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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main(void){
    int n;
    int i;
    int tmp = 0;
    int plus = 0;
    int minus = 0;
    int a[200000];
    long long int result;
    long long int sum = 0;
    long long int ct = 1;
    scanf("%d", &n);
    for ( i = 1; i <= n; i++ ) {
        scanf("%d", &a[i-1]);
        tmp += (a[i-1]-i);
        sum += abs(a[i-1]-i);
        if ( a[i-1]-i > 0 ) {
            plus += 1;
        } else if ( a[i-1]-i < 0 ) {
            minus += 1;
        }
    }
    result = sum;
    if ( tmp < 0 ) {
        while ( 1 ) {
            sum = 0;
            for ( i = 1; i <= n; i++ ) {
                sum += abs(a[i-1]-i+ct);
            }
            ct++;
            if ( result <= sum ) {
                break;
            } else {
                result = sum;
            }
        }
    } else if ( tmp > 0 ) {
        while ( 1 ) {
            sum = 0;
            for ( i = 1; i <= n; i++ ) {
                sum += abs(a[i-1]-i-ct);
            }
            ct++;
            if ( result <= sum ) {
                break;
            } else {
                result = sum;
            }
        }
    }
    printf("%lld\n", result);
    return 0;
}