HDU 1050 move tables

最新推荐文章于 2024-07-26 09:15:00 发布

最新推荐文章于 2024-07-26 09:15:00 发布 · 73 阅读

本文介绍了一个通过统计方法找到一群奶牛中“最平均”奶产量的问题解决方案。任务是找出一个牛奶产出量，使得一半的奶牛产奶量不低于这个数值，而另一半则不高于该数值。文中提供了一段C语言实现的示例代码。

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

Sample Output

有人用贪心做，但是我觉得用统计更简单，详见

代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int NumOfTest,pair;
    int i,Max;
    int a[201];
    int b,c,n1,n2;
    scanf("%d",&NumOfTest);
    while (NumOfTest--)
    {
        scanf("%d",&pair);
        memset(a,0,sizeof(a));
        while (pair--)
        {
            scanf("%d%d",&b,&c);
            if(b>c){b^=c,c^=b,b^=c;}
            if(b%2==1)
                n1 = (b+1)>>1;
            else
                n1 = b/2;
            if(c%2==1)
                n2 = (c+1)>>1;
            else
                n2 = c/2;
            if(n1!=n2)
                for(i=n1;i<=n2;i++)
                    a[i]++;
            else
                a[n1]++;
        }
        Max = -1;
        for(i=1;i<=200;i++)
            if(Max<a[i])
                Max = a[i];
        printf("%d\n",10*Max);
    }
    return 0;
}