HDU 1326 Box of Bricks

http://acm.hdu.edu.cn/showproblem.php?pid=1326

 

Box of Bricks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
Total Submission(s): 2113 Accepted Submission(s):978

Problem Description

Little Bob likes playing with his box of bricks. He puts thebricks 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 wouldhave 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 numberof bricks moved. Can you help?

 

Input

The input consists of several data sets. Each set begins witha line containing the number n of stacks Bob has built. The nextline contains n numbers, the heights hi of the n stacks. You mayassume 1 <= n <= 50 and 1<= hi <= 100.

The total number of bricks will be divisible by the number ofstacks. Thus, it is always possible to rearrange the bricks suchthat all stacks have the same height.

The input is terminated by a set starting with n = 0. This setshould not be processed.

 

Output

For each set, first print the number of the set, as shown inthe sample output. Then print the line ``The minimum number ofmoves is k.'', where k is the minimum number of bricks that have tobe 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 75 0

 

Sample Output

Set #1 Theminimum number of moves is 5.

 

Source

Southwestern Europe 1997

 

Recommend

Ignatius.L

 

题目大意：给一堆积木，每次移动一个积木到另外一个积木上，问最少多少步可以让他们高度一致。

分析：这个之前见过，最少的步数求法为：用每一个积木减去平均值，取绝对值相加之后除以2.

代码如下：

#include<iostream>
#include<math.h> 
using namespacestd;   
int main()
{ 
   int n,ave,sum;
   int s[51];
   int i;
   int count=0;
  while(cin>>n,n)
   {
   ave=sum=0;
   for(i=0;i<n;i++)
 {
  cin>>s[i];
  ave+=s[i];
 }
 ave/=n;
   for(i=0;i<n;i++)
  sum+=abs(s[i]-ave);
 sum/=2;
 cout<<"Set#"<<++count<<endl;
 cout<<"The minimumnumber of moves is"<<sum<<"."<<endl<<endl;//万恶的格式。
   }
    return0; 
}