In today's AWC, problem E, which should be the hardest, can be formalized to this (If you don't want to read the redundant statement that is likely generated by LLM):

You're given a set $$$U = \lbrace 1, 2, 3, \ldots, n \rbrace$$$, and you needs to partition $$$U$$$ into several sets $$$S_1, S_2, \ldots, S_k (1 \le k \le n)$$$ such that:

• $$$\bigcup_{i=1}^{k}S_i = U$$$
• for all $$$1 \le i \lt j \le n$$$, $$$S_i \cap S_j = \emptyset$$$

You're also given another $$$m$$$ sets $$$T_1, T_2, \ldots, T_m$$$. Let the infection function be:

Find the minimum value of $$$\lvert\bigcup_{i=1}^{k}\bigcup_{j=1}^{m} f(S_i, T_i)\rvert$$$.

Given the constraints that $$$n \le 15$$$ and $$$m \le 500$$$, the expected solution should have been SOS dp. And it is the official editorial.

However, because the number of partition $$$k$$$ is not limited, we can simply assign each value $$$i$$$ to to separate group $$$S_i$$$, so that $$$f(S, T)$$$ is not empty if and only if $$$T$$$ contains only one element. And if $$$T$$$ contains only one element, $$$f(S, T)$$$ is always not empty.

Thus, we prove the intuitive approach correct. And the remaining is extremely simple.

It is very likely that the problems of AWC are generated by LLM. The statement are extremely long and explain all the ‌minutia. The editorial itself is also generated by LLM. However, none of the LLMs find the simplest approach. Therefore, I think the workflow might be giving LLM a topic and asking it to generate both statement and editorial with no human review. Because the AWC is still in beta phase, it is still acceptable. However, this case shows that LLM can focus on the topic and ignore the very simple solution, so if it becomes an official contest, human review is a neccesity.