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 m, n;
  int a[4001];
  vector<long long> dfs(int l, int r) {
    if (r == l) return {0};
    vector<long long> ans(r - l + 1, LLONG_MIN);
    int pos = min_element(a + l, a + r) - a;
    auto L = dfs(l, pos), R = dfs(pos + 1, r);
    for (int i = 0; i <= pos - l; ++i) {
      for (int j = 0; j <= r - pos - 1; ++j) {
        for (int op = 0; op <= 1; ++op) {
          ans[i + j + op] =
              max(ans[i + j + op],
                  L[i] + R[j] -
                      1LL * (2 * i * j + op * (2 * i + 2 * j + 1 - m)) * a[pos]);
        }
      }
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    cout << dfs(1, n + 1)[m];
  }
  