zoukankan      html  css  js  c++  java
  • Box of Bricks

    Box of Bricks

    Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1450 Accepted Submission(s): 335
    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
    Set #1
    The minimum number of moves is 5.
    • Code Render Status : Rendered By HDOJ G++ Code Rander Version 0.01 Beta
    #include <iostream>
    #include "string.h"
    using namespace std;
    int main()
    {
      int set=0,num,hi[101],total,mv;
        while( cin>>num&&num!=0)
        {
            mv = 0 ;
            total = 0 ;
            memset(hi,0,sizeof(hi));
    
             for(int i = 0 ; i < num ; i++)
          {
              cin>>hi[i];
              total +=hi[i];
          }
          int k = total/num;
          for(int i = 0 ; i < num ; i++)
          {
              if(hi[i]>k) mv += (hi[i]-k);
          }
          set++;
          cout<<"Set #"<<set<<endl;
          cout<<"The minimum number of moves is "<<mv<<"."<<endl<<endl;
        }
          return 0;
    }
  • 相关阅读:
    转!!javaMail使用网易163邮箱报535 Error: authentication failed
    银行卡验证(验证是否存在,卡号类型,归属行)
    Navicat已经成功连接,密码忘记的解决方法
    Inline&IAT Hook原理
    x64dbg尝鲜
    C# 通过Dynamic访问System.Text.Json对象
    dotnet5将asp.net webapi宿主到wpf
    Asp.Net5 MVC with Vue.js
    在 Visual Studio 中使用跟踪点将信息记录到“输出”窗口中
    WPF带阴影的无边框窗体
  • 原文地址:https://www.cnblogs.com/newpanderking/p/2114932.html
Copyright © 2011-2022 走看看