PAT甲级——A1117 Eddington Number【25】

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

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 using namespace std;
 5 int n;
 6 int main()
 7 {
 8     cin >> n;
 9     vector<int>v(n);
10     for (int i = 0; i < n; ++i)
11         cin >> v[i];
12     sort(v.begin(), v.end());
13     int k = 0;
14     for (k = 0; k < n; ++k)
15         if (v[k] > n - k)
16             break;
17     cout << n - k;
18     return 0;
19 }

 

posted @ 2019-09-04 20:46  自由之翼Az  阅读(156)  评论(0编辑  收藏  举报