出：</strong>9\n<strong>解释：</strong>所有三元组都符合题目要求，因为 1^2 = 1 * 1\n类型 1：(0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2), nums1[i]^2 = nums2[j] * nums2[k]\n类型 2：(0,0,1), (1,0,1), (2,0,1), nums2[i]^2 = nums1[j] * nums1[k]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>nums1 = [7,7,8,3], nums2 = [1,2,9,7]\n<strong>输出：</strong>2\n<strong>解释：</strong>有两个符合题目要求的三元组\n类型 1：(3,0,2), nums1[3]^2 = nums2[0] * nums2[2]\n类型 2：(3,0,1), nums2[3]^2 = nums1[0] * nums1[1]\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre><strong>输入：</strong>nums1 = [4,7,9,11,23], nums2 = [3,5,1024,12,18]\n<strong>输出：</strong>0\n<strong>解释：</strong>不存在符合题目要求的三元组\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums1.length, nums2.length <= 1000</code></li>\n\t<li><code>1 <= nums1[i], nums2[i] <= 10^5</code></li>\n</ul>\n",
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