Box of Bricks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 63   Accepted Submission(s) : 20
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 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? [center][img]../../data/images/1326-1.gif[/img][/center]
 
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.
 
Source
Southwestern Europe 1997

 #include <stdio.h>
#include <stdlib.h> int main()
{
int n,a[],i,sum,sign,j=;
scanf("%d",&n);
while()
{
for(i=,sum=;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
sum/=n;
for(i=,sign=;i<n;i++)
{
if(a[i]>sum)
{
sign+=a[i]-sum;
}
}
printf("Set #%d\nThe minimum number of moves is %d.\n\n",j,sign);
j++;
scanf("%d",&n);
if(n==)
break;
}
return;
}

Box of Bricks的更多相关文章

  1. HDOJ 1326. Box of Bricks 纯水题

    Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  2. Box of Bricks最小移动砖块数目

    Description Little Bob likes playing with his box of bricks. He puts the bricks one upon another and ...

  3. [POJ1477]Box of Bricks

    Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19503   Accepted: 7871 Description Litt ...

  4. HDOJ(HDU) 2088 Box of Bricks(平均值)

    Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one upon ano ...

  5. HDOJ 1326 Box of Bricks(简单题)

    Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one upon ano ...

  6. 591 - Box of Bricks

     Box of Bricks  Little Bob likes playing with his box of bricks. He puts the bricks one upon another ...

  7. HDU 2088 Box of Bricks

    http://acm.hdu.edu.cn/showproblem.php?pid=2088 Problem Description Little Bob likes playing with his ...

  8. HDU 2088 Box of Bricks(脑洞)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=2088 Box of Bricks Time Limit: 1000/1000 MS (Java/Oth ...

  9. zoj 1251 Box of Bricks

    Box of Bricks Time Limit: 2 Seconds      Memory Limit: 65536 KB Little Bob likes playing with his bo ...

随机推荐

  1. 基于TCP协议的网络编程

    TCP通信协议是一种可靠的传输层协议,它在通信的两端各建立一个Socket,从而在通信的两端之间形成虚拟网络链路.一旦建立了虚拟的网络链路,两端的程序就可以通过虚拟链路进行通信.Java使用Socke ...

  2. sql 将某列转换成一个字符串 for xml path用法

    declare @test table( name varchar(10)) insert into @test values('a') insert into @test values('b') i ...

  3. Maven+STS工程中Maven Dependencies 文件夹丢失问题

    在我们使用Maven+sts工程中偶尔会出现这种情况: Maven Dependencies文件夹在新打开的工程中丢失,造成 web project 自动编译出错,缺少必要的库文件: 如下图所示的情况 ...

  4. MVC创建

    [1]创建ASP.NET_MVC应用程序   1>新建项目>web>Visual Studio 2012>ASP.NET MVC4 Web应用程序   2>填写名称,位置 ...

  5. python之~ 序列化与反序列化

    sy1.proto文件 syntax = "proto2"; message stuff { required int32 stuff_ID = ; required ; opti ...

  6. ios打开系统自带APP

    打开系统自带app 打开系统设置: [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root= ...

  7. String类之endsWith方法--->检测该字符串以xx为结尾

    endsWith(XX)方法是java内置类String类的一个内置方法,我们直接拿来用即可了,下边是api说明:检测该字符串以xx为结尾,结果返回布尔值 public class Demo { pu ...

  8. Spring Security(10)——退出登录logout

    要实现退出登录的功能我们需要在http元素下定义logout元素,这样Spring Security将自动为我们添加用于处理退出登录的过滤器LogoutFilter到FilterChain.当我们指定 ...

  9. 【Sort】Merge Sort归并排序

    归并排序运行时间O(N log N),但是由于需要线性附加内存,所以很少用于主存排序. 算法核心在于以下三条语句,分治递归,分别对左半边和右半边的数组进行排序,然后把左右半边的数组一一进行比较放入数组 ...

  10. bzoj1336: [Balkan2002]Alien最小圆覆盖

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1336 1336: [Balkan2002]Alien最小圆覆盖 Time Limit: 1 ...