qau-国庆七天乐——A

A - Who's in the Middle

//先吐槽一波vj，不知道怎么回事，bits/stdc++.h不能用了。。。坑爹

7225007	upstar	A	Compile Error			C++	335	2016-10-11 19:51:41
7224999	upstar	A	Compile Error			GCC	335	2016-10-11 19:51:15
7224991	upstar	A	Compile Error			G++	335	2016-10-11 19:49:50

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 2388

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:

#include<iostream>
#include <algorithm>
using namespace std;
#define M 10500
int main()
{
    int n,i;
    int st[M];
    while(cin>>n){
        for(i=;i<n;i++){
            cin>>st[i];
        }
        sort(st,st+n);
        if(n&)
        cout<<st[n/]<<endl;
        else
            cout<<(st[n/]+st[n/+])/<<endl;
    }
    return ;
}