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 (GCC 5.4.1) (judge verdict: Accepted)


#include <stdio.h>
#include <stdlib.h>

int asc(const void *a, const void *b) {
  return *(long int *)a - *(long int *)b;
}

int main(void)
{
	//変数の宣言
	int n;
	signed long int array[200010];
	long int mid;
	int i;
	long int ans=0;
	
	//データの読み込み
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%ld",&array[i]);
	}
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	
	
	//実際の処理
	
	for(i=0;i<n;i++){
		array[i]=array[i]-i-1;
	}
	
	qsort(array, n, sizeof(long int), asc);
	
	mid=array[(n-1)/2];
	
	for(i=0;i<n;i++){
		array[i]=array[i]-mid;
		array[i]=abs(array[i]);
		ans=ans+array[i];
	}
		
//	printf("計算部分終了\n");
	
	//出力
	
	printf("%ld",ans);
	
	
//	printf("結果の出力終了\n");
	
	return 0;
}