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(){
  long long int n;
  cin>>n;
  vector<long long int> a(n);
  long long int p=0;
  for(int i=0;i<n;i++){
    cin>>a[i];
    a[i]-=(i+1);
  }
  long long int ans=1e9;

  long long int l=-1e9;
  long long int r=1e9;
  while(abs(r-l)>2){
    long long int mid=(l*2+r)/3;
    long long int mid2=(l+r*2)/3;
    long long int tmp=0;
    for(int i=0;i<n;i++)tmp+=abs(a[i]+mid);
    long long int tmp2=0;
    for(int i=0;i<n;i++)tmp2+=abs(a[i]+mid2);
    //cout<<tmp<<" "<<tmp2<<endl;
    if(tmp<tmp2){
      r=mid2;
    }else{
      l=mid;
    }
    ans=min(ans,min(tmp,tmp2));
  }

  cout<<ans<<endl;

}