MikeMirzayanov's blog

Testing Round #6
By MikeMirzayanov, 12 years ago, translation, In English

Testing Round 6 starts on May 11, 2013, 20:00 (UTC). Our goal is to test the platform after recent improvements. All of them are in the Codeforces backend, but they affect many lines of the code.

I invite you to take part in the round. It will be Div. 2 + unofficials from Div. 1. It will contain four-five obsolescent problems. But I think it will be interesting for many of you. The problems contain very weak pretests to force more hacks. It will be unrated round.

Many thanks to participants!

P.S. As it is testing round, we do not guarantee stable work and so on.

Announcement of Testing Round 6
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12 years ago, # |
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Good to see that resource is developing. What I can't say about topcoder.

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12 years ago, # |
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So if this round is unrated for both divisions then what the difference between official participation of this round and unofficial participation? as you say "Div. 2 + unofficials from Div. 1"

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12 years ago, # |
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development of resource is good for cf.

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12 years ago, # |
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what is the difference between before and after improvements ?

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12 years ago, # |
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Hope to see those recent improvements as soon as possible!

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12 years ago, # |
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Pleased to see that Codeforces is improving. :)

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12 years ago, # |
  Vote: I like it +28 Vote: I do not like it

Success! =)

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12 years ago, # |
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After weak pretests, there were some strong enough testcases to BLOW you..

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12 years ago, # |
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for problem C how can the answer for test 3 3 1 be 12 ? isn't it 6?

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    12 years ago, # ^ |
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    When will have only three days first day will be always good, second day will be always bad and third day will be always good.
    The only bad event in the test will take place during the second day.
    Now we have 3 events. There must be at least one event in each day, so there 3 options to choose first event of the first day and 2 options to choose first event of the third day. Now there is only one good event left. It must happen on the first day or on the third day. This is 2 more options. 3 * 2 * 2 = 12

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12 years ago, # |
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Can anyone give a hint for problem D? :)

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

    Such polygon exists for all n >= 5. You can take a regular polygon and move its sides preserving the angles. Or knowing the angle pi*(n-2)/n, take some set of distinct segments and try to construct a polygon of them, going counter-clockwise.

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