滴答滴答---題目連結 

There are many homeless cats in PKU campus. They are all happy because the students in the cat club of PKU take good care of them. Li lei is one of the members of the cat club. He loves those cats very much. Last week, he won a scholarship and he wanted to share his pleasure with cats. So he bought some really tasty fish to feed them, and watched them eating with great pleasure. At the same time, he found an interesting question:

There are m fish and n cats, and it takes ciminutes for the ith cat to eat out one fish. A cat starts to eat another fish (if it can get one) immediately after it has finished one fish. A cat never shares its fish with other cats. When there are not enough fish left, the cat which eats quicker has higher priority to get a fish than the cat which eats slower. All cats start eating at the same time. Li Lei wanted to know, after x minutes, how many fish would be left.

Input

There are no more than 20 test cases.

For each test case:

The first line contains 3 integers: above mentioned m, n and x (0 < m <= 5000, 1 <= n <= 100, 0 <= x <= 1000).

The second line contains n integers c1,c2 … cn,  ci means that it takes the ith cat ci minutes to eat out a fish ( 1<= ci <= 2000).

Output

For each test case, print 2 integers p and q, meaning that there are p complete fish(whole fish) and q incomplete fish left after x minutes.

Sample Input

2 1 1
1
8 3 5
1 3 4
4 5 1
5 4 3 2 1

Sample Output

1 0
0 1
0 3

問 x 分鐘後，剩下幾條完整的魚，幾條魚沒吃完。

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=5001;
int a[maxn];
int s[maxn];
int main()
{
    int n,m,x;
    while(scanf("%d%d%d",&m,&n,&x)!=EOF)
    {
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int ans=0;
        if(m<=n)
        {
            for(int i=0; i<m; i++)
                if(a[i]>x)
                    ans++;
            printf("0 %d\n",ans);
            continue;
        }
        memset(s,0,sizeof(s));
        for(int i=0; i<x; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(s[j]==0&&m!=0)
                {
                    s[j]=a[j];
                    m--;
                }
            }
            for(int j=0; j<n; j++)
                if(s[j]!=0)
                    s[j]--;
        }
        for(int i=0; i<n; i++)
            if(s[i]>0)
                ans++;
            printf("%d %d\n",m,ans);
    }
    return 0;
}