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 n, M;
  long long a[4000];
  void calc(long long l, long long r, vector<long long> &res) {
    if (r < l) {
      res.push_back(0);
      return;
    }
    long long m = l;
    for (long long i = l + 1; i <= r; ++i)
      if (a[i] < a[m]) m = i;
    vector<long long> res1, res2;
    calc(l, m - 1, res1);
    calc(m + 1, r, res2);
    res.resize(r - l + 2, -1e18);
    for (long long i = 0; i < (long long)res1.size(); ++i)
      for (long long j = 0; j < (long long)res2.size(); ++j) {
        res[i + j] = max(res[i + j], res1[i] + res2[j] - 2 * i * j * a[m]);
        res[i + j + 1] =
            max(res[i + j + 1], res1[i] + res2[j] - 2 * i * j * a[m] +
                                    (M - 1) * a[m] - 2 * (i + j) * a[m]);
      }
  }
  int32_t main() {
    cin >> n >> M;
    for (long long i = 0; i < n; ++i) cin >> a[i];
    vector<long long> res;
    calc(0, n - 1, res);
    cout << res[M];
  }
  