HDU 1050 move tables

本文介绍了一个关于寻找一群奶牛中最平均产奶量的方法。面对一个包含奇数个元素的集合,文章通过示例解释了如何确定中位数产奶量,确保一半以上的奶牛产奶量不低于此值。

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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

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;
}

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