1117 Eddington Number (25 分)

1117 Eddington Number (25 分)

 

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

说实话题意没咋看懂，代码猜的一发就过了。

#include <bits/stdc++.h>
using namespace std;
int an[200000], n;
bool cmp(int a, int b){
    return a > b;
}
int main(){
    cin >> n;
    for(int i = 1; i <= n ; i ++){
        cin >> an[i];
    }
    sort(an+1, an+1+n, cmp);
    for(int i = 1; i <= n; i++){
        if(an[i] <= i){
            cout << i-1 <<endl;
            return 0;
        }
    }
    cout << n << endl;
    return 0;
}