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;
  long long dp[4001][4001];
  long long a[4001];
  vector<long long> solve(int l, int r, int m) {
    if (l == r) {
      return {0, a[l] * m - a[l]};
    }
    int mid = dp[l][r];
    vector<long long> le, ri;
    if (mid == r) {
      le = solve(l, r - 1, m), ri = solve(r, r, m);
    } else {
      le = solve(l, mid, m), ri = solve(mid + 1, r, m);
    }
    vector<long long> ans(r - l + 2);
    for (int eval = 1; eval <= r - l + 1; eval++) {
      int mn = eval - ri.size() + 1, mx = le.size() - 1;
      mn = max(mn, 0);
      mx = min(mx, eval);
      for (int x = mn; x <= mx; x++) {
        ans[eval] = max(ans[eval], le[x] + ri[eval - x] -
                                       x * 1LL * (eval - x) * 1LL * a[mid] * 2);
      }
    }
    return ans;
  }
  int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        if (i == j)
          dp[i][j] = i;
        else {
          dp[i][j] = dp[i][j - 1];
          if (a[j] < a[dp[i][j - 1]]) {
            dp[i][j] = j;
          }
        }
      }
    }
    cout << solve(0, n - 1, m)[m] << "\n";
  }
  