1117 Eddington Number

British astronomer Eddington liked to ride a bike. It is said that in order to show off his skill, he has even defined an "Eddington number", E -- that is, the maximum integer E such that it is for E days that one rides more than E miles. Eddington's own E was 87.

Now given everyday's distances that one rides for N days, you are supposed to find the corresponding E (≤).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances of everyday.

Output Specification:

For each case, print in a line the Eddington number for these N days.

Sample Input:

10
6 7 6 9 3 10 8 2 7 8

Sample Output:

6

#include<stdio.h>
#include<algorithm>

using namespace std;

int seq[100005];
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int num;
    scanf("%d",&num);
    for(int i=0;i<num;i++)
    {
        scanf("%d",&seq[i]);
    }
    sort(seq,seq+num,cmp);
    int out=0;
    for(int i=0;i<num;i++)   //必须从0 这时能判断第一个是否满足 即out=1才能判断 
    {                                //否则out则是从0  到了2  中间1时是否正确无法判断 
        if(seq[i]>i+1)
        {
            out=i+1;   //每次判断下一位的情况  若下一位不满足(不能等号 因为是more than ) 直接break  防止有重复数的判断错误 
        }else
        {
            break;
        }
        
    }
    
    printf("%d",out);
    
 }