zoj3778 Talented Chef

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

 题目大意：给出n（表示有n道菜）、m（表示能同时煮m道菜），给出每道菜的时间，求最少需要煮多久。

 思路：直接比较时间最长那道菜和总时间除以m的值哪个最大，取最大的即可。因为要嘛就是煮完的时候要嘛就是时间最长那道菜的时间，要嘛就是总时间除以m。

 

代码如下

/*
 * Author:  Joshua
 * Created Time:  2014/5/18 13:30:03
 * File Name: c.cpp
 */
#include<cstdio>

void solve()
{
    int n,m,max=0,ans,sum=0;
    int a[40005];
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if (a[i]>max) max=a[i];
        sum+=a[i];
    }
    if ((sum-1)/m+1 >max) printf("%d\n",(sum-1)/m+1);
    else printf("%d\n",max);
}

int main()
{
    int times;
       scanf("%d",&times);
       while (times)
       {
           times--;
           solve();
       }
    return 0;
}