Комментарии
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-1
I really prefer it in the normal ICPC format than adding the hack stuff. As you said, challenging in TopCoder is useless for me in div1.
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0
From what I've seen in the solutions, it's enough to consider all subgroups of n contiguous boxes, wrapping around. However, I have no idea why it's correct, maybe it's related with the fact that there is always an answer.
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+15
As far I can deduce, the first paragraph says that the event starts at 19:00 and people can register during the contest. The second paragraph mentions the people that helped preparing the contest. The last line should say something like "good luck".
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+5
Can you lock a problem without solving it? Maybe I would like to challenge bad solutions for a problem that I don't know how to solve.
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0
I used to be that way, that C++ was the best language and no other deserved my time. Then I learned python in some free time, and realized how wrong I was. Now I'm trying to learn Haskell, but I have to way until another free-time slot.
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I did something similiar, for each pair of equal numbers, see if there is a repetition beginning at the first with the second at the middle. To check that all the interval numbers are equal, I used suffix array and RMQ to do that in O(1).
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Yes, that part is the same in my solution, but I used all that stuff to get the intervals. I suppose there is an easier way.
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I used Suffix Arrays, Longest Common Prefix and Range Minimum Query to solve this problem.
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+1
But if there is no penalty for unsuccessful hacking, then people might start making hacks recklessly without analyzing the solutions.
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+4
@MikeMirzayanov: You forgot to translate 2 балла into 2 points in the table.
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0
Thanks for the rules, I had fun reading them, specially the "(epic fail)". After reading that, the format looks really interesting. When will be the first public contest with this format?
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0
I know that part, what I want to know if it it's possible to lock a problem without getting it accepted with the pretests.
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+1
Also, can you lock a problem without submitting a solution? What about submitting but without getting it accepted?
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0
Cool =), that makes it even more interesting.
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0
When you lock your code, you lock all problems or only one?
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0
Thanks :), I always preferred pasting the code. Also, my status during the match dissapeared, and now it only shows the practice submissions. |
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Is it intentional that now the only way to submit is to select a file instead of pasting the code? I made a pair or mistakes because of that, because using the current way forces me to lose the tab with the problem.
Is it possible to restore the submit pasting functionality back? |
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0
It's similar to his first proposal, I decomposed C completely while he only checked the necessary factors. |
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0
It can be patched to work in that case. First, decompose c in it's factors, p1^e1 * p2^e2 * ... Then solve for each pi^ei independently and get the final result by merging all individual results using the Chinese Remainder Theorem.
Let mod = p^k, and tot = p^k - p^(k-1). To solve the equation (a^b) % mod is equivalent to solve ((a%mod) ^ (b%tot)) % mod, except when a is divisible by prime p. To patch this is easier. Count how many times p can divide a, let call this number x. Then, in a^b the prime p will appear exactly x*b times. If x*b >= k, return 0, otherwise, there are not enough primes to make the mod 0, so just do the normal thing. Here is the relevant snippet with this part: http://ideone.com/goy8b |