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.
Source: Southwestern Europe 1997
分析:
大水题。
题意:
输入一串数字,求和,取平均数。小于平均数的数字与平均数的差全部加起来得到的和就是答案。
ac代码:
#include <iostream>
#include<cstdio>
#include<string>
using namespace std;
int a[52];
int main()
{
int n,i;
int c=0;
while(scanf("%d",&n)&&n!=0)
{
c++;
int sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
int average=sum/n;
int num=0;
for(i=0;i<n;i++)
{
if(a[i]<average)
num+=average-a[i];
}
printf("Set #%d\nThe minimum number of moves is %d.\n\n",c,num);
}
return 0;
}
砖墙问题求解
本文介绍了一道关于砖块堆叠的问题,旨在寻找使所有砖堆高度相等所需的最小移动次数。通过计算平均高度并累加低于平均高度的砖堆与平均高度之间的差值来得出解决方案。
1万+

被折叠的 条评论
为什么被折叠?



