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题意：
找出一个最大的数字E，使其满足有E个大于E的数字。
思路：
将数组从大到小进行排列，随着index的增加E的值在减小，当index与E相等的时候就是所要找的最大值。 将样例进行排序可得10， 9， 8，8， 7， 7， 6， 6， 3， 2.
大于等于v[i] 的有i+1个， 即大于v[i] - 1的有i + 1个，即大于 v[i] 的有 i + 2个。根据题意可知要使超过的英里数大于等于天数即：v[i] >= i + 2; 也即：v[i] > i + 1。（这部分推导确实有点……）
Code :
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() { 6 int n, t = 0; 7 cin >> n; 8 vector<int> v(n); 9 for (int i = 0; i < n; ++i) cin >> v[i]; 10 sort(v.begin(), v.end(), greater<int>()); 11 while (t < n && v[t] > t + 1) t++; 12 cout << t << endl; 13 return 0; 14 }
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