UVA 591 ---Box of Bricks 模拟枚举

Box of Bricks

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description

  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  and .

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 

65 2 4 1 7 50

Sample Output 

Set #1The minimum number of moves is 5.

---

Miguel A. Revilla
1998-03-10

#include <stdio.h>#include <string.h>#include <math.h>int main(){    int s[55],sum,n,c = 0;    while(scanf("%d",&n),n)    {        sum = 0;        for(int i = 0;i < n;i++)        {            scanf("%d",&s[i]);            sum += s[i];        }        int ave = sum / n;        int cnt = 0;        for(int i = 0;i < n;i++)        {            cnt += fabs(ave - s[i]);//找平均数和每摞砖的差距，求模然后相加，这便是移动的砖块数的两倍，那第一摞一个放到第二摞其实是算一个，我这里都加进去了        }        printf("Set #%d\n",++c);        printf("The minimum number of moves is %d.\n",cnt / 2);//最后要除以2        printf("\n");    }    return 0;}