maroonrk's blog

AtCoder Regular Contest 111 Announcement
By maroonrk, history, 3 years ago, In English

We will hold AtCoder Regular Contest 111.

The point values will be 300-400-600-600-800-1000.

We are looking forward to your participation!

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3 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

For those interested, I'm planning to do a stream afterwards on Twitch.

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3 years ago, # |
  Vote: I like it +20 Vote: I do not like it

I wish good luck to all participants able to solve a problem in this contest. :/

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

I give up!

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

How to solve E?

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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

How to solve A?

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    3 years ago, # ^ |
    ← Rev. 4   Vote: I like it 0 Vote: I do not like it

    If b divides a, then we can write (a/b) % p = a%(p*b) / b . Since we have to mod the floor value, we can ignore the condition above( b|a ). I used big mod to calculate a%(p*b) My submission

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      3 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      how did you know (a/b) % p = a%(p*b) / b it should be p*b??

      any source or explanation?

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

    Find modulo result of (10^N) with (M^2). The result floor divided by M is the answer.

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      why, can you explain ? and also 10^N is also too large to fit in the memory ?

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3 years ago, # |
  Vote: I like it +20 Vote: I do not like it

To me, B was easier then A. How to solve A?

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    $$$ Answer =\frac {(10^N\%M^2 - 10^N\%M)}{M} $$$
    as we know the property $$$ \lfloor \frac{\lfloor \frac{A}{B} \rfloor }{C} \rfloor = \lfloor \frac{A}{B*C} \rfloor $$$
    Basically you want to calculate A
    $$$A = \lfloor \frac{10^N}{M} \rfloor - \lfloor \frac{10^N}{M * M} \rfloor * M$$$
    let $$$\lfloor \frac{10^N}{M} \rfloor = Q $$$ and $$$\lfloor \frac{10^N}{M*M} \rfloor = Q^\prime $$$
    we know that
    $$$ 10^N = Q * M + R $$$
    $$$ Q = \frac {10^N - R}{M}$$$

    $$$ 10^N = Q^\prime * M^2 + R^\prime $$$
    $$$Q^\prime = \frac {10^N - R^\prime}{M^2}$$$

    So, $$$A = Q - Q^\prime * M$$$
    $$$ A = \frac{(10^N - R) - (10^N - R^\prime) }{M}$$$ $$$ = \frac{R^\prime - R} {M} $$$
    $$$ =\frac {(10^N\%M^2 - 10^N\%M)}{M} $$$

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

      Will you please explain your solution.

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        3 years ago, # ^ |
          Vote: I like it +35 Vote: I do not like it

        Imagine $$$10^n$$$ in base $$$m$$$ and you can see the answer is the second last digit.

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        3 years ago, # ^ |
          Vote: I like it +19 Vote: I do not like it

        $$$10^n = m*k + r_1$$$

        $$$(10^n - r_1)/m = k$$$

        $$$k = \left \lfloor \frac{10^n}{m} \right \rfloor$$$

        $$$r_2 = k \mod m$$$

        $$$k = m*k_2 + r_2$$$

        $$$(10^n - r_1)/m = m*k_2 + r_2$$$

        $$$10^n = (m^2)*k_2 + r_2*m + r_1$$$

        $$$r = r_2*m + r_1$$$

        $$$10^n = (m^2)*k_2 + r$$$

        $$$r = r_2*m + r_1$$$

        $$$answer = r_2 = (r - r_1)/m$$$

        $$$answer = r_2 = ((10^n)\%(m^2) - (10^n)\%m)/m$$$

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        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it
        $$$\frac{10^n}{m}\bmod m=\frac{10^n\bmod m^2+k*m^2}{m}\bmod m$$$

        Meanwhile,

        $$$\frac{k*m^2}{m}\bmod m=0$$$

        Then the answer is

        $$$\frac{10^n\bmod m^2}{m}\bmod m$$$
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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    let y be the answer(0 <= y < M), then the condition is: ( 10N — (10N)%M — y*M)%(M2) = 0

    my proof

    Now how to solve B??

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

      Ty, now i think im able to understand.

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      3 years ago, # ^ |
      ← Rev. 4   Vote: I like it 0 Vote: I do not like it

      Create a graph of colors and keep deleting vertex with degree eguals to 1. In the end the answer is the number of vertex deleteds plus the remainders with degree greater then 1.

      Any vertex wich has degree equals to one, has only one chance to be chosen. In the end of the process of retiring these vertex, vertexes that have degree bigger then 1 can be put in a cycle that make they be chosen.

      I don't know how explain better, my poor english.

      My submission: 19286958

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

        your idea was very clear of showing up the color that occurs once(degree 1) and remove then from graph now consider others. but please would you like to explain what editorial said i am unable to relate it with this.

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          3 years ago, # ^ |
          ← Rev. 3   Vote: I like it 0 Vote: I do not like it

          Every conected component with the number of edges greater or equals than the number of vertex has at least one cycle, so in the process of deleting vertex of degree 1 in this component , it will remain just vertexes with degree greater than 1, so all vertex of the component can be chosen.

