HDU2088 Box of Bricks
来源:互联网 发布:客户满意率的数据来源 编辑:程序博客网 时间:2024/05/29 14:07
Box of Bricks
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15483 Accepted Submission(s): 5131
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.
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.
Output a blank line between each set.
Sample Input
65 2 4 1 7 50
Sample Output
5
输出格式太坑了。
#include <stdio.h>int a[55];int main(){int n,flag = 0;while(~scanf("%d",&n) && n){int i,aver = 0,ans = 0;for(i = 0; i < n; i ++){scanf("%d",&a[i]);aver += a[i];}if(flag)putchar('\n');flag = 1;aver /= n;for(i = 0; i < n; i ++){if(a[i] < aver){ans += (aver - a[i]);}}printf("%d\n",ans);}return 0;}
0 0
- hdu2088 Box of Bricks
- HDU2088 Box of Bricks
- HDU2088 Box of Bricks【水题】
- hdu2088 Box of Bricks(C语言)
- Box of Bricks
- 1477 Box of Bricks
- 1477:Box of Bricks
- 1477 Box of Bricks
- 591 - Box of Bricks
- HDOJ1326 Box of Bricks
- HDU1326:Box of Bricks
- 591 - Box of Bricks
- 591 - Box of Bricks
- UVA591- Box of Bricks
- BNU Box of Bricks
- Box of Bricks
- hdu1326 Box of Bricks
- Box of Bricks
- Mybatis 查询Bean不包含字段的处理
- Redis系列之Redis配置(一)
- IsBackground对线程的重要作用
- 熟悉这几个常用命令,你就是Linux/Unix的vi高手了。
- 1433. 数码问题
- HDU2088 Box of Bricks
- html5学记笔记(三)新的布局
- LeetCode-Binary Tree Level Order Traversal
- Android 侧栏A-Z的快速滑动搜索(一)
- how to remain a TObject after the corresponding TFile is closed
- 工业无线通信网络步入LTE 时代
- ActiveX控件(ATL篇)
- ubuntu 下pip的卸载,安装,更新与使用
- Android ImageView