Qingyu's blog

XXII Open Cup, Grand Prix of Xi'an
By Qingyu, history, 3 years ago, In English

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

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3 years ago, # |
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How to solve Div.2 M(Colors and Pearls) and P(Number Sequence)?

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    3 years ago, # ^ |
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    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 $$$\Theta(n \sqrt n)$$$.

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    N: 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.

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    O 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.

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3 years ago, # |
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Thanks to the problemsetters, the problems were really nice (D in particular). How to solve C, G, J and K?

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    3 years ago, # ^ |
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    There's an editorial prepared by the problemsetters.

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      3 years ago, # ^ |
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      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.

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        3 years ago, # ^ |
          Vote: I like it +26 Vote: I do not like it

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

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    3 years ago, # ^ |
      Vote: I like it -75 Vote: I do not like it

    baaki sab ban gaya bhai?

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3 years ago, # |
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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):

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10 9 100 3 -63 4

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.

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    3 years ago, # ^ |
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    As the problem setter, thanks for caring about problem F!

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3 years ago, # |
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The contest is available in GYM now :)

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