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 <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b){return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b){return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 200001
#define MOD 1000000007

typedef long long int lli;

int main(void) {
  int N;
  scanf("%d", &N);
  int A[MAX];
  rep(i, 0, N) scanf("%d", &A[i]);

  int sum = 0;
  rep(i, 0, N) {
    sum += A[i] - (i + 1);
  } 
  double b1 = (double)sum / (double)N;
  double b2;
  if ( b1 > 0 ) b2 = b1 + 1;
  else if ( b1 < 0 ) b2 = b1 - 1;
  else b2 = b1;

  int ans1 = 0, ans2 = 0;
  rep(i, 0, N) {
    ans1 += abs(A[i]- ((int)b1 + i + 1));
    ans2 += abs(A[i]- ((int)b2 + i + 1));
  }
  printf("%d\n", min(ans1, ans2));
  return 0;
}