Who\\\'s in the Middle HDU - 1157

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.

#include<stdio.h>
void sort(int p[],int n){
	int i,j,temp;//排序函数，如果是c++的话可以用函数库
	for(i=0;i<n;i++){
		for(j=0;j<n-i-1;j++){
			if(p[j+1]<p[j]){
				temp = p[j+1];
				p[j+1] = p[j];
				p[j] = temp;
			}
		}
	}
}
int main()
{
	int n;
	//记住是输入多组数据，之间一直没注意
	while(scanf(\"%d\",&n)!=EOF){
	int p[10000];//也可以用malloc去申请空间然后记得free就好了
	int i;
	for(i=0;i<n;i++){
		scanf(\"%d\",&p[i]);
	}
	sort(p,n);//排序
		printf(\"%d\\n\",p[n/2]);//不用加1或者减1的原因是数组的第一个是0，a[0]
	}
	return 0;
 }