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>
#define MAX 500111
#define mod 1000000007
const long long  inf = 2000000000000;
using namespace std;

long long a[MAX],n;
bool can(long long x){
    long long ans = 0;
    for(int i = 1; i <= n ; i++)ans += (x - a[i]);
    return ans <= 0;
}
long long get(long long x){
    long long ans= 0 ;
    for(int i = 1; i <= n; i++) ans += abs(x - a[i]);
    return ans;
}
int main(){ 
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);  
    cin>>n;
    for(int i = 1; i  <= n ; i++){cin>>a[i];a[i] -= i;}
    long long l = -inf ,h = inf;
    while(h - l > 1){
        long long mid = (l+h)>>1;
        if(can(mid))
            l = mid;
        else  h = mid;
    }
    long long mn = inf;
    for(long long i = l - 2; i < h + 3 ; i++)mn = min(get(i),mn);
    cout<<mn;
    return 0;
}