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 (≤).
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤), 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
艾丁顿数,求大于N的骑车天数大于N
#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { int N, e = 0; scanf("%d", &N); vector<int> v(N); for(int i = 0; i < N; i++) scanf("%d", &v[i]); sort(v.begin(), v.end(), greater<int>()); while(e < N && v[e] > e+1) e++; printf("%d", e); return 0; }