存量题库数据更新

leetcode/problem/maximum-performance-of-a-team.html

@@ -2,12 +2,12 @@
 
 <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>
 
-<p>The performance of a team is the sum of their engineers&#39; speeds multiplied by the minimum efficiency among their engineers.</p>
+<p>The performance of a team is the sum of its engineers&#39; speeds multiplied by the minimum efficiency among its engineers.</p>
 
 <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>
 
 <p>&nbsp;</p>
-<p><strong>Example 1:</strong></p>
+<p><strong class="example">Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
@@ -16,7 +16,7 @@
 We 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.
 </pre>
 
-<p><strong>Example 2:</strong></p>
+<p><strong class="example">Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
@@ -25,7 +25,7 @@ We have the maximum performance of the team by selecting engineer 2 (with speed=
 </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.
 </pre>
 
-<p><strong>Example 3:</strong></p>
+<p><strong class="example">Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4