Who's in the Middle - POJ 2388 水题

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本文介绍了如何通过排序找到一组奇数数量奶牛中乳制品产量的中位数，即一半奶牛产量大于或等于中位数，另一半则不大于或等于中位数。

Who's in the Middle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30908		Accepted: 17934

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

Sample Output

题意+思路：排序找出中间的数。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int num[10010];
int main()
{ int n,i,j,k;
  scanf("%d",&n);
  for(i=1;i<=n;i++)
   scanf("%d",&num[i]);
  sort(num+1,num+1+n);
  printf("%d\n",num[(n+1)/2]);
}