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;
  inline long long read() {
    char ch = getchar();
    long long nega = 1;
    while (!isdigit(ch)) {
      if (ch == '-') nega = -1;
      ch = getchar();
    }
    long long ans = 0;
    while (isdigit(ch)) {
      ans = ans * 10 + ch - 48;
      ch = getchar();
    }
    if (nega == -1) return -ans;
    return ans;
  }
  void print(vector<long long> x) {
    for (long long i = 0; i < (long long)x.size(); i++)
      printf("%lld%c", x[i], " \n"[i == (long long)x.size() - 1]);
  }
  long long a[4005], n, m;
  long long val[4005], f[4005][4005], vis[4005];
  signed main() {
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++) {
      f[i][i] = a[i];
      for (long long j = i + 1; j <= n; j++) f[i][j] = min(f[i][j - 1], a[j]);
    }
    long long ans = 0;
    for (long long i = 1; i <= m; i++) {
      long long maxn = -0x3f3f3f3f3f3f3f3f, maxid = 0;
      for (long long j = 1; j <= n; j++)
        if (!vis[j]) {
          long long w = val[j] * 2 + (m - 1) * a[j];
          if (w > maxn) maxn = w, maxid = j;
        }
      vis[maxid] = 1;
      ans += maxn;
      for (long long i = 1; i < maxid; i++) val[i] -= f[i][maxid];
      for (long long i = maxid + 1; i <= n; i++) val[i] -= f[maxid][i];
    }
    cout << ans << endl;
    return 0;
  }
  