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: Wrong Answer)

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

int main(void){
    int n;
    scanf("%d",&n);
    int set1[n];                           /*初期数列*/
    int i,sum;
    for(i=0;i<n;i++){
        scanf("%d",&set1[i]);
        set1[i]-=(i+1);           /*数列からその要素の番号を引いた数列を用意(正負の区別アリ)*/
    }
    
    int b=0;
    int p,q;                                /*2要素の比較用引数*/
    while(1){                               /*無限ループ*/
        p=q=0;
        for(i=0;i<n;i++){
            p+=abs(set1[i]-b);             /*絶対値の和をbとb+1について求める*/
            q+=abs(set1[i]-(b+1));         
        }
        if(p<q){
            break;                            /*要素数を増やして数が増加した瞬間にbreak;*/
        }
        b++;                                  /*そうでなければ増やす*/
    }
    if(b!=0){
        printf("%d\n",p);
        return 0;
    }
    while(1){                                   /*マイナスについても考える*/
        p=q=0;
        for(i=0;i<n;i++){
            p+=abs(set1[i]-b);             /*絶対値の和をbとb-1について求める*/
            q+=abs(set1[i]-(b-1));
        }
        if(p<q){
            break;
        }
        b--;
    }
    printf("%d\n",p);
    return 0;
}