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
这是做过的最水的25’题了； 1 #include<iostream>
2 #include<vector>
3 #include<algorithm>
4 using namespace std;
5 int main(){
6 int n, i;
7 cin>>n;
8 vector<int> v(n);
9 for(i=0; i<n; i++)cin>>v[i];
10 sort(v.begin(), v.end());
11 bool flag=true;
12 for(i=0; i<n; i++){
13 if(v[i]>n-i){cout<<(n-i); flag=false; break;}
14 }
15 if(flag) cout<<0<<endl;
16 return 0;
17 }
原文地址：https://www.cnblogs.com/mr-stn/p/9160712.html