HDU 2088 Box of Bricks

http://acm.hdu.edu.cn/showproblem.php?pid=2088

 

Box of Bricks

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5277    Accepted Submission(s): 1838

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

6 5 2 4 1 7 5 0

 

 

Sample Output

5

 

 

Author

qianneng

 

 

Source

冬练三九之二

 

 

Recommend

lcy

 

分析：简单题，轻轻地AC之。。

代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
int abs(int x)
{
  return x>0?x:-x;
}
int main()
{
      int n,i,hei,ans;
      int w[51];
      int count=0;
      while(scanf("%d",&n),n)
      {
        if(count!=0) printf("\n");
        ans=hei=0;
        for(i=0;i<n;i++)
          {
            scanf("%d",&w[i]);
            hei+=w[i];
          }
          hei=hei/n;
              for(i=0;i<n;i++)
                ans+=abs(hei-w[i]);
          printf("%d\n",ans/2);
          count++;
      }
  return 0;
}