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.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        long sum = 0;
        int[] p = new int[n];
        long pp = 0;
        for (int i=0; i<n; i++) {
            int x = sc.nextInt();
            p[i] = x;
            arr[i] = x - (i+1);
            pp += arr[(int)i];
            sum += Math.abs(arr[(int)i]);
        }
        long val = 0;
        if (pp == 0 || sum == 0) {
        	val = 0;
        	long xx = 0;
            for (int i=0; i<n; i++) {
                xx += Math.abs(p[i] - (val+i+1));
            }
            System.out.println(xx);
        }
        //System.out.println(0);
        else if (sum > 0) {
            val = (long)Math.ceil(sum / (1.0*n));
            long val1 = (long)Math.floor(sum / (1.0*n));
            // if (sum % n == 0) val = 0;
            //System.out.println(sum);
            //System.out.println(val);
            //val = 0;
            //val  =-3;
            long xx = 0;
            if (pp < 0) val = -val;
            //val = -3;
            //val = 1;
            for (int i=0; i<n; i++) {
                xx += Math.abs(p[i] - (val+i+1));
            }
            long xx1 = 0;
            for (int i=0; i<n; i++) {
                xx1 += Math.abs(p[i] - (val1+i+1));
            }
            long xx2 = 0;
            for (int i=0; i<n; i++) {
                xx2 += Math.abs(p[i] - ((val-1)+i+1));
            }
            long xx3 = 0;
            for (int i=0; i<n; i++) {
                xx3 += Math.abs(p[i] - ((val+1)+i+1));
            }
            System.out.println(Math.min(xx,Math.min(xx1,Math.min(xx2,xx3))));
        } else {
            val = (int)Math.ceil(sum / (1.0*n));
            val = -val;
            long xx = 0;
            for (int i=0; i<n; i++) {
                xx += Math.abs(p[i] - (val+i+1));
            }
            System.out.println(xx);
        }
    }
}