Talented Chef ZOJ - 3778 (思维+贪心)

Talented Chef

ZOJ - 3778

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：有一个人要做n道菜，给出每道菜所需步骤数。已知这个人每次同时能做m道菜的1步。问最少需要几次把这些菜做完。

题目思路：贪心。
①当n <= m : 令n道菜每次减少1，所以次数为maxn（最多步骤的那道菜）

②当n > m考虑两种情况：

1. 每次如果恰好使得m个数-1.那么次数为sum / m。如果最后一次不是m个数，则需 + 1

   eg : 1 2 3 4 5 6 7 8 9 10

   

2. 如果未取到最后一次，需要做的菜 < m，则 结果为 maxn（最多步骤的那道菜）

   eg: 1 1 1 1 1 1 1 1 1 10
   

code:

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int num[40005];
int main(){
    int t;
    cin >> t;
    while(t--)
    {
        int n,m,sum = 0,maxn = 0;
        cin >> n >> m;
        for (int i = 0; i < n; i++)
        {
            cin >> num[i];
            maxn = max(maxn,num[i]);
            sum += num[i];
        }
        int ret = sum / m;
        if (sum % m) ret++;
        cout << max(ret,maxn) << endl;      
    }   
    return 0;
}