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

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

     return ;
 }