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;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int n, k;
int main() {
    cin >> n >> k;
    vector<long long> a(n, 0);
    for(int i = 0; i < n; ++i) cin >> a.at(i);

    vector<long long> aup;
    long long start = a.at(0)+1;
    for(int i = 1; i < n; ++i) aup.push_back(a.at(i));
    int uplen = aup.size();
    k--;

    long long mincost = 1e18;
    for(int bit = 0; bit < (1<<uplen); ++bit) {
        vector<int> ss(uplen, 0);
        int cnt = 0;
        for(int i = 0; i < uplen; ++i) {
            if(bit & (1<<i)) {
                ss.at(i) = 1;
                cnt++;
            }
        }

        if(cnt == k) {
            long long cstart = start;
            long long sum = 0;
            for(int j = 0; j < uplen; ++j) {
                if(ss.at(j) == 1) {
                    if(cstart < aup.at(j)) {
                        sum += 0;
                        cstart = aup.at(j);
                    }else {
                        sum += (cstart - aup.at(j));
                        cstart++;
                    }
                }
            }
            chmin(mincost, sum);
        }
    }
    cout << mincost << endl;
}