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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x : v) {
      in >> x;
    }
    return in;
  }
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    function<vector<long long>(int, int)> dfs = [&](int l, int r) {
      if (l > r) {
        return vector<long long>{0};
      }
      int p = l;
      for (int i = l + 1; i <= r; i++) {
        if (a[i] < a[p]) {
          p = i;
        }
      }
      auto L = dfs(l, p - 1);
      auto R = dfs(p + 1, r);
      vector<long long> dp(L.size() + R.size() + 1);
      for (int i = 0; i < (int)L.size(); i++) {
        for (int j = 0; j < (int)R.size(); j++) {
          dp[i + j] = max(dp[i + j], L[i] + R[j] - (long long)a[p] * i * j * 2LL);
          dp[i + j + 1] =
              max(dp[i + j + 1], L[i] + R[j] + (long long)m * a[p] -
                                     a[p] * (2LL * (i + 1) * (j + 1) - 1));
        }
      }
      return dp;
    };
    cout << dfs(0, n - 1)[m] << "\n";
  }
  