### AlexSkidanov's blog

By AlexSkidanov, 9 years ago,

Hello everyone!

The second round of MemSQL start[c]up will take place on August, 3rd, 10:00am PDT. There will be two contests running simultaneously, one for people who participate onsite, and one for everybody else who advanced to the round two. Both rounds share the problemset and are rated based on the combined scoreboard.

Onsite participants will have special prizes for first three places. All onsite participants as well as the top 100 in the online contest will receive a start[c]up t-shirt.

People who have not advanced to the round two can participate in the round unofficially. Unofficial participation will be rated.

The contest will be 3 hours long, and will feature 6 problems. The score distribution is 500-1000-1000-2000-2500-3000.

The problem set has been developed by MemSQL engineers pieguy, nika, exod40, SkidanovAlex and dolphinigle.

Good luck and happy coding!

UPDATE: Editorial is up!

Announcement of MemSQL start[c]up Round 2

• +138

 » 9 years ago, # |   +28 are other people allowed to join unofficially for fun? maybe in a separate room?
•  » » 9 years ago, # ^ |   +4 Will be div2 edition?
•  » » » 9 years ago, # ^ |   +14 Div2 participants are welcome to register for the round unofficially, but the problemset might be too hard for them. Expect problems A and B to be comparable to Div2-C and problem C to be as hard as Div2-D or E.
•  » » » » 9 years ago, # ^ |   -29 It's ok, I think most of Div.2 participants could solve at least two problem in this contest.
•  » » 9 years ago, # ^ |   +26 Yes, it will be possible to participate unofficially.
 » 9 years ago, # |   0 is it rated for Div2 participants?
•  » » 9 years ago, # ^ |   +4
•  » » » 9 years ago, # ^ |   0 That's not exactly the answer to the Haghani's question. AlexSkidanov only stated, that round will be rated for those not advanced from round 1
•  » » » 9 years ago, # ^ |   0 I hope all the commentators could give your reasonable comments. I don't know why this comment(just contains a "Yes", and it's a fact) got two "down"s.
•  » » 9 years ago, # ^ |   +26 It will be rated for all the users.We welcome Div2 users to participate, but if you register for the round, please be advised that this problemset is not designed for Div2 participants, and you might end up with 0 problems, which will result in a significant loss of rating.
•  » » » 9 years ago, # ^ | ← Rev. 4 →   -6 if i register but don't attempt any problem, then will i loss point? is there pretest as usual cf?Edit: (i asked about rating, cz, there was a notification while registering, it said something like:- "it will not easy for div2, u may loose point, do u still wish to register?" )
•  » » » » 9 years ago, # ^ |   +4 u won't...but the rating isn't the most important thing.Just have fun. at least you can do better than me,LOL
 » 9 years ago, # |   0 Hello I have a question.How can I register the online contest officially if I have advanced to the round two?According to the statement above, both official and unofficial online participators need to register the online version. I assume the system will automatically recognize the official particpator (who passed the round 1), am I correct?Thank you
•  » » 9 years ago, # ^ |   +3 Yes, it automatically registers those who have not advanced to round 2 for the unofficial track. After you register, please double check, than in the list of registrants you do not have a star symbol next to your handle (star symbols means registered for the unofficial track).
 » 9 years ago, # |   +13 Are unofficial participants competing for T-shirts?
•  » » 9 years ago, # ^ |   +15 No
 » 9 years ago, # |   0 Hope a good contest for div 2 participants ;)
•  » » 9 years ago, # ^ |   +21 I'm sure it will be too hard for many participants from Div2.
 » 9 years ago, # |   0 What does the "Unofficial participation" mean? What kind of participants will be considered as unofficial participants.
