HDU2088JAVA2

Box of Bricks

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14836    Accepted Submission(s):
4912

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.

 

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.

 

Sample Input

6

5 2 4 1 7 5

0

 

Sample Output

5

import java.util.Scanner;
public class Main2088 {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
int flag=0;
while(cin.hasNext()){
int n=cin.nextInt();
if(n==0)
break;
if(flag==1)
System.out.println();
int []a=new int [n];
int sum=0;
for(int i=0;i<n;i++){
a[i]=cin.nextInt();
sum+=a[i];
}
sum=sum/n;
int s=0;
for(int i=0;i<n;i++){
if(a[i]>sum){
s+=(a[i]-sum);
}
}
System.out.println(s);
flag=1;

}
}

}

这是一个水题，注意一下格式的输出就行。