题解：Cats and Fish（模拟）

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 ci minutes 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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 5000 + 10;
int a[maxn],eat[maxn];  //eat值为一代表正在吃，0代表未吃
int main()
{
    int m,n,x;
    while(scanf("%d%d%d",&m,&n,&x) == 3)
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&a[i]);
        int q = 0;  //残缺的鱼
        int p = m;  //完整的鱼
        sort(a,a + n);
        memset(eat,0,sizeof(eat));
        //枚举时间轴
        for(int i = 1; i <= x; i++)
        {
            if(!p) break;//都吃完了 
            for(int j = 0;j< n; j++)
            {
                if(!p) break;//都吃完了 
                if(i % a[j] == 0)//到时间了 
                {
                    if(eat[j])//正在吃 
                    {
                        eat[j] = 0;
                        q--;
                    }
                    else p--;//c=1 
                }
                else
                {
                    if(eat[j]) continue;
                    else
                    {
                        eat[j] = 1;
                        q++;
                        p--;
                    }
                }
            }
        }
        printf("%d %d\n",p,q);
    }
    return 0;
}

来源：https://blog.csdn.net/weixin_43540515/article/details/99694159