zoj 3778 Talented Chef

最新推荐文章于 2019-05-26 21:44:00 发布

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本文介绍了一个烹饪问题的算法解决方案，旨在帮助厨师在有限时间内完成尽可能多的菜肴制作。通过合理分配每分钟内可处理的不同菜肴数量，算法能够计算出完成所有菜肴所需的最短时间。

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Talented Chef

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

Author: JIANG, Kai
Source: The 11th Zhejiang Provincial Collegiate Programming Contest

题目大意：需要做n道菜，第i道菜Ai步，每一步可以同时做M道不同的菜，问最少需要多少时间能做完所有的菜。

思路：如果每次都可以选择m道菜，那么答案就是向上取整(sum/m)，如果最大值大于ans，则答案肯定为最大值

#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define FOR(i, s, t) for(int (i)=(s); (i)<(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 40000 + 20;

int main() {
    int T;

    scanf("%d", &T);
    while(T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        int sum = 0;
        int tmax = 0;
        for(int i=0; i<n; i++) {
            int t;
            scanf("%d", &t);
            sum += t;
            tmax = max(t, tmax);
        }
        int ans = sum / m;
        if(sum % m) ans += 1;
        if(ans < tmax) {
            ans = tmax;
        }
        printf("%d\n", ans);
    }
    
    return 0;
}