1117. Eddington Number(25)

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本文介绍了一种计算Eddington数的方法，这是一种衡量骑行者骑行能力的指标，通过输入连续N天的骑行距离来计算相应的Eddington数。

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1117. Eddington Number(25)
时间限制 250 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>

using namespace std;
const int MaxN = 100010;
int main() {
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG
    std::ios::sync_with_stdio(false);

    int n, data[MaxN]; cin >> n;
    for (int i = 1; i <= n; ++i)cin >> data[i];

    int ans = 0;
    sort(data + 1, data + n + 1, [](int a, int b) {return a > b; });
    for (int i = 1; i <= n && data[i] > i; ++i)++ans;
    cout << ans;
    return 0;
}