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0
Maybe consider sacrificing some pieces of paper in favor of making round even more ideal? :-) |
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+28
Lol, spent last 33 minutes of the round debugging Div.1B, only to understand after the round that coordinates are not guaranteed to be sorted. In my opinion, such problems problems should contain coordinates already in adequate order. Div.1B/Div2.D should check participant's ability to invent and implement solution, not ability to carefully read problem statement... Otherwise, problem is quite nice |
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0
After AB->BA replaces some new occurrences of AA or BB may appear. But if you will repeat these 2 sets of substitutions until none could be applied any more, then, I guess, this will be correct. |
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+1
Is there any editorial? |
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+10
Is there any upsolving for this finals? Or when and where would it be published? |
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0
Math education also means that you know methods of proving math theorems. Theorems by themselves are mostly useless for CP, but same methods and ideas are used in solving hard CP problems. It's possible to do CP without math. But once you'll reach the limit, because you won't be able to invent solutions for hard problems. |
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+2
My way was to have good math education in school and spend a lot of time using different methods to prove some hard math theorems. Then same methods help to create ideas for solutions during programming contests. |
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0
Oh, shit! Thanks! I was so close to achieving red color and a T-shirt... |
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0
I passed pretest 2 when realized that t must be non-negative. So I took t = 0 if f(t) is minimum at negative t (but I got wa56 then and I have to clue about it) |
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0
I think, that in your solution there is no square that is beaten by 4 initial rooks. |
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On
MikeMirzayanov →
2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred), 8 years ago
+5
Looks like it could be done without bitsets. I've implemented this idea and got AC with time complexity O(N^2 + sum(deg[i]^2)) which is not worse than O(N * M). |
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0
Because programming is math + implementation. And the second part is much easier. |
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+7
Seems like I've seen it before... |
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-43
It will be as easy as IOI |
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+13
I was always interested in how do you choose amont all authors of the contest the one who will post the announcement. The one with lower contribution? Or simply by tossing the coin? How it is done generally and how it was done in this round?) |
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0
Had you found this test by yourself? Or is there a way to see tests in atcoder (after the contest of course)? |
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+10
Great solution! But it looks like we should delete points with x mod 2 = 1 in case when d mod 4 = 2 and delete poits with (x + y) mod 2 = 1 in case when d mod 4 = 1. |
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+11
May be jury doesn't have their own solutions for all of the problems and they are now trying to understand the code of the guys who solved them. |
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+10
Actually, there should be no difference for you: in which round to participate. Except the fact, that if you'll write official one, you'll have chances to pass to the next round. |
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+5
It's never written about ACM, so I think it will be standart CF |
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+18
And after every contest there is a dilemma: go to sleep or wait until raitings are updated... |
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+10
He isn't a great cheater if he is still in div2 |
