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 <stdio.h>
long int min(a, b){return a < b ? a : b;}
int max(a, b){return a > b ? a : b;}
int main(void){
    int n, k;
    scanf("%d%d", &n, &k);
    int a[15];
    int i, j;
    for(i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    long int ans = 999999999999999;
    int bit;
    int a_temp[15];
    int can_see = 0;
    int maxh = 0;
    int cost;
    for(bit = 0; bit < (1 << n); ++bit){
        cost = 0;
        can_see = 0;
        for(i = 0; i < n; i++){
            a_temp[i] = a[i];
        }
        for(i = 0; i < n; i++){
            if(bit & (1 << i)){
                can_see++;
                //0番目から(i-1)番目までの建物の高さの最大値を求める
                //ただしi = 0のときは0
                maxh = 0;
                for(j = 0; j <= i - 1; j++){
                    maxh = max(maxh, a_temp[j]);
                }
                //i番目の高さをmaxh+1以上にする
                if(a_temp[i] < maxh + 1){
                    cost += maxh + 1 - a_temp[i];
                    a_temp[i] = maxh + 1;
                }
            }
        }
        if(can_see >= k){
            ans = min(ans, cost);
        }
    }
    printf("%ld\n", ans);
    return 0;
}