•  » » 9 years ago, # ^ |   0 People who have not advanced to the round two can participate in the round unofficially. Unofficial participation will be rated. If you register for an online round, but you have not advanced to round 2, your participation will be unofficial (in the dashboard and the scoreboard such participants are marked with a star).
•  » » » 9 years ago, # ^ |   -31 Thank you.And Will there be round-3? If so, a man who have not advanced to the round 2 but get a high rank in round 2 unoffically, can he participant in round 3 as an offical participant?
 » 9 years ago, # |   -50 is it for sql language,or we can participate in any language
•  » » 9 years ago, # ^ |   +5 You can use one of these languages.
•  » » 9 years ago, # ^ |   +26 Can you solve problems using sql? XD
•  » » » 9 years ago, # ^ |   -32
•  » » » » 9 years ago, # ^ |   -19 either this is not funny joke or you don't know what is sql
•  » » » » » 9 years ago, # ^ |   +41 Chuck can solve all these problems even using HTML.
•  » » » » » 9 years ago, # ^ |   +29 or you don't know who is chuck
 » 9 years ago, # | ← Rev. 2 →   +2 Anyone know why profile page says "The page is temporary blocked by administrator." ?Edit: after system tests it keeps msg as before..
•  » » 9 years ago, # ^ |   +10 Because when a contest is running they blocked some pages that are not necessary at this moment.
•  » » 9 years ago, # ^ |   +2 I think it block it until the rates change completely
 » 9 years ago, # |   +26 when is the rating changed
 » 9 years ago, # | ← Rev. 3 →   0 Could anybody tell me some smaller input so that I could figure out the mistake in my code for B. It was hacked on some large random test case when it was possible to have a subsequence of length 100. http://codeforces.com/contest/335/submission/4223108In that case my code prints out some non printable character. I am unable to figure out why ?? Any help would be appreciated !!
•  » » 9 years ago, # ^ |   +5 Small inputs: "a" "aa" "aaa" "aaaa" "ab" "abc" "aba" "abac" "abcbabcc". (This isn't meant as a joke, I really used that to debug my solution :D)Just write a random generator and make your own input(s). It's hard to find a tricky case here, anyway, so any random string should do.
•  » » » 9 years ago, # ^ | ← Rev. 3 →   0 Oh, it was a silly kind of mistake. string t; t += s1; t += s2; where s1 and s2 are two characters is not equivalent to t += s1 + s2;
•  » » 9 years ago, # ^ |   +7 "abababababababababababababababababababababababababcbababababababababababababababababababababababababa"(101 characters in the string, 100 characters in the solution)
 » 9 years ago, # |   +3 Why doesn't multiset in STL have such common operations? Just to find the maximum and the second largest number?
•  » » 9 years ago, # ^ | ← Rev. 2 →   0 What's wrong with *--st.end() and *----st.end() ?
•  » » » 9 years ago, # ^ |   0 Multiset, not the set. And many numbers will be same.But I want the second largest number strictly. For example, 5 5 4 4 3 2 1 MAX 5 The Second Largest 4
•  » » » » 9 years ago, # ^ |   +6 You can use map with values equal to the number of occurences of each key.
•  » » » » » 9 years ago, # ^ |   0 Good idea! Thx
•  » » » » 9 years ago, # ^ |   0 I din't think you need element strictly lower.In that case you may use *--st.lower_bound(*--st.end()).
•  » » » 9 years ago, # ^ |   -9 So we must use BBT
•  » » 9 years ago, # ^ |   -8 It does. set st; ... set::iterator it=st.end(); int mx=*(--it); // max int mx2=*(--it); // second largest 
 » 9 years ago, # |   -19 no one solved problem F. I am curious to know the approach of this problem. btw it was very gud contest. :) :)
 » 9 years ago, # |   +7 When will we be able to register for practice?
•  » » 9 years ago, # ^ |   -47 i dont know why this comment was ratting +.while my comment is usefull more
