maroonrk's blog

AtCoder Grand Contest 070 Announcement
By maroonrk, history, 32 hours ago, In English

We will hold AtCoder Grand Contest 070. This contest counts for GP30 scores.

The point values will be 600-1000-1000-1800-1800.

We are looking forward to your participation!

  • Vote: I like it
  • +89
  • Vote: I do not like it

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8 hours ago, # |
  Vote: I like it +25 Vote: I do not like it

Because of some serious problems, this contest might be the last atcoder contest this year. So please treasure this precious oppotunity!

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    8 hours ago, # ^ |
    ← Rev. 2   Vote: I like it -13 Vote: I do not like it

    pAxgjA0.png

    I found that it was your first time participating in AGC as a rated contestant. It showed that you actually attached importance to the contest. Wish you good luck!

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    8 hours ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    lol at first i thought last atcoder contest forever

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4 hours ago, # |
  Vote: I like it +17 Vote: I do not like it

Wow, very cute A! Good old $$$1/7=0.(142857)$$$ :D (but with 1/1019 this time)

B seems great too, though I couldn't figure it out even though I got a lot of good observations. I can prove that the result without forbidding anything is $$$2n^n$$$, but it still seems like a long way to go...

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4 hours ago, # |
  Vote: I like it +23 Vote: I do not like it

What made you think of decimal expansion when solving A? I OEIS'ed it :(

I brute forced when X and 2*X have a lot of digits in common, and searched the digit sequence in OEIS. Found decimal expansion of 19

It also referred me here, to try 383 after 19: magic-square-something

Then I wrote the checker to see if all multiples are present, and indeed 19 and 383 worked.

Next item on that list was 32327, which is too big. So I just guessed that maybe the first primes after 1000 would work instead. 1009 and 1013 didn't work, but 1019 worked. Proof by AC!

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4 hours ago, # |
  Vote: I like it +24 Vote: I do not like it

who is gblaasnicse?

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3 hours ago, # |
  Vote: I like it +5 Vote: I do not like it

Problem C can also be solved somewhat easily if you google the right keywords and dig out this paper.

Solution

Thanks for the contest! This is probably the highest delta I've ever received for solving a single problem.

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3 hours ago, # |
  Vote: I like it +7 Vote: I do not like it

I have a question for the editorial of the problem B. Suppose $$$n = 2, p_2 = 1$$$ and $$$G = \{1 \rightarrow 2, 2 \rightarrow 2\}$$$. Then $$$f(G, \emptyset) = 1, f(G, \{1\}) = 0, f(G, \{2\}) = 1, f(G, \{1, 2\}) = 1$$$ and the sum will be $$$3$$$ instead of $$$2$$$, what am I missing ? For $$$S = \{1, 2\}$$$ "consists solely of several vertex-disjoint cycles." does not work.

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    3 hours ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Thanks! The editorial has been corrected.

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114 minutes ago, # |
  Vote: I like it +14 Vote: I do not like it

Thanks a lot for the round!

My screencast with mumbling and a short recap of the solutions at the end. Unfortunately my docking station stopped working about 20 minutes before the end, so there are a few minutes that are cut out, and afterwards there is no screencast, only camera.

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