Talented Chef ZOJ - 3778

最新推荐文章于 2020-05-18 12:03:52 发布

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最新推荐文章于 2020-05-18 12:03:52 发布

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原文链接：http://www.cnblogs.com/Asimple/p/7384080.html

本文介绍了一种关于多任务并行处理的问题——烹饪问题，并提供了一个简洁高效的算法解决方案。该问题涉及如何在有限时间内完成多道菜品的制作，每道菜包含若干步骤，且同一时间可以同时处理多个菜品的一步骤。文章通过示例展示了如何计算完成所有菜品所需的最短时间。

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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 Mdifferent 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 Tindicating 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：一个人可以在一分钟同时进行m道菜的一个步骤，共有n道菜，每道菜各有xi个步骤，求做完的最短时间。

思维题，以平均值来做菜，所用的时间总是最短。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define mod 7
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 20000+5;
int n, m, T, len, cnt, num;

void input() {
    cin >> T;
    while( T -- ) {
        cin >> n >> m;
        int Max = 0;
        int sum = 0;
        for(int i=0; i<n; i++) {
            cin >> num;
            sum += num;
            Max = max(Max, num);
        }
        cnt = sum/m;
        if( sum%m ) cnt ++;
        if( cnt < Max ) cnt = Max;
        if( m >= n ) cout << Max << endl;
        else cout << cnt << endl;
    }
}

int main() {
    input();
    return 0;
}