hdu 1157 Who's in the Middle

Who's in the Middle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9594    Accepted Submission(s): 4614

Problem 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.

题目大意：给一个数n表示有n头奶牛，然后下边的n行有奶牛的产奶量，求出产奶量中等的是多少 思路：快排，输出中间数 真不知道题目要干嘛，没说n的奇偶有什么影响，也没有产奶量一样怎么办，想多了，原来这么水  2014,10,28 

#include
#include
using namespace std;
int main(){
	int n,a[11000],i;
	while(scanf("%d",&n)!=EOF){
		for(i=0;i