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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,a[N],f,zero,z;
vector<int>v1,v2;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),a[i]-=i;
    ll ans=0;
    for(int i=1;i<=n;i++) ans+=abs(a[i]);
    for(int i=1;i<=n;i++)
        if(a[i]<0) v1.push_back(-a[i]);
        else if(a[i]>0) v2.push_back(a[i]);
        else zero++;
    sort(v1.begin(),v1.end());
    sort(v2.begin(),v2.end());
    if(v1.size()>v2.size()+zero)
    {
        int tep=v2.size()+zero,thi=v1.size(),res=0;
        for(int i=0;i<v1.size();i++)
        {
            if(thi<=tep) break;
            ans-=(ll)(thi-tep)*(v1[i]-res);
            thi--;tep++;
            res=v1[i];
        }
    }
    else if(v2.size()>v1.size()+zero)
    {
        int tep=v1.size()+zero,thi=v2.size(),res=0;
        for(int i=0;i<v2.size();i++)
        {
            if(thi<=tep) break;
            ans-=(ll)(v2[i]-res)*(thi-tep);
            thi--;tep++;
            res=v2[i];
        }
    }
    printf("%lld\n",ans);
}