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)

//c.c C-Linear Approximation
#include<stdio.h>

static void iswap(int *a, int i, int j) {
    int x = a[i]; 
    a[i] = a[j]; 
    a[j] = x; 
}

void qs(int *a, int i, int j) { 
    if(j <= i) { 
        return; 
    }
    int s = i, pivot = a[j]; 
    for(int k = i; k < j; ++k) {
        if(a[k] < pivot) { 
            iswap(a, s++, k); 
            } 
    }
    iswap(a, j, s); 
    qs(a, i, s-1); 
    qs(a, s+1, j); 
}

void quicksort(int *a, int n) { 
    qs(a, 0, n-1); 
}


int iabs(int x){
    if(x<0){
        return -x;
    }
    return x;
}

int main(void){
    int n;
    scanf("%d",&n);
    int a[n];
    for(int i=0;i<n;++i){
        scanf("%d",&a[i]);
    }

    int b[n];
    for(int i=0;i<n;++i){
        b[i] = a[i] - i - 1;
    }

    quicksort(b,n);

    int ansnum1,ansnum2,ansnum,ans1=0,ans2=0,ans=0;
    if(n%2!=0){
        ansnum = n/2;
        for(int i=0;i<n;++i){
            ans += iabs(a[i]-b[ansnum]-(i+1));
        }
    }else{
        ansnum1 = n/2 - 1;
        ansnum2 = n/2;
        for(int i=0;i<n;++i){
            ans1 += iabs(a[i]-b[ansnum1]-(i+1));
            ans2 += iabs(a[i]-b[ansnum2]-(i+1));
        }
        if(ans1>ans2){
            ans = ans2;
        }else{
            ans = ans1;
        }
    }
    printf("%d\n",ans);
    return 0;
}