Who's in the Middle

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本文为博主胡乱瞎写，如有需要，随便转载，顺带给个momo哒。

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本文介绍了一道经典的算法题目，旨在找到一组奶牛中产奶量处于中间位置的那只奶牛。通过快速排序算法对奶牛的产奶量进行排序，并找出中位数作为最平均的奶牛。

Who's in the Middle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23050 Accepted Submission(s): 9789

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

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.

注：这道题需要多组输入，虽然从题面上没有看出来

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int a[10010],n;
void quicksort(int left,int right)//快排
{
    int i,j,t,temp;
    if(left>right)
        return;
    temp=a[left];
    i=left;
    j=right;
    while(i!=j)
    {
        while(a[j]>=temp&&j>i)
            j--;
        while(a[i]<=temp&&i<j)
            i++;
        if(i<j)
        {
            t=a[i];
            a[i]=a[j];
            a[j]=t;
        }

    }
    a[left]=a[i];
    a[i]=temp;

    quicksort(left,i-1);
    quicksort(i+1,right);
}



int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        quicksort(1,n);
        printf("%d\n",a[(n+1)/2]);
    }
    return 0;
}