HDU 1.3.8 Who's in the Middle

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本文介绍了一个通过排序算法快速找到一组数据中位数的方法。具体应用场景为：在一个包含奇数个元素的数据集中找到中位数，使得一半的元素不大于该数值，另一半不小于该数值。使用了C++实现这一算法。

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

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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.

Source
USACO 2004 November

Recommend
mcqsmall

题目有点坑，没说清楚有多组测试数据。。其实就是水题一道。

#include<stdio.h>  
#include"algorithm"  
using namespace std;  
int cowOutput[10000];  
int cmp(int a, int b)  
{  
    return a < b;  
}  
int main()  
{  
    int cowAmount;  
    while (scanf("%d", &cowAmount) != EOF)  
    {  
        for (int i = 0; i < cowAmount; i++)  
            scanf("%d", &cowOutput[i]);  
        sort(cowOutput, cowOutput + cowAmount, cmp);  
        printf("%d\n", cowOutput[cowAmount / 2]);  
    }  
    return 0;  
}