Uva 591 Box of Bricks

最新推荐文章于 2017-03-17 17:53:19 发布

HelloWorld10086

最新推荐文章于 2017-03-17 17:53:19 发布

阅读量868

点赞数

CC 4.0 BY-SA版权

分类专栏：数学文章标签： Uva Box of Bricks

本文链接：https://blog.youkuaiyun.com/helloworld10086/article/details/38032411

数学专栏收录该内容

28 篇文章

订阅专栏

本文提供了一种算法，帮助Bob通过最小的积木移动次数，将不同高度的积木堆调整为相同的高度。通过计算平均高度并累加超出平均高度的积木数量来确定最少的移动次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Box of Bricks

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and buildsstacks of different height. ``Look, I've built a wall!'', he tells his older sister Alice. ``Nah, you shouldmake 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 allstacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimumnumber 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 stacksBob has built. The next line contains n numbers, the heights h_i of the n stacks. You may assume $1 \le n \le 50$ and $1 \le h_i \le 100$ .

The total number of bricks will be divisible by the number of stacks. Thus, it is always possibleto 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 isk.'', where k is the minimum number of bricks thathave 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.

题目大意：

给你用积木摞起来的不同高度的柱子，问弄成一样高时，最少的移动次数。

分析：简单题。求出平均数，把超过平均数的加和即可。

#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;
const int N = 10000;
int num[N];
int main() {
	int n,t;
	t=1;
	while (cin >> n) {
		if(n == 0)
			break;
		
		int sum = 0;
		int avg;
		
		for(int i=0;i<n;i++) {
			cin >> num[i];
			sum += num[i];	
		}
		avg = sum/n;
		
		int mov=0;
		for(int i=0;i<n;i++) {
			if(num[i]>avg) {
				mov += (num[i]-avg);	
			}	
		}
		cout<<"Set #"<<t++<<endl;
		cout<<"The minimum number of moves is "<<mov<<"."<<endl<<endl; 
	}
	return 0;
}