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
6 1 6 2 5 3 7
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);
}
}
Balancer - CodeForces 440B,布布扣,bubuko.com
原文：http://blog.csdn.net/kimi_r_17/article/details/37908823