ZOJ 3778 Talented Chef（数学啊 ）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5017

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

Source: The 11th Zhejiang Provincial Collegiate Programming Contest

题意：

一共有n道菜每道菜都需要固定的时间来完成，一个人最多能同时选m道菜来同时做！

求最少需要多少分钟才能做完所有的菜！

代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i, j;
		int n, m;
		int sum = 0;
		int tt, maxx = 0;
		scanf("%d%d",&n,&m);
		for(i = 0; i < n; i++)
		{
			scanf("%d",&tt);
			sum+=tt;
			if(tt > maxx)
			{
				maxx = tt;
			}
		}
		int ans = sum/m;
		if(sum%m != 0)
		{

			ans+=1;
		}
		if(ans < maxx)
		{
			ans = maxx;
		}
		printf("%d\n",ans);
	}
	return 0;
}
/*
3 999
2 2 2
10 999
1 2 3 4 5 6 7 8 9 10
*/