文章标题 HDU 1157 ： Who's in the Middle（水）

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本文介绍了一个简单的算法问题——寻找一组牛奶产量数据的中位数。通过排序数组并选取中间元素的方式，快速找到使得一半数量的奶牛产量不低于该值且另一半不高于该值的数据点。

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Who’s in the Middle

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.
题意：水题，找中位数，直接排一下序，取中间的数就行了
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n;
int output[10005];
int main ()
{
    while (scanf ("%d",&n)!=EOF){
        for (int i=0;i<n;i++){
            scanf ("%d",&output[i]);
        }
        sort(output,output+n);
        printf ("%d\n",output[n/2]);
    }
    return 0;
}