华中农业大学第五届程序设计大赛网络同步赛-K

K．Deadline

There are N bugs to be repaired and some engineers whose abilities are roughly equal. And an engineer can repair a bug per day. Each bug has a deadline A[i].

Question: How many engineers can repair all bugs before those deadlines at least? 1<=n<= 1e6. 1<=a[i] <=1e9

Input

Description There are multiply test cases. In each case, the first line is an integer N , indicates the number of bugs. The next line is n integers indicates the deadlines of those bugs.

Output

Description There are one number indicates the answer to the question in a line for each case.

Input

4 1 2 3 4

Output

1

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1000005;
int a, book[N], day[N];

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        int len = 0;
        memset(book, 0, sizeof(book));
        memset(day, 0, sizeof(day));
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a);
            if(a <= n)
              book[a]++;
            if(a < n && a > len)len = a;
        }
        int ans = 1;
        for(int i = 1; i <= len; i++)
        {
            while(book[i]){
                bool fg = false;
                for(int j = 1; j <= ans; j++){
                    if(day[j] < i){
                        book[i]--;
                        day[j]++;
                        fg = true;
                        break;
                    }
                }
                if(!fg){
                    ans++;
                    day[ans]++;
                    book[i]--;
                }
            }
        }
        printf("%d
", ans);
    }

    return 0;
}