1117 Eddington Number(25)

Posted 2020-11-14 mr-stn

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 }