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 n, m;
  const int maxN = 4005;
  long long a[maxN];
  vector<long long> solve(int l, int r) {
    if (l > r) {
      return {0LL};
    }
    int where = l;
    for (int k = l; k <= r; k++) {
      if (a[k] < a[where]) where = k;
    }
    auto t1 = solve(l, where - 1);
    auto t2 = solve(where + 1, r);
    for (int z = 0; z < t1.size(); z++) {
      t1[z] += a[where] * z * z;
    }
    for (int z = 0; z < t2.size(); z++) {
      t2[z] += a[where] * z * z;
    }
    vector<long long> ans(r - l + 1 + 1, -1e18);
    int pt1 = 0;
    int pt2 = 0;
    for (int z = 0; z < t1.size() + t2.size() - 1; z++) {
      ans[z] = t1[pt1] + t2[pt2] - z * z * a[where];
      if (z != (int)ans.size() - 2) {
        if (pt1 + 1 < (int)t1.size() &&
            (pt2 == (int)t2.size() - 1 ||
             t1[pt1 + 1] - t1[pt1] > t2[pt2 + 1] - t2[pt2])) {
          pt1++;
        } else {
          pt2++;
        }
      }
    }
    assert(pt1 == (int)t1.size() - 1 && pt2 == (int)t2.size() - 1);
    for (int i = (int)ans.size() - 2; i >= 0; i--) {
      ans[i + 1] =
          max(ans[i + 1], ans[i] + (m - 1) * a[where] - 2 * i * a[where]);
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<long long> t = solve(1, n);
    cout << t[m] << '\n';
    return 0;
  }
  