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1117 Eddington Number

Statement

Metadata

  • 作者: CHEN, Yue
  • 单位: 浙江大学
  • 代码长度限制: 16 KB
  • 时间限制: 250 ms
  • 内存限制: 64 MB

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 (\le N).

Input Specification

Each input file contains one test case. For each case, the first line gives a positive integer N (\le 10^5), 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

Solution

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, a[N];

int main() {
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        sort(a + 1, a + 1 + n, [&](int x, int y) {
            return x > y;
        });
        int res = 0;
        for (int i = 1; i <= n; ++i) {
            if (a[i] > i) {
                res = i;
            }
        }
        printf("%d\n", res);
    }
    return 0;
}
Last update: May 4, 2022
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