prabowo's blog

Invitation to TOKI Regular Open Contest #13
By prabowo, history, 4 years ago, In English

Dear Codeforces community,

We are excited to invite you to TOKI Regular Open Contest (TROC) #13!

Key details:

Scoring distribution:

  • Div 2: 100 — 200 — 300 — 400 — 500
  • Div 1: 100 — 200 — 300 — 500 — 500

We would like to thank hocky, fushar as organizers, rapel, jt3698 as testers, and Yoshiyuki as proofreader.

Please register to the contest, and we hope you will enjoy TROC #13!

UPD1: We postponed the contest by two hours so that it no longer clashes with AtCoder

UPD2: Contest is over! Our top 5:

Div 1:

  1. Benq
  2. wiwitrifai
  3. KrK
  4. Egor
  5. hitonanode

Div 2:

  1. neal
  2. tourist
  3. zscoder
  4. maroonrk
  5. risujiroh

Editorial is available here (English on page 8).

You can upsolve the problems here.

Thank you for participating and see you on the next contest!

  • Vote: I like it
  • +147
  • Vote: I do not like it

Comments Comments (29)
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4 years ago, # |
← Rev. 2   Vote: I like it +22 Vote: I do not like it

Woah, a TROC that doesn't conflict with other contest sites? This is rare.

Hopefully I didn't just jinx it.

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    4 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    Welp, nevermind.

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

    Look what you have done? Now it clashes with Atcoder Beginner contest 170!

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

    Jinxed

    :)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

    (read: visible depression)

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +13 Vote: I do not like it

    looks fine :')

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    4 years ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    bruh why did you have to say that

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

Is it rated?

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4 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Yesssss!!!! I've been waiting for this >:D

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

Would you consider rescheduling it as it conflicts with the ABC contest?

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

Auto comment: topic has been updated by prabowo (previous revision, new revision, compare).

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4 years ago, # |
← Rev. 3   Vote: I like it +8 Vote: I do not like it

Friendly bump, the contest starts in 3 hours!

We hope you enjoy our contest and problems!

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

This contest starts in 20min.

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

How tf is neal, tourist, zscoder, maroonrk, yosupo etc. in div2...

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

Auto comment: topic has been updated by prabowo (previous revision, new revision, compare).

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

Can someone prove or at least tell why, in problem Div1C it must hold p % c == 0?

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    4 years ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    C is the number of nails that will become a part of the polygon, alternatively the number of vertices of the polygon. A polygon that has C vertices have C edges. Since we want a regular polygon where each side has the same length, the perimeter must be divided into C exact parts, thus the perimeter must be divisible by C.

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

      Since we want a regular polygon where each side has the same length, the perimeter must be divided into C exact parts

      What do you mean by this? In what way can this be visualized? We just came to the conclusion that a regular polygon with integer length sides and an integer length perimeter must necessarily have a perimeter divisble by c, but why?

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        4 years ago, # ^ |
          Vote: I like it +16 Vote: I do not like it

        In Euclidean geometry, a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length). - Wikipedia

        Since it is equilateral, therefore a regular polygon with $$$C$$$ vertices must have perimeter of length $$$C \cdot S$$$ where $$$S$$$ is an integer (since the length is an integer). Therefore, the perimeter must be divisible by $$$C$$$

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

          Therefore, when looking for the answer with $$$C$$$ vertices, one must first find the $$$S$$$. Afterward, we are then required to determine if the circular array can be into $$$C$$$ parts where each part is of length $$$S$$$.

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

          Since it is equilateral, therefore a regular polygon with C vertices must have perimeter of length C⋅S

          Yeah, this is the part the is making trouble for me. Why exactly does this held true just cause all the sides are of equal length? Is this really that obvious because I still don't see it lol.

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

          Actually, what if you have a regular polygon with 6 vertices and a side length of, say, 2. The perimeter of this polygon is 4 (the perimeter is (side length / sin(180degrees / n) = 2 / sin (30) = 4).

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

            Side length is the length of a side. A polygon with 6 vertices has 6 sides. If each side has length 2, the length of all the sides (also known as the perimeter) would be 2+2+2+2+2+2=12. As you can see, 6 is one of the factors of 12.

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

              Yeah, I misunderstood stuff, thought the perimeter was 2 * radius and didn't see that the arches were given instead of distances between points... Thanks.

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

          Amm I misread and misunderstood everything hah, I thought the perimeter is 2 * r while it's the path around a polygon and I thought that the distances between points are given and not the arches... Thanks.

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

      Actually, what if you have a regular polygon with 6 vertices and a side length of, say, 2. The perimeter of this polygon is 4 (the perimeter is (side length / sin(180degrees / n) = 2 / sin (30) = 4).

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

      Amm I misread and misunderstood everything hah, I thought the perimeter is 2 * r while it's the path around a polygon and I thought that the distances between points are given and not the arches... Thanks.

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