For the first subtask, it suffices to follow the definition of a good triplet. Firstly, there are ~\mathcal{O}(N^3)~ ways to select ~(a, b, c)~. Secondly, consider the condition that the origin is strictly inside the triangle. This is equivalent to saying that ~P_a~, ~P_b~, and ~P_c~ divide the circle into ~3~ arcs, and each arc is strictly shorter than ~\frac{C}{2}~. This leads to the following approach. Sort the positions so that ~P_a \le P_b \le P_c~, then check that

~P_b - P_a < \frac{C}{2}~, ~P_c - P_b < \frac{C}{2}~, and ~(P_a + C) - P_c < \frac{C}{2}~.

For example, the first of these checks can be implemented in code as P[b]-P[a] < (C+1)/2 in C++ or Java, or P[b]-P[a] < (C+1)//2 in Python 3.

For the second subtask, the goal is to find a solution that works well when ~C~ is small. Let ~L_x~ be the number of points drawn at location ~x~. If three locations ~i, j, k~ satisfy the second condition, we want to add ~L_i \times L_j \times L_k~ to the answer. This approach is ~\mathcal{O}(N + C^3)~ and is too slow. However, we can improve from three nested loops to two nested loops. If ~i~ and ~j~ are chosen already, either ~k~ does not exist, or ~k~ is in an interval. It is possible to replace the third loop with a prefix sum array. This algorithm's time complexity is ~\mathcal{O}(N + C^2)~.

It is possible to optimize this approach to ~\mathcal{O}(N + C)~ and earn ~15~ marks. The idea is to eliminate both the ~j~ and ~k~ loop. Start at ~i = 0~ and compute ~L_j \times L_k~ in ~\mathcal{O}(C)~ time. The next step is to transition from ~i = 0~ to ~i = 1~, and to maintain ~L_j \times L_k~ in ~\mathcal{O}(1)~ time. This requires strong familiarity with prefix sum arrays. Care with overflow and off-by-one errors is required.