          In the case of there's no cycle in the connected component (when number of edges is the number of vertex minus 1), one vertex will remain without be chosen.

          I think this is the way the two solutions are linked.

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

      Amazing proof!!

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

      Make a graph with numbers on cards as nodes and edges as cards. For each connected components, count the number of edges it has (repeated is counted twice, vertices of components only once). If number of edges for a component is greater than equal to the number of vertices for the component, then we can take all vertices as numbers on cards. Otherwise we can only take (number of vertices — 1) for the component as contribution to the answer.

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    3 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    I felt A took more thinking than B and C and D

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

Wow, I spent an hour trying to figure out how to calculate the sum of floors in E, and it turns out that Atcoder contestlib has a function to do exactly that???

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3 years ago, # |
  Vote: I like it -8 Vote: I do not like it

can anyone please explain what is wrong with this solution to D.

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    3 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    consider a single loop of nodes. your set of edges would contain 1 extra edge from the dfs root to one of its children

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

I am wondering why my solution to B is incorrect.

My idea: build a graph using $$$N$$$ nodes as cards and some other nodes as colors. Connect an undirected edge between two kinds of nodes when a card has a color on its either side.

Then I tried Dinic but got TLE. I thought of a linear method but it got 13 WAs. My code is Here, could anyone found what the problem is?

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3 years ago, # |
  Vote: I like it +164 Vote: I do not like it

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    3 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    Could you elaborate? This is wa, isn't it?

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      3 years ago, # ^ |
        Vote: I like it +18 Vote: I do not like it

      Probably judge solution is incorrect. I guessed some common pitfall and got AC.

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        3 years ago, # ^ |
          Vote: I like it +89 Vote: I do not like it

        Ok I got it, it's like you answer the question in the title, ok haha funny

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +45 Vote: I do not like it

    We found this during the contest :)

    Maybe you were glad when you found the title "Do you like query problems?", then opened it, and ...

    Thank you for participation!

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      3 years ago, # ^ |
        Vote: I like it +25 Vote: I do not like it

      xd

      I liked it even though it was not a query problem. Thank you!

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3 years ago, # |
  Vote: I like it -11 Vote: I do not like it

Can anyone please explain the time complexity of D without the constraint that guarantees the existence of a solution to me? From the editorial of D, it is $$$O(N+M+\frac{N(N+M)}{\sigma})(\sigma:wordsize)$$$ but I don't know why.

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    3 years ago, # ^ |
    ← Rev. 5   Vote: I like it +6 Vote: I do not like it

    Soln remains the same. The additional step is checking if the found soln is correct. Given a directed graph, count no of nodes reachable from a particular node. That wordsize is a hint for bitsets.

    Spoiler

    Upd: To the ones downvoting above comment. It wasn't obvious. I had the same doubt when I first read this.

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

      I completely forgot that this task gave $$$c_i$$$ for every $$$1 \le i \le N$$$ so that I was always thinking about the way to check if the graph was strongly connected. That's why I didn't understand this time complexity.

      Anyway, thanks a lot for your help!

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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Can somebody explain how to get the idea that problem B can be solved by transforming it into a graph problem. I could never think that the problem relates to graph. Thank you.

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

    For me, at first I had a wrong idea. But eventually got the solution. First, I thought for every card I can choose only one color — (a or b). May be I could solve it using bicoloring. So, started drawing a graph. And eventually realized it wasn't a bicoloring problem and found the solution that I need to check only if there's any cycle.

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

      why to apply graph i dont understand we have to find max color showing on one side so can not it be like this https://atcoder.jp/contests/arc111/submissions/19280968

      i know its wrong but why what i am not getting

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        3 years ago, # ^ |
        ← Rev. 3   Vote: I like it 0 Vote: I do not like it

        Check this input

        • 3
        • 1 2
        • 1 3
        • 1 3

        Your solution gives 2. But you can take colors 2, 1, 3. So ans should be 3.

        • 4
        • 1 2
        • 1 3
        • 1 4
        • 3 4

        AC Ouput: 4 (2, 3, 1, 4)

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          3 years ago, # ^ |
          ← Rev. 4   Vote: I like it 0 Vote: I do not like it

          Can you give some test cases which fails my code. I try to solve it using SET but failed 2 tests. But could not figure out what cases it fails.

              N = int(input())
    s = set()
    for i in range(N):
        a, b = list(map(int, stdin.readline().split()))
        s.add(a)
        s.add(b)
    
    
    print(min(N, len(s)))
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3 years ago, # |
  Vote: I like it +86 Vote: I do not like it

Problem E's maker has no wood!

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3 years ago, # |
← Rev. 3   Vote: I like it +5 Vote: I do not like it

I just want to note that the final %m in the solution given at the end of the editorial for problem A is unnecessary.

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

can somebody explain why is this getting runtime error?

https://atcoder.jp/contests/arc111/submissions/19387114

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

in my opinion, problem F should not have queries type 2

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