description: You are given a strictly convex polygon with n vertices.
  
  You will make k cuts that meet the following conditions: 
  
    * each cut is a segment that connects two different nonadjacent vertices; 
    * two cuts can intersect only at vertices of the polygon. 
  
  
  
  Your task is to maximize the area of the smallest region that will be formed by the polygon and those k cuts.
  
  Input
  
  The first line contains two integers n, k (3 ≤ n ≤ 200, 0 ≤ k ≤ n-3).
  
  The following n lines describe vertices of the polygon in anticlockwise direction. The i-th line contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^8) — the coordinates of the i-th vertex.
  
  It is guaranteed that the polygon is convex and that no two adjacent sides are parallel.
  
  Output
  
  Print one integer: the maximum possible area of the smallest region after making k cuts multiplied by 2.
  
  Examples
  
  Input
  
  
  8 4
  -2 -4
  2 -2
  4 2
  1 5
  0 5
  -4 4
  -5 0
  -5 -1
  
  
  Output
  
  
  11
  
  
  Input
  
  
  6 3
  2 -2
  2 -1
  1 2
  0 2
  -2 1
  -1 0
  
  
  Output
  
  
  3
  
  Note
  
  In the first example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (-2, -4) and (4, 2), 
    * (-2, -4) and (1, 5), 
    * (-5, -1) and (1, 5), 
    * (-5, 0) and (0, 5). 
  
  <image> Points (-5, -1), (1, 5), (0, 5), (-5, 0) determine the smallest region with double area of 11. 
  
  In the second example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (2, -1) and (0, 2), 
    * (2, -1) and (1, 0), 
    * (-1, 0) and (0, 2). 
  
  <image> Points (2, -2), (2, -1), (-1, 0) determine one of the smallest regions with double area of 3. 
solution: #include <bits/stdc++.h>
  using namespace std;
  struct pnt {
    long long x, y;
    pnt(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
  } P[300];
  long long operator*(pnt A, pnt B) { return A.x * B.y - A.y * B.x; }
  long long area(pnt A, pnt B, pnt C) { return A * B + B * C + C * A; }
  pair<int, long long> dp[300][300];
  long long sz[300][300];
  int main() {
    int n = 0, k = 0;
    scanf("%d%d", &n, &k);
    k++;
    for (int i = 1; i <= n; i++) scanf("%lld%lld", &P[i].x, &P[i].y);
    for (int l = 1; l <= n; l++) {
      for (int r = l + 1; r <= n; r++) sz[l][r] = sz[l][r - 1] + P[r - 1] * P[r];
      for (int r = l + 1; r <= n; r++) sz[l][r] += P[r] * P[l];
    }
    long long S = 0;
    for (int i = 1; i < n; i++) S += P[i] * P[i + 1];
    S += P[n] * P[1];
    long long L = 0, R = S / k;
    while (L < R) {
      long long x = (L + R + 1) >> 1;
      memset(dp, 0, sizeof(dp));
      for (int len = 3; len <= n; len++) {
        for (int l = 1; l + len - 1 <= n; l++) {
          int r = l + len - 1;
          if (sz[l][r] < x) continue;
          dp[l][r] = make_pair(1, 0);
          for (int k = l + 1; k <= r - 1; k++) {
            pair<int, long long> val = make_pair(
                dp[l][k].first + dp[k][r].first,
                dp[l][k].second + dp[k][r].second + area(P[l], P[k], P[r]));
            if (val.second >= x) val.first++, val.second = 0;
            dp[l][r] = max(dp[l][r], val);
          }
        }
      }
      if (dp[1][n].first >= k)
        L = x;
      else
        R = x - 1;
    }
    printf("%lld", L);
  }
  