Problem:
Max Score: 350 Points

Problem Statement

There are N buildings along the line. The i-th building from the left is colored in color i, and its height is currently a_i meters.

Chokudai is a mayor of the city, and he loves colorful thigs. And now he wants to see at least K buildings from the left.

You can increase height of buildings, but it costs 1 yens to increase 1 meters. It means you cannot make building that height is not integer.

You cannot decrease height of buildings.

Calculate the minimum cost of satisfying Chokudai's objective.

Note: "Building i can see from the left" means there are no j exists that (height of building j) ≥ (height of building i) and j < i.

Input Format

N K
a_1 a_2 a_3 ... a_N

Output Format

Print the minimum cost in one line. In the end put a line break.

Constraints

1 ≤ K ≤ N ≤ 15

1 ≤ a_i ≤ 10^9

Scoring

Subtask 1 [120 points]

N = K

Subtask 2 [90 points]

N ≤ 5

a_i ≤ 7

Subtask 3 [140 points]

There are no additional constraints.

Sample Input 1

5 5
3949 3774 3598 3469 3424

Sample Output 1

1541

The optimal solution is (height of buildings from the left) = [3949, 3950, 3951, 3952, 3953].

Sample Input 2

5 3
7 4 2 6 4

Sample Output 2

7

The optimal solution is (height of buildings from the left) = [7, 8, 2, 9, 4].

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>
#include <climits>
#include <iomanip>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
int main(){
  ll n,k;
  cin >> n >> k;
  vector<ll> a(n);
  //vector<bool> see(n);
  ll count = 0;
  ll nowmax = 0;
  for(int i = 0;i < n;i++){
    cin >> a[i];
    //see[i] = false;
    if(a[i] > nowmax){
      //see[i] = true;
      count++;
      nowmax = a[i];
    }
  }
  //cout << count << endl;
  if(k - count <= 0){
    cout << 0 << endl;
    return 0;
  }
  ll res = 0;
  res = pow(10,18);
  for(int i = 0;i < (1<<n); i++){//2のa乗回繰り返す(0 ~ 2^a- 1)
    ll nowcost = 0;
    ll keta = 0;
    ll purasu = 0;
    ll maxhigh = 0;
    for(int j = 0;j < n;j++){
      //if(!see[j]){
        if(i & (1<<j)){//小さい方からj桁目(0から)が1であったら
          if(0 <= maxhigh + 1 - a[j])purasu++,nowcost+= maxhigh + 1 - a[j];
          //if(i == 31)cout << maxhigh << endl;
          //if(i == 11)cout << "j" << j << " " << nowcost << " " << maxhigh << endl;
              maxhigh = max(a[j],maxhigh + 1);
        }
      //}
      else maxhigh = max(maxhigh,a[j]);
      if(purasu + count >= k){
        res = min(res,nowcost);
        //cout << nowcost << endl;
        //if(nowcost == 8)cout << "i" << i << endl;
      }
    }
  }
  cout << res << endl;
}