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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
int main () {
  int n;
  cin >> n;
  vector<long> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> a.at(i);
    b.at(i) = a.at(i)-(i+1);
  }
  long x = 0, p = 0, q = 0, r = 0;
  for (int i = 0; i < n; i++) {
    x += b.at(i);
  }
  x /= n;
  for (int i = 0; i < n; i++) {
    p += abs(b.at(i)-(x-1));
    q += abs(b.at(i)-x);
    r += abs(b.at(i)-(x+1));
  }
  long ans;
  if (p <= q && p <= r) ans = p;
  else if (q < p && q <= r) ans = q;
  else ans = r;
  cout << ans << endl;
}