zoj 1251
来源:互联网 发布:五冠淘宝店铺年收入 编辑:程序博客网 时间:2024/04/30 20:52
Box of Bricks
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.
代码:
#include <stdio.h>int main(){int n,i,avg,sum=0,num=0,count=1;int h[51];while(1){scanf("%d",&n);if (!n){break;}for(i=0;i<n;i++){scanf("%d",&h[i]);}for(i=0;i<n;i++){sum=sum+h[i];}avg=sum/n;for(i=0;i<n;i++){if (h[i]>avg){num=num+h[i]-avg;}}printf("Set #%d\nThe minimum number of moves is %d.\n\n",count++,num);sum=0;num=0;}return 0;}
- zoj 1251
- ZOJ-1251
- ZOJ 1251
- ZOJ 1242 1251
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- ZOJ
- eclipse远程调试java程序简单示例
- android程序启动画面之Splash总结
- Delphi 编译指令的使用方法
- POJ2375 Cow Ski Area——求强连通分量
- 关于Android的nodpi,xhdpi,hdpi,mdpi,ldpi
- zoj 1251
- Eclipse快捷键
- javaScript中URL编码转换
- 管理自己的time
- 将Array、Dictionary等集合类的序列化和反序列化
- 提高编程技巧的十大方法
- JS 获取上传文件大小的方法
- IT人员迅速提升自我效率的十大方法
- JDBC Class.forName(); 原代码讲解