1.2.2 Box of Bricks

Problem Description

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 consideration, 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 hi of the n stacks. You may assume 1≤n≤50 and 1≤hi≤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, print the minimum number of bricks that have to be moved in order to make all the stacks the same height.
Output a blank line between each set.

 

Sample Input

65 2 4 1 7 50

 

Sample Output

Set #1The minimum number of moves is 5.

#include "stdio.h"#include "stdlib.h"int main(){int stack,n=0;while(scanf("%d",&stack)!=EOF&&stack!=0){int sum=0;int a[stack+1];int b[stack];a[0]=stack;while(stack){scanf("%d",&a[stack]);sum=sum+a[stack];stack--;}stack=a[0];a[0]=sum/stack;sum=0;int i,j;for(i=0;i<stack;i++){j=a[0]-a[i+1];b[i]=abs(j);sum=sum+b[i];}n++;printf("Set #%d\nThe minimum number of moves is %d.\n\n",n,sum/2);}return 0;}/*取绝对值：#include "math.h" fabs函数是计算浮点数的，他返回一个浮点数，你用%d输出当然不对。如果只是计算整数的话，abs 就行了#include<math.h>int 型int abs(int x);long 型long labs(int x);浮点数 float doubledouble fabs(double x);*/