# 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 (≤N).

## Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤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
``````

## Code:

``````#include <iostream>
#include <vector>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std;
int main()
{
int n;
scanf("%d", &n);
vector<int> vec(n);
for (int i = 0; i < n; i++)
scanf("%d", &vec[i]);
sort(vec.begin(), vec.end(), [](const int n1, const int n2) {
return n1 > n2;
});
int ans = 0;
while (ans < n && vec[ans] > ans+1) ans++;
printf("%d\n", ans);
return 0;
}
``````

## 思路：

https://www.liuchuo.net/archives/2478

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