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 <algorithm>
#include <cassert>
#include <cmath>
#include <cstdlib>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <queue>

using namespace std;

using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
using ldb = long double;

#define pb emplace_back
#define forn(i, x, n) for(auto i = (int)x; i < (int)n; ++i)

#define int long long

struct init {
    init() {
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        cout << fixed << setprecision(10);
    }

    ~init() {
        cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC << "s.\n";
    }
} init;

const int N = 100500 * 2, MOD = 1000 * 1000 * 1000 + 7;
const ldb EPS = 5e-12, HALF = 0.5;

int a[N];

signed main() {
    int n;
    cin >> n;
    int sum = 0;

    forn(i, 0, n) {
        cin >> a[i];
        a[i] -= i + 1;
        sum += a[i];
    }

    int x = sum / 1. / n;
    int mn = 1e18;

    forn(i, x - 100, x + 100) {
        int s = 0;
        forn(j, 0, n) {
            s += abs(a[j] - i);
        }

        mn = min(mn, s);
    }

    cout << mn;
}