1117. Eddington Number(25)

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 (<=N).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N(<=10⁵), 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

找出一个E，有E天，行程超过E miles，先从小到大排序，从后往前看，如果当前看了s个数，而最后一个数又比s大，那么s满足条件，找出最大的s。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int s[100005];
int n,m;
int main()
{
    cin>>n;
    for(int i = 0;i < n;i ++)
    {
        cin>>s[i];
    }
    sort(s,s + n);
    for(int i = n - 1;i >= 0;i --)
    {
        if(s[i] > n - i && m < n - i)m = n - i;
    }
    cout<<m;
}