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 fio ios_base::sync_with_stdio(false); cin.tie(NULL);

typedef long long int lli;

const lli msz=2e5+1;
lli a[msz];
lli n;


lli sad(lli b)
{
	lli val=0;
	for(lli i=1;i<=n;i++)
	{
		val+=abs(a[i]-(b+i));
	}
	return val;
}

lli tSearch(lli lo,lli hi)
{
	while(hi-lo>3)
	{
		lli m1=lo+(hi-lo)/3;
		lli m2=hi-(hi-lo)/3;
		lli sd1=sad(m1);
		lli sd2=sad(m2);
		//cout<<"lo "<<lo<<" hi "<<hi<<" sd1 "<<sd1<<" sd2 "<<sd2<<endl;
		if(sd1>sd2) lo=m1;
		else hi=m2;
	}
	lli ans=2e9+1;
	for(lli i=lo;i<=hi;i++)
	{
		ans=min(ans,sad(i));
	}
	return ans;
}

int main()
{
	fio;
	cin>>n;
	lli mx=0;
	for(lli i=1;i<=n;i++)
	{
		cin>>a[i];
		mx=max(mx,abs(a[i]));
	}
	cout<<tSearch(-mx,mx)<<endl;
	cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n";
	return 0;
}