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;
template <class Itr>
auto Median(Itr first,Itr last){
    typename iterator_traits<Itr>::difference_type n = distance(first, last);
    sort(first,last);
    if(n % 2 == 1){
        return *(first+n/2)*2;
    } else {
        return (*(first+n/2)+*(first+n/2-1)/2);
    }
}

signed main(){
    int n; cin >> n;
    vector<int>v(n);
    int ofs = 1;
    for(auto &i :v)
        cin >> i,i-=ofs++;
    int m = round(Median(v.begin(), v.end())/(double)2);
    int ans = 0;
    for(auto i :v)ans += abs(i-m);
    cout << ans << endl;
}