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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll N,a[200005],b;
ll sum,ans1,ans2,ans3;//考虑到负数 小数点的接近不同 取+-1 
int main()
{
	scanf("%d",&N);
	for(int i=0;i<N;++i)
	{
		scanf("%d",&a[i]);
		a[i]-=(i+1);
	}
	sort(a,a+N);
	if(N&1)b=a[N/2];//n奇为中间 
	else b=(a[N/2]+a[N/2+1])/2;
	for(int i=0;i<N;++i)
	{
		ans1+=abs(a[i]-b);
		ans2+=abs(a[i]-(b+1));
		ans3+=abs(a[i]-(b-1));
	} 
	printf("%lld\n",min(min(ans1,ans2),ans3));
	return 0;
}