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>
int main(void) {
	int n, i, b, ans, sem, c,out,z,y;
	ans = 0;
	out = 0;
	scanf("%d", &n);
	int a[200000];
	int k[200000];
	int x[200000];
	int s[200000];
	for (i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (i = 1; i <= n; i++) {
		k[i] = a[i] - i;
		
	}
	for (i = 1; i <= n-1; i++) {
		if (k[i] > k[i + 1]) out++;
		else;
	}
	while (out >= 1) {
		for (i = 1; i <= n - 1; i++) {
			if (k[i] > k[i + 1]) {
				z = k[i];
				y = k[i + 1];
				k[i] = y;
				k[i + 1] = z;

			}
			else;
			out = 0;
			for (i = 1; i <= n - 1; i++) {
				if (k[i] > k[i + 1]) out++;
				else;
			}

		}
	}
	
	b = n % 2;
	
	if (b = 1) {
		c = (n + 1) / 2;
		for (i = 1; i <= n; i++) {
			x[i] = k[i] - k[c];
			if (x[i] <= 0) s[i] = -x[i];
			else s[i] = x[i];
		}
		for (i = 1; i <= n; i++) {
			ans = ans + s[i];
		}
	}
	else {
		c = n / 2;
		for (i = 1; i <= n; i++) {
			x[i] = k[i] - k[c];
			if (x[i] <= 0) s[i] = -x[i];
			else s[i] = x[i];
		}
		for (i = 1; i <= n; i++) {
			ans = ans + s[i];
		}
	}
	printf("%d", ans);

	return 0;
}