杭电1326-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?



 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

6
5 2 4 1 7 5
0英文题目，所以必须给解释：题目其实很简单就是给你一些高低不平的盒子，怎样才能使他们一样高用最少的移动次数这里有个诀窍：就是不管怎样移动：移动的盒子个数是固定的，所以最少的移动次数也是固定的即：每个位置的盒子与摆放好的平均盒子个数之差的绝对值的和再除以2就是所需最少移动次数下面看AC代码：#include<iostream>
#include<cmath>
const int MAX=300;
int x[MAX];
using namespace std;
int main()
{
int n,m,sum,p=1;
while(cin>>n&&n)
{
m=0;
sum=0;
for(int i=0;i<n;i++)
{
cin>>x[i];
sum+=x[i];
}
sum/=n;
for(int i=0;i<n;i++)
{

m+=abs(x[i]-sum);
}
cout<<"Set #"<<p++<<endl;
cout<<"The minimum number of moves is "<<m/2<<"."<<endl;
cout<<endl;
}
return 0;
}