Talented Chef

最新推荐文章于 2019-04-23 21:43:56 发布

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郭教练是一位才华横溢的厨师，擅长同时烹饪M道菜。为了展示他的烹饪天才给女友，他决定准备N道菜。每道菜包含Ai步骤，郭教练每分钟最多能完成M道菜的一个步骤。本文探讨了如何最高效地完成所有菜品。

Talented Chef

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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Description

As we all know, Coach Gao is a talented chef, because he is able to cook M dishes in the same time. Tonight he is going to have a hearty dinner with his girlfriend at his home. Of course, Coach Gao is going to cook all dishes himself, in order to show off his genius cooking skill to his girlfriend.

To make full use of his genius in cooking, Coach Gao decides to prepare N dishes for the dinner. The i-th dish contains A_i steps. The steps of a dish should be finished sequentially. In each minute of the cooking, Coach Gao can choose at most M different dishes and finish one step for each dish chosen.

Coach Gao wants to know the least time he needs to prepare the dinner.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N and M (1 <= N, M <= 40000). The second line contains N integers A_i (1 <= A_i <= 40000).

Output

For each test case, output the least time (in minute) to finish all dishes.

Sample Input

2
3 2
2 2 2
10 6
1 2 3 4 5 6 7 8 9 10

Sample Output

3
10

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int t,n,m,k;
	scanf("%d",&t);
	int maxn,sum,ans;
	while(t--){
		scanf("%d%d",&n,&m);
		maxn=0;
		sum=0;
		ans=0;
		for(int i=0;i<n;i++){
			scanf("%d",&k);
			sum+=k;
			maxn=max(maxn,k);
		}
		if(sum%m)	ans=sum/m+1;
		else	ans=sum/m;
		ans=max(ans,maxn);
		printf("%d\n",ans);
	}
	return 0;
}