C - Talented Chef

原创于 2018-04-24 17:54:06 发布 · 207 阅读

0 ·

CC 4.0 BY-SA版权

数学题专栏收录该内容

5 篇文章

订阅专栏

本文解析了一道关于高效烹饪多道菜品的问题，通过算法确定完成所有菜品所需的最短时间。问题涉及了如何合理分配烹饪资源以达到最优解。

C - 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

题意：求做完n道菜，所需的最短时间。

依据：最大值 * m < sum （n之和）

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int ca;
    scanf("%d",&ca);
    while(ca--)
    {
        int n,m,a,mm=-1;
        long long sum=0;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a);
            sum+=a;
            mm=max(mm,a);
        }
        long long ans=sum/m;
        if(sum%m)
            ans++;    //mm*m>sum
        if(ans<mm)   //如果ans<最大输入值
            ans=mm;
                       //如果
        printf("%lld\n",ans);
    }
    return 0;
}