UVA 591 - Box of Bricks（数学）

砖块堆叠最小移动数计算

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本博客探讨如何计算将不同高度的砖块堆叠调整为相同高度所需的最少移动次数，通过输入堆叠数量和各堆高度，输出调整所需的具体步骤。

Box of Bricks

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. ``Look, I've built a wall!'', he tells his older sister Alice. ``Nah, you should make all stacks the same height. Then you would have a real wall.'', she retorts. After a little con- sideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?

Input

The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line containsn numbers, the heights h_i of the n stacks. You may assume $1 Ÿ\le n \leŸ 50$ and $1 \leŸ h_i Ÿ\le 100$ .

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height.

The input is terminated by a set starting with n = 0. This set should not be processed.

Output

For each set, first print the number of the set, as shown in the sample output. Then print the line ``The minimum number of moves isk.'', where k is the minimum number of bricks that have to be moved in order to make all the stacks the same height.

Output a blank line after each set.

Sample Input

6
5 2 4 1 7 5
0

Sample Output

Set #1
The minimum number of moves is 5.

Miguel A. Revilla
1998-03-10

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问需要的最少的移动步数，求出平均值，多出平均值的部分加在一起就好了

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int n,a[55],t=1;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        printf("Set #%d\n",t++);
        int sum=0,ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        int ave=sum/n;
        for(int i=0;i<n;i++)
        {
            if(a[i]>ave)
            {
                ans+=(a[i]-ave);
            }
        }
        printf("The minimum number of moves is %d.\n\n",ans);
    }
    return 0;
}