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>
#include <cmath>

const double PI = 3.14159265358979323846;
//using namespace boost::multiprecision;
using namespace std;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
typedef pair<ll, ll> P;
const ll INF = 1e15;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

std::istream &operator>>(std::istream &in, set<int> &o) {
    ll a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ifstream myfile("C:\\Users\\riku\\Downloads\\0_00.txt");
//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
const int mod = 1000000007;
//const ll mod = 1e10;
typedef priority_queue<string, vector<string>, greater<string> > PQ_ASK;

ll pow2(ll a, ll b) {
    ll res = a;
    for (ll i = 1; i < b; i++) {
        res *= a;
    }
    return res;
}

int main() {
    int n;
    cin >> n;
    vector<int> numbers(n);
    rep(i, n) cin >> numbers[i];
    rep(i, n) numbers[i] -= (i + 1);
    sort(numbers.begin(), numbers.end());
    int k = numbers[(n + 1) / 2];
    int ans = 0;
    for (int i : numbers) ans += abs(i - k);
    cout << ans << endl;
}