Who‘s in the Middle

最新推荐文章于 2024-07-15 17:53:06 发布

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这篇博客介绍了如何在给定一组牛奶产量的奇数牛群中找到中位数，确保至少一半的牛产量相同或更高，另一半相同或更低。算法演示了如何通过冒泡排序找出第(n+1)/2个牛的产奶量作为中位数。

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
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.

答案如下

#include<cstdio>
int main()
{
	int n;
	int an[10005]={0};
	scanf("%d",&n);
	for(int i=0;i<n;i++)
    scanf("%d",&an[i]);
    for(int i=0;i<n-1;i++)
    {
    	for(int j=0;j<n-i-1;j++)
    	{
    		if(an[j]>an[j+1])
    		{
    			int temp=an[j];
    			an[j]=an[j+1];
    			an[j+1]=temp;
			}
		}
	}
	printf("%d",an[(n+1)/2-1]);
	return 0;
}