Could not find an official announcement, let's discuss problems here.

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Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1)

03:29:47

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Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1)

03:29:47

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Could not find an official announcement, let's discuss problems here.

Discussion of 2020-2021 Summer Petrozavodsk Camp, Day 6: XJTU Contest (XXII Opencup, Grand Prix of XiAn)

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How to solve Div.2 M(Colors and Pearls) and P(Number Sequence)?

It looks like the problems for Div.2 were taken from some past ICPC Asia online contests...

M: A brute force solution is to denote $$$f_n$$$ as the answer for $$$a_{1} \cdots a_n$$$, then $$$f_n = \min_{0 \leq i < n} (f_i+v^2(i,n))$$$. Note that $$$f_n \leq n$$$, so we only need to keep the states that satisfy $$$v(i,n)\leq \sqrt n$$$. Time complexity is $$$\Theta(n \sqrt n)$$$.

CodeN: For a pile of stones, we have $$$\mathcal G_n = \operatorname{mex}_{i+j<n} ( \mathcal G_i, \mathcal G_{i} \stackrel{*}{+} G_j )$$$, and you can proof $$$\mathcal G_n = *_n$$$ by induction. So the whole game is equivalent to the regular Nim game.

CodeO is just a simple breadth first search problem.

P: We can make each $$$a_i\oplus b_i$$$ become $$$(111\cdots 1)_2$$$ by greedily pairing. This is obviously the optimal answer.

CodeThanks to the problemsetters, the problems were really nice (D in particular). How to solve C, G, J and K?

There's an editorial prepared by the problemsetters.

I had been looking for the editorial since the end of the contest, but couldn't seem to find it anywhere. Do you mind sharing how you found it? Most of the previous GP editorials were usually posted in the respective CF announcements, but this one doesn't seem to have any official announcement post.

Thanks for sharing the editorial btw.

Well... In fact, the contest was used in Petrozavodsk Camp, and it was shared by the problemsetters during the camp.

baaki sab ban gaya bhai?

Tere ban gaye kya?

Thanks to problem F! I learned something new.

Here is a tricky case not in the judge tests (it's OK as I know it's not easy to create a strong cases):

We have $$$E = \{ (3,1), (6,2), (9,3), (12,4), (15,5) \}$$$ and the answer should be 701423582.

My first attempt used quadtree-like DFS but I only checked the center of the ellipse and the boundary lattice points of the square to decide that they have any intersection. It is not correct for thin ellipses.

As the problem setter, thanks for caring about problem F!

The contest is available in GYM now :)