rng_58's blog

CODE FESTIVAL 2017 Qualification Round C
By rng_58, history, 7 years ago, In English

CODE FESTIVAL 2017 Qualification Round C will be held on Sunday (time). The writers are sugim48 and wo_.

Contest Link

Contest Announcement will be posted later.

This is one of the three qualification rounds of CODE FESTIVAL. In total, 20 foreign students will qualify in three rounds (Check the official site for detailed rules). If you are eligible for the onsite contest, please don't forget to fill the form at https://krs.bz/rhd-itm/m/codefes2017_en. Please check the detail of the tournament at http://mirror.codeforces.com/blog/entry/53502.

The contest duration is 2 hours, and there will be 6 problems. The first 4 problems are mainly used for choosing domestic students and much easier than other tournament competitions. However, we added two more problems and we hope these are interesting and challenging enough for choosing 20 qualifiers. Note that there is no time penalty for incorrect submissions. The time penalty is MAX, not SUM.

The point values are 100 — 200 — 400 — 700 — 1600 — 1800. If you are unfamiliar with AtCoder System, 2X-point problem in AtCoder is as hard as TopCoder's d1 X-point problem.

Let's discuss problems after the contest.

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7 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Reminder: The round starts in 24 hours.

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7 years ago, # |
  Vote: I like it +78 Vote: I do not like it

Tokyo! Tokyo! Tokyo! ヾ(≧▽≦*)o

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7 years ago, # |
  Vote: I like it +45 Vote: I do not like it

As always, hard problems were great :).

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7 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Why does wire penetrate exactly A + B + C — 2 blocks (problem E)? Can someone please tell me the proof?

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    7 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    Because gcd(A, B) = gcd(B, C) = gcd(A, C) = 1, the wire will never pass through one of the vertices or edges of the cubes, except the start and the end.

    Basically, if you trace the position of the wire and look at it's position (x, y, z). Every time one of ⌊ x⌋, ⌊ y⌋, ⌊ z⌋ change, it means you've hit a new cube. Because they can't increase at the same time (see the argument above), and we have A - 1 changes on x direction, B - 1 and C - 1 changes on the y and z, plus the cube from the beginning, we'll hit exactly A - 1 + B - 1 + C - 1 + 1 = A + B + C - 2 cubes.

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      7 years ago, # ^ |
        Vote: I like it +15 Vote: I do not like it

      Thank you very much. That is a very nice proof.

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7 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Can someone please provide any hack for this submission

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

    Good day to you,

    Try eohqetj .. imho the answer shall be 7 (is that right?)

    Good Luck & Have Nice Day

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