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;
  const int nax = 4005;
  int n, m;
  long long int a[nax];
  vector<long long int> reku(int L, int R) {
    if (L > R) return {0};
    if (L == R) {
      return {0, a[L] * (m - 1)};
    }
    int pos = L;
    long long int v = a[L];
    for (int i = L; i <= R; i++) {
      if (a[i] < v) {
        v = a[i];
        pos = i;
      }
    }
    auto LF = reku(L, pos - 1);
    auto RG = reku(pos + 1, R);
    vector<long long int> ans(R - L + 2, -2e18);
    for (int i = 0; i < LF.size(); i++) {
      for (int j = 0; j < RG.size(); j++) {
        ans[i + j] = max(ans[i + j], LF[i] + RG[j] - v * i * j * 2);
        ans[i + j + 1] =
            max(ans[i + j + 1],
                LF[i] + RG[j] + v * m - v * i * j * 2 - v - v * (i + j) * 2);
      }
    }
    return ans;
  }
  void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    auto res = reku(1, n);
    cout << res[m] << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int tt = 1;
    while (tt--) solve();
    return 0;
  }
  