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;
#define lo long
const int lmax=1999999999;
const long long lmx=1999999999999999999;
long long k,n,a[200003],sum[200004],ans,k1,k2,ans1,ans2;
int main(){std::ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++)
{
	cin>>a[i];
	a[i]-=i;
	sum[i]+=abs(abs(a[i])+sum[i-1]);
}
ans=lmx;
for(int i=1;i<=n;i++)
{
	k=abs(sum[n]-2LL*sum[i]);
	if(i+1<=n)k=abs(sum[n]-2LL*sum[i])+abs(a[i+1]-a[i]);
	k1=abs(a[i]);
	k2=k1*n;
	ans=min(ans,abs(k-k2));
}
cout<<ans<<endl;
}
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