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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;

int main(){
    int N,K;
    cin >> N >> K ;
    vi a(N);
    for(int i=0;i<N;i++) cin >> a[i];
    long long ans=1e18;
    for(int i=0;i<(1<<N);i++){
        int temp=i;
        int count=0;
        for(int j=0;j<N;j++){
            if(temp%2==1) count++;
            temp/=2;
        }
        if(count!=K) continue;
        vi building;
        int temp2=i;
        for(int j=0;j<N;j++){
            if(temp2%2==1){
                building.push_back(a[j]);
            }
            temp2/=2;
        }
        if(building[0]!=a[0]) continue;
        long long temp_ans=0;
        int now_max=0;
        for(auto x:building){
            if(now_max>=x){
                temp_ans+=now_max-x+1;
                now_max++;
            }else{
                now_max=x;
            }
        }
        ans=min(ans,temp_ans);
    }
    cout << ans << endl;
    
}