leetcode-problemset/leetcode-cn/originData/maximize-score-after-n-operations.json

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            "title": "Maximize Score After N Operations",
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            "content": "<p>You are given <code>nums</code>, an array of positive integers of size <code>2 * n</code>. You must perform <code>n</code> operations on this array.</p>\n\n<p>In the <code>i<sup>th</sup></code> operation <strong>(1-indexed)</strong>, you will:</p>\n\n<ul>\n\t<li>Choose two elements, <code>x</code> and <code>y</code>.</li>\n\t<li>Receive a score of <code>i * gcd(x, y)</code>.</li>\n\t<li>Remove <code>x</code> and <code>y</code> from <code>nums</code>.</li>\n</ul>\n\n<p>Return <em>the maximum score you can receive after performing </em><code>n</code><em> operations.</em></p>\n\n<p>The function <code>gcd(x, y)</code> is the greatest common divisor of <code>x</code> and <code>y</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:\n(1 * gcd(1, 2)) = 1\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,4,6,8]\n<strong>Output:</strong> 11\n<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:\n(1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4,5,6]\n<strong>Output:</strong> 14\n<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:\n(1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 7</code></li>\n\t<li><code>nums.length == 2 * n</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
            "translatedTitle": "N 次操作后的最大分数和",
            "translatedContent": "<p>给你 <code>nums</code> ，它是一个大小为 <code>2 * n</code> 的正整数数组。你必须对这个数组执行 <code>n</code> 次操作。</p>\n\n<p>在第 <code>i</code> 次操作时（操作编号从 <strong>1</strong> 开始），你需要：</p>\n\n<ul>\n\t<li>选择两个元素 <code>x</code> 和 <code>y</code> 。</li>\n\t<li>获得分数 <code>i * gcd(x, y)</code> 。</li>\n\t<li>将 <code>x</code> 和 <code>y</code> 从 <code>nums</code> 中删除。</li>\n</ul>\n\n<p>请你返回 <code>n</code> 次操作后你能获得的分数和最大为多少。</p>\n\n<p>函数 <code>gcd(x, y)</code> 是 <code>x</code> 和 <code>y</code> 的最大公约数。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [1,2]\n<b>输出：</b>1\n<b>解释：</b>最优操作是：\n(1 * gcd(1, 2)) = 1\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [3,4,6,8]\n<b>输出：</b>11\n<b>解释：</b>最优操作是：\n(1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><b>输入：</b>nums = [1,2,3,4,5,6]\n<b>输出：</b>14\n<b>解释：</b>最优操作是：\n(1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 7</code></li>\n\t<li><code>nums.length == 2 * n</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
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