Problem:
Score : 300 points

Problem Statement

Snuke has an integer sequence A of length N.

He will freely choose an integer b.
Here, he will get sad if A_i and b+i are far from each other.
More specifically, the sadness of Snuke is calculated as follows:

abs(A_1 - (b+1)) + abs(A_2 - (b+2)) + ... + abs(A_N - (b+N))

Here, abs(x) is a function that returns the absolute value of x.

Find the minimum possible sadness of Snuke.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible sadness of Snuke.

Sample Input 1

5
2 2 3 5 5

Sample Output 1

2

If we choose b=0, the sadness of Snuke would be abs(2-(0+1))+abs(2-(0+2))+abs(3-(0+3))+abs(5-(0+4))+abs(5-(0+5))=2.
Any choice of b does not make the sadness of Snuke less than 2, so the answer is 2.

Sample Input 2

9
1 2 3 4 5 6 7 8 9

Sample Output 2

0

Sample Input 3

6
6 5 4 3 2 1

Sample Output 3

18

Sample Input 4

7
1 1 1 1 2 3 4

Sample Output 4

6

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int N=scanner.nextInt();
        int[] a=new int[N];
        for(int i=0;i<N;i++){
            a[i]=scanner.nextInt();
        }
        long sum=0;
        for(int i=0;i<N;i++){
            sum+=a[i]-(i+1);
        }
        if(sum%N==0){
            long b=sum/N;
            long result=0;
            for(int i=0;i<N;i++){
                result+=Math.abs(a[i]-(b+i+1));
            }
            System.out.println(result);
            return;
        }else{
            long b1=(long)Math.floor(sum/(double)N);
            long b2=(long)Math.ceil(sum/(double)N);
            long result1=0;
            long result2=0;
            for(int i=0;i<N;i++){
                result1+=Math.abs(a[i]-(b1+i+1));
                result2+=Math.abs(a[i]-(b2+i+1));
            }
            System.out.println(Math.min(result1,result2));

        }

    }
}