1117 Eddington Number

1117 Eddington Number
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
分析：
给定一个人n天每天的跑步长度，现在要求找到最大的N,即有N天跑步长度超过N。
解决方法：
从小到大排序，从后往前查找，记录当前跑步的最大长度减一的值meter,并且记录个数cnt,取min(meter,cnt)为N赋值，如果meter<N循环终止。
参考代码：

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
	int n, temp, maxn = 0, cnt = 0;
	vector<int>v;
	scanf_s("%d", &n);              //使用n之前，先要初始化n
	for (int i = 1; i <= n; i++) {
		scanf_s("%d", &temp);
		v.push_back(temp);
	}
	sort(v.begin(), v.end());
	for (auto it = v.rbegin(); it != v.rend(); it++) {
		temp = *it - 1;
    if (temp < maxn)
			break;
		cnt++;
		if (min(cnt, temp) > maxn)
			maxn = min(cnt, temp);
	}
		cout << maxn;
      return 0;
}