# 1117. Eddington Number(25)-PAT甲级真题

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

6

#### 题目解释

E表示有E天骑车超过E公里，求E的最大整数

#### 分析

• 对数组a进行排序，骑车公里数大于等于a[i]的天数有i+1天，超过a[i]-1的天数有i+1天
• 限制条件为a[e]-1>=i+1=>a[e]>i+1
#include
#include
#include

using namespace std;

int main() {
int num;
int e = 0;
cin >> num;
vector v(num);
for (int i = 0; i < num; i++) {
cin >> v[i];
}
sort(v.begin(), v.end(), greater());
while (e < num && v[e] > e + 1)
e++;
cout << e << endl;
return 0;
}
`