ZOJ 3778 Talented Chef

Talented Chef

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

【思路分析】

   乍一看马上就想起了计算机操作系统里面讲过的时间片轮转调度算法，然后就陷入了这个大坑。。。。。。

代码如下：

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,a;
        int sum = 0, maxn = 0;
        scanf("%d %d",&n,&m);
        for(int i = 0;i < n;i++)
        {
            scanf("%d",&a);
            sum += a;
            maxn = max(maxn,a);
        }
        int ans = sum / m;
        if(sum % m != 0)
            ans++;
        if(ans < maxn)
            ans = maxn;
        if(m >= n)
            printf("%d\n",maxn);
        else
            printf("%d\n",ans);
    }
    return 0;
}