maroonrk's blog

AtCoder Regular Contest 151 Announcement
By maroonrk, history, 20 months ago, In English

We will hold AtCoder Regular Contest 151.

The point values will be 300-500-600-700-700-1000.

We are looking forward to your participation!

  • Vote: I like it
  • +81
  • Vote: I do not like it

Comments Comments (27)
| Write comment?
»
20 months ago, # |
  Vote: I like it +11 Vote: I do not like it

Only hope I won't fall back to 1 Kyu. I climbed to 1 Dan yesterday.

→ Reply
»
20 months ago, # |
  Vote: I like it +24 Vote: I do not like it

rp++!

→ Reply
»
20 months ago, # |
  Vote: I like it +27 Vote: I do not like it

Hope you all have a good time!

→ Reply
»
20 months ago, # |
  Vote: I like it 0 Vote: I do not like it

such no math, so thanks

→ Reply
»
20 months ago, # |
  Vote: I like it +33 Vote: I do not like it

Grundy Number is not needed in C.

→ Reply
  • »
    »
    20 months ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    But it's easier to solve this problem by using Grundy Number.

    → Reply
  • »
    »
    20 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Can you please explain?

    → Reply
    • »
      »
      »
      20 months ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      First we flip all odd squares, then adjacent squares must be the same. For an empty interval sandwiched between two known squares, the parity of operation times is fixed. Let $$$x=a_1-1,y=n-a_m$$$, we have solved the case $$$x=y$$$ because we can ignore the left and right parts, there one can always imitate the other. For the case $$$|x-y|>1$$$ the first player wins. He can use one step to make $$$x=y$$$ and decide the parity of an interval. For the case $$$|x-y|=1$$$ the first wins if the parity(ignore left and right) is even. If $$$x+y=3\bmod 4$$$ the first wins too. Otherwise the second wins.

      → Reply
»
20 months ago, # |
  Vote: I like it +19 Vote: I do not like it

I got 1588 rating. When will I get blue name?

→ Reply
»
20 months ago, # |
  Vote: I like it -44 Vote: I do not like it

I opened my computer at 20:13(utc+8) and want to take part in this contest, because my rating fell in the last contest. But when I registered, I discovered that I was considered unrated immediately. I was very sad, so I didn't do even one problem.

I think it is too early to set unrated automatically (only 5 minutes after contest started). If this can be 15 minutes, I could do the problem happily.

I know that I will receive many downvotes in Codeforces, but I still want to say it.

→ Reply
»
20 months ago, # |
  Vote: I like it +46 Vote: I do not like it

Can anyone share how they solved F? The editorial looks ridiculous.

→ Reply
»
20 months ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it

Can anyone share the proof of Grundy number for different segments in problem C?

→ Reply
»
20 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain me about grundy number or attach some link to follow ? Its a new term for me , thanks in advance

→ Reply
»
20 months ago, # |
  Vote: I like it +5 Vote: I do not like it

For problem C, the editorial say "If the component touches just one digit, the Grundy number is n."

I can't really understand how is this true. Can anyone please help me?

→ Reply
  • »
    »
    20 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Wish you have understood it. If not, you can consider the definition of Grundy number. Let $$$f_x$$$ be the Grundy number when the segment has limit on only one side. Then $$$f_x = \operatorname{mex} ( f_y \operatorname{xor} 1 (y < x), f_y (y \leq x)) $$$

    $$$f_y \operatorname{xor} 1$$$ is for when we put a number same from the limit on the side, it would split into two games where one has length $$$y$$$ and limit on one side, one has the same limits on both sides whose Grundy number is just $$$1$$$. $$$f_y (y \leq x)$$$ is for when we put a number different from the limit, and we can solve it like the situation upside.

    → Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hello, in the editorial of B (link), it says the number of A’s such that A′<A lexicographically and the number of those such that A′>A are equal from the symmetry what is the symmetry here? how are they equal from symmetry? Can someone provide proof that they are equal using symmetry?

→ Reply