POJ 1477:Box of Bricks
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 19949 | Accepted: 8029 |
Description
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 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
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.
Source
Southwestern European Regional Contest 1997
你 离 开 了 , 我 的 世 界 里 只 剩 下 雨 。 。 。
#include <stdio.h>
int main()
{
int n,i,s,aver,a[51],num,j=0;
while(~scanf("%d", &n))
{
if(n == 0) break;
j++,s = 0,num = 0;
for(i = 0; i < n; i++)
{
scanf("%d", &a[i]);
s+=a[i];
}
aver= s/n;
for(i = 0; i < n; i++)
if(a[i] > aver)
num = num + a[i] - aver;
printf("Set #%d\nThe minimum number of moves is %d.\n\n", j, num);
}
return 0;
}
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