You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively.

Choose at most k different engineers out of the n engineers to form a team with the maximum performance.

The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.

Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.

Example 1:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2 Output: 60 Explanation: 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. 

Example 2:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3 Output: 68 Explanation: 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. 

Example 3:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4 Output: 72 

Constraints:

• 1 <= <= k <= n <= 105
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 105
• 1 <= efficiency[i] <= 108

公司有编号为 1 到 n 的 n 个工程师，给你两个数组 speed 和 efficiency ，其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。请你返回由最多 k 个工程师组成的 最大团队表现值 ，由于答案可能很大，请你返回结果对 10^9 + 7 取余后的结果。团队表现值 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」。

• 题目要求返回最大团队表现值，表现值需要考虑速度的累加和，和效率的最小值。即使速度快，效率的最小值很小，总的表现值还是很小。先将效率从大到小排序。从效率高的工程师开始选起，遍历过程中维护一个大小为 k 的速度最小堆。每次遍历都计算一次团队最大表现值。扫描完成，最大团队表现值也筛选出来了。具体实现见下面的代码。

package leetcode import ( "container/heap""sort" ) funcmaxPerformance(nint, speed []int, efficiency []int, kint) int { indexes:=make([]int, n) fori:=rangeindexes { indexes[i] =i } sort.Slice(indexes, func(i, jint) bool { returnefficiency[indexes[i]] >efficiency[indexes[j]] }) ph:=speedHeap{} heap.Init(&ph) speedSum:=0varmaxint64for_, index:=rangeindexes { ifph.Len() ==k { speedSum-=heap.Pop(&ph).(int) } speedSum+=speed[index] heap.Push(&ph, speed[index]) max=Max(max, int64(speedSum)*int64(efficiency[index])) } returnint(max% (1e9+7)) } typespeedHeap []intfunc (hspeedHeap) Less(i, jint) bool { returnh[i] <h[j] } func (hspeedHeap) Swap(i, jint) { h[i], h[j] =h[j], h[i] } func (hspeedHeap) Len() int { returnlen(h) } func (h*speedHeap) Push(xinterface{}) { *h=append(*h, x.(int)) } func (h*speedHeap) Pop() interface{} { res:= (*h)[len(*h)-1] *h= (*h)[:h.Len()-1] returnres } funcMax(a, bint64) int64 { ifa>b { returna } returnb }