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 <bits/stdc++.h>
using namespace std;

int main() {
   int N,K;
   cin >> N >> K;
   int A[N];
   for (int i=0; i<N; i++) {
       cin >> A[i];
   }
   int total=0;
   for (int tmp=0; tmp<(1<<15); tmp++) {
       bitset<15> s(tmp);
       int B[K],C[N],count=0;
       if (!s.test(count)) continue;
       for (int i=0; i<N; i++) {
           if (s.test(i)) count++;
       }
       if (count!=K) continue;
       count=0;
       for (int i=0; i<N; i++) {
           if (s.test(i) && count!=K) {
               B[count]=i;count++;
           }
           else C[i]=A[i];
       }
       C[0]=A[0];
       int n=A[0];
       count=0;
       for (int i=1; i<K; i++) {
           int a=B[i-1];
           int b=B[i];
           for (int j=a; j<b; j++) {
               if (n<C[j]) n=C[j];
               if (j==b-1 && n<A[b]) C[b]=A[b];
               else if (j==b-1) {
                   C[b]=n+1;
                   count+=n+1-A[b];
               }
           }
       }
       if (total>count) total=count;
   }
   cout << total << endl;
}