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 }