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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int main(){
  int n,k,i,j,r;
  long long a[18];
  long long b[18];
  long long sum;
  long long max = -9999;
  long long maxnow;
  int cnt;
  int use[18];
  int f;
  long long min = 99999999999;

  scanf("%d %d", &n, &k);

  for(i = 0;i < n;i++)
    scanf("%lld", &a[i]);

  for(i = 0;i < (1 << n);i++){
    sum = 0;
    for(j = 0;j < n;j++){
      use[j] = 0;
      b[j] = a[j];
    }
    cnt = 0;

    for(j = 0;j < n;j++){
      if((i >> j) & 1){
        cnt++;
        use[j] = 1;
      }
    }

    if(cnt != k) continue;

    maxnow = -9999;
    for(j = 0;j < n;j++){
      if(use[j] == 1){
        for(r = j;r >= 0;r--){
          if(b[r] > maxnow) maxnow = b[r];
        }

        sum += maxnow - b[j];
        b[j] = maxnow;
        maxnow++;
      }
    }

    if(min > sum) min = sum;
  }

  printf("%lld\n", min);

  return 0;
}