1117 Eddington Number (25 分)

1117 Eddington Number (25 分)

 

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

说实话题意没咋看懂，代码猜的一发就过了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int an[200000], n;
 4 bool cmp(int a, int b){
 5     return a > b;
 6 }
 7 int main(){
 8     cin >> n;
 9     for(int i = 1; i <= n ; i ++){
10         cin >> an[i];
11     }
12     sort(an+1, an+1+n, cmp);
13     for(int i = 1; i <= n; i++){
14         if(an[i] <= i){
15             cout << i-1 <<endl;
16             return 0;
17         }
18     }
19     cout << n << endl;
20     return 0;
21 }