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 N = 4005;
  int n, m, a[N];
  long long dp[N][N];
  int solve(int l, int r) {
    if (l > r) return 0;
    int mid = min_element(a + l, a + r + 1) - a;
    int ls = solve(l, mid - 1);
    int rs = solve(mid + 1, r);
    int s1 = min(m, mid - l), v1 = a[ls] - a[mid];
    int s2 = min(m, r - mid), v2 = a[rs] - a[mid];
    for (int i = 1; i <= s1; ++i) dp[ls][i] += 1ll * v1 * i * (m - i);
    for (int i = 1; i <= s2; ++i) dp[rs][i] += 1ll * v2 * i * (m - i);
    for (int i = 0; i <= s1; ++i) {
      for (int j = 0; j <= s2 && i + j <= m; ++j) {
        long long val = dp[ls][i] + dp[rs][j];
        dp[mid][i + j] = max(dp[mid][i + j], val);
        dp[mid][i + j + 1] = max(dp[mid][i + j + 1], val);
      }
    }
    return mid;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    int u = solve(1, n);
    printf("%lld\n", dp[u][m]);
    return 0;
  }
  