ZOJ3778（Talented Chef）水

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.
【ZOJ3778（Talented Chef）】To make full use of his genius in cooking, Coach Gao decides to prepare N dishes for the dinner. The i-th dish contains Ai 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 Ai (1 <= Ai <= 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

首先可以确定在m>=n，输出的就是maxn

因为在sum足够的情况下，每天必定要减少m，所以算出平均需要的天数，ans = sum/m，在不能整除的情况下，ans还要+1

算出的平均如果小于maxn，那么结果当然还是maxn，否则就是答案了

#include #include #include using namespace std; int main() { int t,n,m,maxn,sum,a,i; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); maxn = sum = 0; for(i = 0; i=n) printf("%d\n",maxn); else printf("%d\n",ans); }return 0; }