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leetcode-cn/originData/maximum-performance-of-a-team.json

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             "titleSlug": "maximum-performance-of-a-team",
             "content": "<p>You are given two integers <code>n</code> and <code>k</code> and two integer arrays <code>speed</code> and <code>efficiency</code> both of length <code>n</code>. There are <code>n</code> engineers numbered from <code>1</code> to <code>n</code>. <code>speed[i]</code> and <code>efficiency[i]</code> represent the speed and efficiency of the <code>i<sup>th</sup></code> engineer respectively.</p>\n\n<p>Choose <strong>at most</strong> <code>k</code> different engineers out of the <code>n</code> engineers to form a team with the maximum <strong>performance</strong>.</p>\n\n<p>The performance of a team is the sum of its engineers&#39; speeds multiplied by the minimum efficiency among its engineers.</p>\n\n<p>Return <em>the maximum performance of this team</em>. Since the answer can be a huge number, return it <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 = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2\n<strong>Output:</strong> 60\n<strong>Explanation:</strong> \nWe have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3\n<strong>Output:</strong> 68\n<strong>Explanation:\n</strong>This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4\n<strong>Output:</strong> 72\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>speed.length == n</code></li>\n\t<li><code>efficiency.length == n</code></li>\n\t<li><code>1 &lt;= speed[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= efficiency[i] &lt;= 10<sup>8</sup></code></li>\n</ul>\n",
             "translatedTitle": "最大的团队表现值",
-            "translatedContent": "<p>给定两个整数 <code>n</code> 和 <code>k</code>，以及两个长度为 <code>n</code> 的整数数组 <code>speed</code> 和<code> efficiency</code>。现有 <code>n</code> 名工程师，编号从 <code>1</code> 到 <code>n</code>。其中 <code>speed[i]</code>&nbsp;和 <code>efficiency[i]</code>&nbsp;分别代表第 <code>i</code>&nbsp;位工程师的速度和效率。</p>\n\n<p>从这 <code>n</code> 名工程师中最多选择 <code>k</code> 名不同的工程师，使其组成的团队具有最大的团队表现值。</p>\n\n<p><strong>团队表现值</strong>&nbsp;的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。</p>\n\n<p>请你返回该团队的​​​​​​最大团队表现值，由于答案可能很大，请你返回结果对 <code>10^9 + 7</code> 取余后的结果。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2\n<strong>输出：</strong>60\n<strong>解释：</strong>\n我们选择工程师 2（speed=10 且 efficiency=4）和工程师 5（speed=5 且 efficiency=7）。他们的团队表现值为 performance = (10 + 5) * min(4, 7) = 60 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3\n<strong>输出：</strong>68\n<strong>解释：\n</strong>此示例与第一个示例相同，除了 k = 3 。我们可以选择工程师 1 ，工程师 2 和工程师 5 得到最大的团队表现值。表现值为 performance = (2 + 10 + 5) * min(5, 4, 7) = 68 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4\n<strong>输出：</strong>72\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 10^5</code></li>\n\t<li><code>speed.length == n</code></li>\n\t<li><code>efficiency.length == n</code></li>\n\t<li><code>1 &lt;= speed[i] &lt;= 10^5</code></li>\n\t<li><code>1 &lt;= efficiency[i] &lt;= 10^8</code></li>\n</ul>\n",
+            "translatedContent": "<p>给定两个整数 <code>n</code> 和 <code>k</code>，以及两个长度为 <code>n</code> 的整数数组 <code>speed</code> 和<code> efficiency</code>。现有 <code>n</code> 名工程师，编号从 <code>1</code> 到 <code>n</code>。其中 <code>speed[i]</code>&nbsp;和 <code>efficiency[i]</code>&nbsp;分别代表第 <code>i</code>&nbsp;位工程师的速度和效率。</p>\n\n<p>从这 <code>n</code> 名工程师中最多选择 <code>k</code> 名不同的工程师，使其组成的团队具有最大的团队表现值。</p>\n\n<p><strong>团队表现值</strong>&nbsp;的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。</p>\n\n<p>请你返回该团队的最大团队表现值，由于答案可能很大，请你返回结果对 <code>10^9 + 7</code> 取余后的结果。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2\n<strong>输出：</strong>60\n<strong>解释：</strong>\n我们选择工程师 2（speed=10 且 efficiency=4）和工程师 5（speed=5 且 efficiency=7）。他们的团队表现值为 performance = (10 + 5) * min(4, 7) = 60 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3\n<strong>输出：</strong>68\n<strong>解释：\n</strong>此示例与第一个示例相同，除了 k = 3 。我们可以选择工程师 1 ，工程师 2 和工程师 5 得到最大的团队表现值。表现值为 performance = (2 + 10 + 5) * min(5, 4, 7) = 68 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4\n<strong>输出：</strong>72\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 10^5</code></li>\n\t<li><code>speed.length == n</code></li>\n\t<li><code>efficiency.length == n</code></li>\n\t<li><code>1 &lt;= speed[i] &lt;= 10^5</code></li>\n\t<li><code>1 &lt;= efficiency[i] &lt;= 10^8</code></li>\n</ul>\n",
             "isPaidOnly": false,
             "difficulty": "Hard",
             "likes": 152,