Box of Bricks
2015-08-01 16:13
302 查看
Box of Bricks
Description
![](http://uva.onlinejudge.org/components/com_onlinejudge/images/button_pdf.png)
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?
![](http://7xjob4.com1.z0.glb.clouddn.com/3805abeeb5d1fbbdfa6d13a0c67464d3)
n of stacks Bob has built. The next line contains n numbers, the heights
hi of the n stacks. You may assume
![](http://7xjob4.com1.z0.glb.clouddn.com/fc7403cb0c19cc6dc52e951779c088b8)
and
![](http://7xjob4.com1.z0.glb.clouddn.com/1c0dcd12206666bb16efed77d28fccc6)
.
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 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.
Miguel A. Revilla
1998-03-10
大意:
求出将所有的积木排成相同高度所需的最小搬运数;
要点:#include <stdio.h>
int main(){
int n[55];
int num;
int count = 1;
while(scanf ("%d", &num) && num != 0){
getchar();
int sum, m, move, d;
d = num;
sum = m = move = 0;
while (num--){
scanf ("%d", &n[m]);
getchar();
sum += n[m];
m++;
}
int h = sum / d;
for (int i = 0; i < m; i++){
if (n[i] > h)
move += n[i] - h;
}
printf ("Set #%d\nThe minimum number of moves is %d.\n\n", count, move);
count ++;
}
return 0;
}
每次输出之后跟一个空行,又跪了啊;
代码:
Description
![](http://uva.onlinejudge.org/components/com_onlinejudge/images/button_pdf.png)
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 numbern of stacks Bob has built. The next line contains n numbers, the heights
hi of the n stacks. You may assume
and
.
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.
Miguel A. Revilla
1998-03-10
大意:
求出将所有的积木排成相同高度所需的最小搬运数;
要点:#include <stdio.h>
int main(){
int n[55];
int num;
int count = 1;
while(scanf ("%d", &num) && num != 0){
getchar();
int sum, m, move, d;
d = num;
sum = m = move = 0;
while (num--){
scanf ("%d", &n[m]);
getchar();
sum += n[m];
m++;
}
int h = sum / d;
for (int i = 0; i < m; i++){
if (n[i] > h)
move += n[i] - h;
}
printf ("Set #%d\nThe minimum number of moves is %d.\n\n", count, move);
count ++;
}
return 0;
}
每次输出之后跟一个空行,又跪了啊;
代码:
相关文章推荐
- 二叉树的最大深度
- SPRING配置事务回滚
- Web列表分页--(单页web)容易上手的方法
- 面向互联网的文本信息处理,语音和音乐搜索技术的发展现状【搜集资料时学习所得,未详加整理】
- HTTP请求与数据下载
- 正式开通CSDN博客!
- Date类型数据的获取与转换与单选框按钮数据的获取
- hdoj.5115 Dire Wolf【区间DP】 2015/08/01
- HDU 5335(2015 ACM多校训练第四场1009)
- hdu-4970--Killing Monsters--数组的巧妙
- java技术经理面试题
- C语言编程入门——数组(上)
- Tom and paper
- Servlet实现文件上传,可多文件上传
- nyoj 九的余数
- Scala入门到精通——第十四节 Case Class与模式匹配(一)
- ise错误Cannot access memory * directly;Cannot assign memory * directly
- 最近点对问题(欧几里得算法)
- 静态初始化与构造函数,先执行静态初始化,再执行构造函数,总是先执行父类的,再执行子类的
- PsCreateSystemThread