G - Who's in the Middle

        这个题主要是找中位数,因为题目给的是奇数个数据,所以按照从小到大排序后找第(n-1)/2个元素即可。值得注意的一点是,使用<algorithm>的sort函数,结束的元素是个开区间,也就是说要加1,这里被坑了一下,AC的代码非常精炼。


Description

FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. 

Given an odd number of cows N (1 <= N < 10,000) and their milk output (1..1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less. 
 

Input

* Line 1: A single integer N 

* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow. 
 

Output

* Line 1: A single integer that is the median milk output. 
 

Sample Input

5
2
4
1
3
5
 

Sample Output

3

Hint

 INPUT DETAILS: Five cows with milk outputs of 1..5 OUTPUT DETAILS: 1 and 2 are below 3; 4 and 5 are above 3.


下面是我的代码:


#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=10000;
int cows[maxn];
int n;
int main()
{
    while(cin>>n)
    {
        for(int i=0;i<n;i++)cin>>cows[i];
        sort(cows, cows+n);
        cout<<cows[(n-1)/2]<<endl;
    }
    return 0;
}