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>

// #define fi first
// #define se second
#define rep(i, n) for (ll i=0;i<(n); i++)
#define ALL(v) (v).begin(), (v).end()
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pi;

//const ll M=1000000007ll;
//const ll MOD=1000000007ll;
const ll MOD=2019ll;
const ll MAX=100000001ll;
//const ll M = 1000000000+7;

//const int mod = 1000000007;
const int mod = 2019;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
   return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(ll n){
  if(n==0) return 1;
  mint ans = f(n/2);
  ans *= ans;
  if(n%2==1) ans*= 2;
  return ans;
}

mint g(ll n, ll x){
  if(x==0) return 1;
  mint ans = g(n, x/2);
  ans *= ans;
  if(x%2==1) ans*= n;
  return ans;  
}

mint choose(ll n, ll k){
  mint x=1, y=1;
  rep(i,k){
    x *= (n-i);
    y *= (i+1);
  }
  return x / y;
}

struct BIT {
  private:
   vector<ll> bit;
   ll N;
 
  public:
   BIT(ll size) {
     N = size;
     bit.resize(N + 1);
   }
 
   // 一点更新です
   void add(ll a, ll w) {
     for (ll x = a; x <= N; x += x & -x) bit[x] += w;
   }
 
   // 1~Nまでの和を求める。
   int sum(ll a) {
     ll ret = 0;
     for (ll x = a; x > 0; x -= x & -x) ret += bit[x];
     return ret;
   }
 };
 
ll fff(ll n, vector<ll>& v, ll i, ll j, ll rl){
  ll can = 0;
  ll can_i = 0;

  cout << v[0]<<endl;
  
  // right
  if(rl == 0){
    rep(k, n){
      if(v[k]*(j-k)>can){
	can = v[k]*(j-k);
	can_i = k;
      }
      //cout << can<<endl;
    }
    v[can_i] = 0;
    return can;
  }
  // left
  else{
    rep(k, n){
      if(v[k]*(k-i)>can){
	can = v[k]*(k-i);
	can_i = k;
      }
      //cout << can<<endl;

    }
    v[can_i] = 0;
    return can;
  }
  
}

ll ff(ll x){
  return x * (x+1) / 2;
}

void dfs(string s, ll ma, ll n){
  //cout << ma<<endl;
  if(s.size()==n){
    cout << s <<endl;
    return;
  }
  rep(i, ma+1){
    dfs(s + char('a'+i), max(ma, i+1), n);
  }
}

int main(){
  ll n;
  cin>>n;

  ll ans = 0;
  vector<ll> v(n);
  rep(i, n){
    cin >> v[i];
    ans += (v[i]-i-1);
  }

  ans /= n;

  ll hoge = 0;
  rep(i, n){
    hoge += abs(v[i]- (ans + i));
  }
  ll hoge1 = 0;
  rep(i, n){
    hoge1 += abs(v[i]- (ans + 1 + i));
  }
  ll hoge2 = 0;
  rep(i, n){
    hoge2 += abs(v[i]- (ans + 2 + i));
  }

  //cout << hoge <<hoge2<<endl;
  if(hoge2 < hoge) ans++;
  
  cout << min(hoge, min(hoge1, hoge2)) << endl;
    
}