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Обсуждение задачи 1152. Кривые зеркала

How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано Miguel Angel 18 апр 2002 03:24
Thanks :)
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано zrp 21 дек 2008 07:01
maybe you need to DP
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано SevenEleven [Tartu U] 23 дек 2008 17:11
I used bitwise operations (we can represent each subset as number between 1 and 2^n - 1), this AC'ed with 0.031s, may be you can further improve it ..
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано Artem Khizha [DNU] 27 июл 2011 20:14
Well, I used DP on subsets, but can't get faster than 0.093. Is there any special trick?
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано Ainan Ahmed 9 июн 2017 22:18
I used bitmask... AC on 0.015
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано imaginary friend 17 авг 2018 02:15
probably they just upgraded the machine from 2011, that's why you are faster than others :P
Re: How to get AC with 0.02 s (mine is 0.7 s)??(-)
Послано Viktor Krivoshchekov`~ 28 ноя 2019 15:20
I get 0.015 and used queue for each recount