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 <iostream>
#include <algorithm>
using namespace std;
int main(void){
   int N;
   cin>>N;
   int a[N],b[N];
   for(int i=0;i<N;i++){
       cin>>a[i];
       b[i]=a[i]-(i+1);
   }
   sort(b,b+N);
   int gap,gap1,gap2;
   if(N%2==1) {gap=b[(N-1)/2];
   int max=0;
   for(int i=0;i<N;i++){
       max+=abs(b[i]-gap);
   }
   cout<<max;}
   else {
       gap1=(b[N/2]+b[N/2-1])/2;
        gap2=(b[N/2]+b[N/2-1])/2+1;
       int max1=0,max2=0;
       for(int i=0;i<N;i++){
           max1+=abs(b[i]-gap1);
       }
       for(int i=0;i<N;i++){
           max2+=abs(b[i]-gap2);
       }
       if(max1>=max2) cout<<max2;
       else cout<<max1;
   }
   return 0;
}