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;
typedef long long ll;

int main(){
    ll n;
    cin>>n;
    ll a[200005];
    for(int i=1;i<=n;i++){
        scanf("%lld",a+i);
    }
    ll ans=1e18;
    ll l=0,r=1e18;
    while(l+100<r){
        ll mid=(l+r)>>1;
        ll lmid = (l+mid)>>1;
        ll rmid = (mid+r)>>1;
        ll tmpmid=0,tmplmid=0,tmprmid=0;
        for(int i=1;i<=n;i++){
            tmpmid+=abs(a[i]-mid-i);
            tmplmid+=abs(a[i]-lmid-i);
            tmprmid+=abs(a[i]-rmid-i);
        }
        ans=min(ans,tmpmid);
        ans=min(ans,tmplmid);
        ans=min(ans,tmprmid);
        if(tmplmid<=tmpmid && tmpmid<=tmprmid){
            r=lmid;
        }
        else if(tmplmid>=tmpmid && tmpmid>=tmprmid){
            l=rmid;
        }
        else{
            l=lmid;r=rmid;
        }
    }
    for(int b=l;b<r;b++){
        ll tmp=0;
        for(int i=1;i<=n;i++){
            tmp+=abs(a[i]-b-i);
        }
        ans=min(ans,tmp);
    }cout<<ans;
    return 0;
}