1411061519-hd-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): 9627    Accepted Submission(s): 4631

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

524135

 

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
       是给定多组数据，求每一组数据的中间值而不是求其平均值。
解题思路
        存数组，快排序，输中值
代码
 

 

#include<stdio.h>#include<algorithm>using namespace std;int s[11000];int main(){int n;int i,j;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)    scanf("%d",&s[i]);sort(s,s+n);printf("%d\n",s[n/2]);}return 0;}