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)

// last.cpp : コンソール アプリケーションのエントリ ポイントを定義します。
//
#include<iostream>
#include <cmath>
#include <string>
#include <float.h>
using namespace std;

int main()
{
int l;
cin>>l;
int* num = new int[l];
int sum=0;
int a[8];
int fa=0;
int ca=0;
int min;
for (int i = 0;i<l;i++){
	cin>>num[i];
	sum=sum+num[i]-i;
}
double ans = double(sum)/double(l);
int f = floor(ans);
int c = ceil(ans);
for(int i = 0;i<l;i++){
	a[5]=a[5]+abs(num[i]-i-f);
	a[4]=a[4]+abs(num[i]-i-c);
	a[0]=a[0]+abs(num[i]-i-f-1);

	a[1]=a[1]+abs(num[i]-i-f-2);
	a[2]=a[2]+abs(num[i]-i-f+1);
	a[3]=a[3]+abs(num[i]-i-f+2);
}
min=a[0];
for(int i=0;i<6;i++){

	if(min>a[i]){
		min=a[i];
	}
}
cout<<min;
delete num;

}