Uva 591 Box of Bricks

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 assumeand .

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.

---

Miguel A. Revilla 1998-03-10

 

#include<stdio.h>
#include<string.h>

int main()
{
    int i, n, stack[52], average, ans, cnt = 0;
    while(scanf("%d", &n) != EOF && n)
    {
        for(i=average=0; i<n; ++i)
        {
            scanf("%d", &stack[i]);
            average += stack[i];
        }
        average = average/n;
        ans = 0;
        for(i=0; i<n; ++i)
            if(stack[i]>average) ans += stack[i] - average;
        printf("Set #%d\n", ++cnt);
        printf("The minimum number of moves is %d.\n\n", ans);
        
    }
    return 0;
}

解题思路：
这题无比的简单，我就想不明白Uva上那么低的通过率，题目的意思是说有几组“良莠不齐”砌成淑状的砖块，总数是能够被组数整除的，所有每一组都能分成平等的砖块数（只要你想）而现在题目就要问你是每一组的砖块数相同至少需要移动多少块砖

而只要知道平均每组砖的数量然后再遍历所有组的砖数，然后将突出的砖“削平”并累加就行了