591 - Box of Bricks

最新推荐文章于 2020-02-11 10:41:28 发布

原创最新推荐文章于 2020-02-11 10:41:28 发布 · 936 阅读

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本文介绍了一个有趣的编程问题：如何将不同高度的砖堆调整到相同的高度，且移动的砖块数量最少。通过算法分析和代码实现，展示了问题的解决过程。

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 contains n 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 is k.'', 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.

这道题很水，就是说把n个不同高度的砖块排成同样的高度。问最少要移动多少砖块。

wa了一次，大家要注意UVa对输出的格式判的很严格，我因为多输出了一个回车就wa了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[60];
int main ()
{
    int n,i,t=1,ans=0,sum=0;
    while(cin>>n)
    {
        if (n==0) break;
        for (i=1; i<=n; i++)
        {
            scanf("%d",a+i);
            sum+=a[i];
        }
        int ave=sum/n;
        for (i=1; i<=n; i++)
            if (a[i]>ave)
                ans+=a[i]-ave;
        printf("Set #%d\n",t);
        printf("The minimum number of moves is %d.\n\n",ans);
        t++;
        ans=0;
        sum=0;
    }
    return 0;
}