Balancer - CodeForces 440B



Description

Petya has k matches, placed in n matchboxes lying in a line from left to right. We know that k is divisible by n. Petya wants all boxes to have the same number of matches inside. For that, he can move a match from its box to the adjacent one in one move. How many such moves does he need to achieve the desired configuration?

Input

The first line contains integer n (1 ≤ n ≤ 50000). The second line contains n non-negative numbers that do not exceed 10⁹, the i-th written number is the number of matches in the i-th matchbox. It is guaranteed that the total number of matches is divisible by n.

Output

Print the total minimum number of moves.

Sample Input

Input

6
1 6 2 5 3 7

Output

12

题意：一步步的移火柴使火柴盒里的火柴数相等所需要的最短步数。

思路：第一个不够的都从第二个拿，第二个不够的都从第三个拿，出现负数没关系，可以先后面拿过来。

AC代码：

import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		long m[]=new long[n];
		long sum=0;
		for(int i=0;i<n;i++){
			m[i]=scan.nextLong();
			sum+=m[i];
		}
		
		long av=sum/n;
		for(int i=0;i<n;i++){
			m[i]-=av;
		}
		long ans=Math.abs(m[0]);
		for(int i=1;i<n;i++){
			m[i]+=m[i-1];
			ans+=Math.abs(m[i]);
		}
		System.out.println(ans);
	}

}