ZOJ_1251_Box of Bricks

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ACM解题报告（简单题）同时被 2 个专栏收录

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本文介绍了一道关于砖块堆叠的问题，旨在寻找使所有砖堆高度相等所需的最小移动次数。通过计算各堆砖块的高度并与平均高度比较，确定需要调整的砖块数量。

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Box of Bricks

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 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, first print the number of the set, as shown in the sample output. Then print the line ``The minimum number of moves is k.'', 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

6
5 2 4 1 7 5
0

Sample Output

Set #1
The minimum number of moves is 5.

题意和思路：这个题目英文挺好懂一看完首先知道肯定是要跟平均值比，平均值就是高度，还有就是只有比平均值高的才需要动，v[i]-avg的和实际上就是移动的次数

Sample Program Here

/******************************************
**
**      Author: Wan KaiMing
**      Date:   2012-11-04-21.24 星期日
**
*******************************************/


#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){
   int n=0; //砖块数目
   int count=0; //测试例
   while(cin>>n){

      if(n==0) continue;
      vector<int> v;
      int i=0;
      int sum=0; //求和
      int k=n;  //n后面有用，所以定义个k来保存
      while(k--){   //初始化n个数
        cin>>i;
        v.push_back(i);
        sum+=i;
      }

      int avg=sum/n; //要弄成墙每列需要的砖块数（平均值）
      int moves=0;// 记录移动次数
      for(i=0;i<n;i++){
        if(v[i]>avg)
        moves+=v[i]-avg;
      }

      cout<<"Set #"<<++count<<endl;
      cout<<"The minimum number of moves is "<<moves<<'.'<<endl<<endl;
   }

   return 0;
}