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); } }
Balancer - CodeForces 440B,布布扣,bubuko.com