poj 2388 Quick_sort 求中间值

最新推荐文章于 2021-08-26 23:21:51 发布

翻译最新推荐文章于 2021-08-26 23:21:51 发布 · 358 阅读

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#快速排序

本文介绍了一道经典的算法题目，即寻找一群奶牛中产奶量处于中位数的那头奶牛。通过给出的样例代码，展示了如何使用快速排序算法来找到中位数，确保至少一半的奶牛产奶量不低于此数值。

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Who's in the Middle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42058		Accepted: 24335

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

//下面是代码，题意就是排好序后求中间的那个数
#include<stdio.h>
#define MAX 10005
int a[MAX];
void quick_sort(int s[], int l, int r)
{
	if (l < r)
	{
		int i = l, j = r, x = s[l];
		while (i < j)
		{
			while(i < j && s[j] >= x) // 从右向左找第一个小于x的数
				j--;  
			if(i < j) 
				s[i++] = s[j];

			while(i < j && s[i] < x) // 从左向右找第一个大于等于x的数
				i++;  
			if(i < j) 
				s[j--] = s[i];
		}
		s[i] = x;
		quick_sort(s, l, i - 1); // 递归调用 
		quick_sort(s, i + 1, r);
	}
}
int main()
{
int T;
scanf("%d",&T);
for(int i=0;i<T;i++)
{
scanf("%d",&a[i]); 
}
quick_sort(a,0,T-1); 
printf("%d\n",a[T/2]);
	return 0;
}