leetcode-problemset/leetcode-cn/originData/number-of-beautiful-pairs.json

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            "title": "Number of Beautiful Pairs",
            "titleSlug": "number-of-beautiful-pairs",
            "content": "<p>You are given a <strong>0-indexed </strong>integer array <code>nums</code>. A pair of indices <code>i</code>, <code>j</code> where <code>0 &lt;=&nbsp;i &lt; j &lt; nums.length</code> is called beautiful if the <strong>first digit</strong> of <code>nums[i]</code> and the <strong>last digit</strong> of <code>nums[j]</code> are <strong>coprime</strong>.</p>\n\n<p>Return <em>the total number of beautiful pairs in </em><code>nums</code>.</p>\n\n<p>Two integers <code>x</code> and <code>y</code> are <strong>coprime</strong> if there is no integer greater than 1 that divides both of them. In other words, <code>x</code> and <code>y</code> are coprime if <code>gcd(x, y) == 1</code>, where <code>gcd(x, y)</code> is the <strong>greatest common divisor</strong> of <code>x</code> and <code>y</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,5,1,4]\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> There are 5 beautiful pairs in nums:\nWhen i = 0 and j = 1: the first digit of nums[0] is 2, and the last digit of nums[1] is 5. We can confirm that 2 and 5 are coprime, since gcd(2,5) == 1.\nWhen i = 0 and j = 2: the first digit of nums[0] is 2, and the last digit of nums[2] is 1. Indeed, gcd(2,1) == 1.\nWhen i = 1 and j = 2: the first digit of nums[1] is 5, and the last digit of nums[2] is 1. Indeed, gcd(5,1) == 1.\nWhen i = 1 and j = 3: the first digit of nums[1] is 5, and the last digit of nums[3] is 4. Indeed, gcd(5,4) == 1.\nWhen i = 2 and j = 3: the first digit of nums[2] is 1, and the last digit of nums[3] is 4. Indeed, gcd(1,4) == 1.\nThus, we return 5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [11,21,12]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> There are 2 beautiful pairs:\nWhen i = 0 and j = 1: the first digit of nums[0] is 1, and the last digit of nums[1] is 1. Indeed, gcd(1,1) == 1.\nWhen i = 0 and j = 2: the first digit of nums[0] is 1, and the last digit of nums[2] is 2. Indeed, gcd(1,2) == 1.\nThus, we return 2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 9999</code></li>\n\t<li><code>nums[i] % 10 != 0</code></li>\n</ul>\n",
            "translatedTitle": "美丽下标对的数目",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。如果下标对 <code>i</code>、<code>j</code> 满足 <code>0 ≤ i &lt; j &lt; nums.length</code> ，如果&nbsp;<code>nums[i]</code> 的 <strong>第一个数字</strong> 和 <code>nums[j]</code> 的 <strong>最后一个数字</strong> <strong>互质</strong> ，则认为 <code>nums[i]</code> 和 <code>nums[j]</code> 是一组 <strong>美丽下标对</strong> 。</p>\n\n<p>返回 <code>nums</code> 中 <strong>美丽下标对</strong> 的总数目。</p>\n\n<p>对于两个整数 <code>x</code> 和 <code>y</code> ，如果不存在大于 1 的整数可以整除它们，则认为 <code>x</code> 和 <code>y</code> <strong>互质</strong> 。换而言之，如果 <code>gcd(x, y) == 1</code> ，则认为 <code>x</code> 和 <code>y</code> 互质，其中 <code>gcd(x, y)</code> 是 <code>x</code> 和 <code>k</code> <strong>最大公因数</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,5,1,4]\n<strong>输出：</strong>5\n<strong>解释：</strong>nums 中共有 5 组美丽下标对：\ni = 0 和 j = 1 ：nums[0] 的第一个数字是 2 ，nums[1] 的最后一个数字是 5 。2 和 5 互质，因此 gcd(2,5) == 1 。\ni = 0 和 j = 2 ：nums[0] 的第一个数字是 2 ，nums[2] 的最后一个数字是 1 。2 和 5 互质，因此 gcd(2,1) == 1 。\ni = 1 和 j = 2 ：nums[1] 的第一个数字是 5 ，nums[2] 的最后一个数字是 1 。2 和 5 互质，因此 gcd(5,1) == 1 。\ni = 1 和 j = 3 ：nums[1] 的第一个数字是 5 ，nums[3] 的最后一个数字是 4 。2 和 5 互质，因此 gcd(5,4) == 1 。\ni = 2 和 j = 3 ：nums[2] 的第一个数字是 1 ，nums[3] 的最后一个数字是 4 。2 和 5 互质，因此 gcd(1,4) == 1 。\n因此，返回 5 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [11,21,12]\n<strong>输出：</strong>2\n<strong>解释：</strong>共有 2 组美丽下标对：\ni = 0 和 j = 1 ：nums[0] 的第一个数字是 1 ，nums[1] 的最后一个数字是 1 。gcd(1,1) == 1 。\ni = 0 和 j = 2 ：nums[0] 的第一个数字是 1 ，nums[2] 的最后一个数字是 2 。gcd(1,2) == 1 。\n因此，返回 2 。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 9999</code></li>\n\t<li><code>nums[i] % 10 != 0</code></li>\n</ul>\n",
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