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;
  const int maxn = 210;
  int n, K;
  long long x[maxn], y[maxn];
  pair<int, long long> f[maxn][maxn];
  int main() {
    scanf("%d %d", &n, &K), K++;
    for (int i = 1; i <= n; i++) {
      scanf("%lld %lld", &x[i], &y[i]);
    }
    auto cross = [&](int i, int j, int k) {
      long long x1 = x[j] - x[i], y1 = y[j] - y[i];
      long long x2 = x[k] - x[i], y2 = y[k] - y[i];
      return x1 * y2 - x2 * y1;
    };
    auto chk = [&](long long x) {
      for (int len = 3; len <= n; len++) {
        for (int l = 1, r = len; r <= n; l++, r++) {
          f[l][r] = {-1e9, 0};
          for (int k = l + 1; k < r; k++) {
            int a = f[l][k].first + f[k][r].first;
            long long b = f[l][k].second + f[k][r].second + cross(l, k, r);
            if (b >= x) a++, b = 0;
            f[l][r] = max(f[l][r], make_pair(a, b));
          }
        }
      }
      return f[1][n].first >= K;
    };
    long long l = 0, r = 1e17, ans;
    while (l <= r) {
      long long mid = (l + r) / 2;
      chk(mid) ? l = (ans = mid) + 1 : r = mid - 1;
    }
    printf("%lld\n", ans);
    return 0;
  }
  