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 {
    /**
     * @param args
     */
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] A = new int[N];
        for (int i=0; i<N; i++) {
            A[i] = sc.nextInt();
        }

        int ans = solve3(N, A);

        System.out.println(ans);
    }

    private static int solve(int N, int[] A) {
        int sad = Integer.MAX_VALUE;

        int minB = A[0]-1, maxB = A[0]-1;
        for (int i=1; i<N; i++) {
            int b = A[i] - (i+1);
            minB = Math.min(minB, b);
            maxB = Math.max(maxB, b);
        }


        for (int b=minB; b<=maxB; b++) {
            sad = Math.min(sad, calcSad(N, A, b));
        }

        return sad;
    }

    private static int solve2(int N, int[] A) {
        int minSad = Integer.MAX_VALUE;

        for (int i=0; i<N; i++) {
            int b = A[i] - (i+1);
            minSad = Math.min(minSad, calcSad(N, A, b));
        }

        return minSad;
    }

    private static int solve3(int N, int[] A) {
        for (int i=0; i<N; i++) {
            A[i] -= i+1;
        }
        Arrays.sort(A);

        int minSad = calcSad2(N, A, A[N/2]);

        return minSad;
    }

    private static int calcSad(int N, int[] A, int b) {
        int sad = 0;

        for (int i=0; i<N; i++) {
            sad += Math.abs(A[i] - (b+i+1));
        }

        return sad;
    }

    private static int calcSad2(int N, int[] A, int b) {
        int sad = 0;

        for (int i=0; i<N; i++) {
            sad += Math.abs(A[i] - b);
        }

        return sad;
    }
}