description: Alice has an integer sequence a of length n and all elements are different. She will choose a subsequence of a of length m, and defines the value of a subsequence a_{b_1},a_{b_2},…,a_{b_m} as $$$∑_{i = 1}^m (m ⋅ a_{b_i}) - ∑_{i = 1}^m ∑_{j = 1}^m f(min(b_i, b_j), max(b_i, b_j)), where f(i, j) denotes \min(a_i, a_{i + 1}, \ldots, a_j)$$$.
  
  Alice wants you to help her to maximize the value of the subsequence she choose.
  
  A sequence s is a subsequence of a sequence t if s can be obtained from t by deletion of several (possibly, zero or all) elements.
  
  Input
  
  The first line contains two integers n and m (1 ≤ m ≤ n ≤ 4000).
  
  The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i < 2^{31}).
  
  Output
  
  Print the maximal value Alice can get.
  
  Examples
  
  Input
  
  
  6 4
  15 2 18 12 13 4
  
  
  Output
  
  
  100
  
  
  Input
  
  
  11 5
  9 3 7 1 8 12 10 20 15 18 5
  
  
  Output
  
  
  176
  
  
  Input
  
  
  1 1
  114514
  
  
  Output
  
  
  0
  
  
  Input
  
  
  2 1
  666 888
  
  
  Output
  
  
  0
  
  Note
  
  In the first example, Alice can choose the subsequence [15, 2, 18, 13], which has the value 4 ⋅ (15 + 2 + 18 + 13) - (15 + 2 + 2 + 2) - (2 + 2 + 2 + 2) - (2 + 2 + 18 + 12) - (2 + 2 + 12 + 13) = 100. In the second example, there are a variety of subsequences with value 176, and one of them is [9, 7, 12, 20, 18].
solution: #include <bits/stdc++.h>
  using namespace std;
  int N, M;
  int A[4040];
  vector<long long> ret;
  void calc(int L, int R) {
    if (R - L == 0) {
      ret.clear();
      ret.push_back(0);
      return;
    }
    int mid = L;
    for (int i = L + 1; i < R; i++)
      if (A[mid] > A[i]) mid = i;
    vector<long long> l, r;
    calc(L, mid);
    l.swap(ret);
    calc(mid + 1, R);
    r.swap(ret);
    ret.clear();
    ret.assign(l.size() + r.size(), -(long long)1e18);
    for (int i = 0; i < l.size(); i++)
      for (int j = 0; j < r.size(); j++) {
        ret[i + j] = max(ret[i + j], l[i] + r[j] - (long long)A[mid] * i * j * 2);
      }
    for (int i = ret.size() - 2; i >= 0; i--)
      ret[i + 1] = max(ret[i + 1], ret[i] + (long long)(M - 1 - 2 * i) * A[mid]);
  }
  int main() {
    cin >> N >> M;
    for (int i = 0; i < N; i++) cin >> A[i];
    calc(0, N);
    cout << ret[M] << endl;
  }
  