 » 9 years ago, # | ← Rev. 3 →   +22 i solved B in O(n^2) by using the fact that there are only 26 variety of characters. if n >= 3000 then there is a character that has occurred more than (3000/26) >= 100 times.so I print a string with length 100 with only this character if n < 3000 then the problem would be the same as LPS (longest palindrome subsequence) which is solvable in O(n^2) if don't think this was the intended solutionand problem C is very similar to problem SGU 328. but with lower constraints (it is solvable in O(R))
•  » » 9 years ago, # ^ |   0 I think your solution for problem B is the intended solution, that's why the 100 constraint is there.
•  » » » 9 years ago, # ^ |   +5 Probably. I didn't come up with that solution but a different one with complexity O(100n) that doesn't depend on the number of types of characters.
•  » » » » 9 years ago, # ^ | ← Rev. 2 →   0 Sounds like your idea is better than quadratic. Because with 500 types of symbols it would be necessary to process whole length of string (50000). Could you explain in short words?
•  » » » » » 9 years ago, # ^ | ← Rev. 2 →   +3 Let F(i, j) is the maximum index k > i such that we can obtain a palindrome of length 2 * j from characters in [1, i] and [k, n] (j characters in each part). Actually, we also need to know the maximum index of occurrences that is less than k for each type of characters. In my submission 4225240, I construct a table in O(26n) for finding in O(1). In case the number of character types is large, this can be handled in O(log(n)).So, overall complexity is O(100n + 26n) or O(100n log(n)).
•  » » 9 years ago, # ^ |   +9 Ken Ichi Kawarabayashi would say: "100 is a constant, alphabet's size is a constant, so B can be solved in constant time" :P
 » 9 years ago, # |   +19 Thanks a lot for these nice, short and clear problems.
 » 9 years ago, # |   +20 Why the problem-setters' nicks point to topcoder profiles instead of codeforces profiles?
 » 9 years ago, # |   +41 Thank you for onsite competition!It was perfect!And congratulations to Seikang who won poker tournament!
 » 9 years ago, # |   +20 Thanks for the contest and the onsite event! (great problems, awesome people, and tasty pies) Thanks for the poker lessons. I finally learnt! (and for the prize XD)
•  » » 9 years ago, # ^ |   +34 I'm glad you liked the pies (that was the true prize, of course).
•  » » » 9 years ago, # ^ |   +5 +1 key lime pie, ping pong and beer :) Thanks MemSQL !
•  » » 9 years ago, # ^ |   +12 Damn! I missed the pies while walking around San-Francisco :(BTW, I also enjoy the event very much but I still think that my performance was quite poor and I got the prize spot only by luck.
•  » » 9 years ago, # ^ |   +12 The onsite event was great indeed! Thanks a lot.
•  » » » 9 years ago, # ^ |   0 can you give me some hint on problem F?
•  » » 9 years ago, # ^ |   +28 Yes, it was great! I played the poker first time and it was funny :)
 » 9 years ago, # |   -10 it‘s so geili(this is from china ,means:great)
 » 9 years ago, # |   +6 Will any editorial be published? I would gladly read solutions to E and F.
•  » » 9 years ago, # ^ |   +6 any idea to problem F?
•  » » 9 years ago, # ^ |   +19 It will be published on Monday.
 » 9 years ago, # |   0 Thanks to MemSQL engineers for your contest. I was an offical participant (passed round 1) and finish the online contest with rank 74 (solved 2 problems). So when and how can I receive a T-shirt?
•  » » 9 years ago, # ^ |   +34 We will be sending them out shortly. Someone will contact you to confirm your t-shirt size and address very soon.
 » 9 years ago, # |   +6 when would the editorial be published？
•  » » 9 years ago, # ^ |   +18 Yes, dolphinigle is polishing it up right now and will publish it as soon as it is ready.
 » 9 years ago, # |   +12 Any news about the T-shirt?
 » 9 years ago, # |   +10 is there anybody knows something abouth T-shirts?
 » 9 years ago, # |   +8 Has anyone received their t-shirt yet?
•  » » 8 years ago, # ^ |   +3 and now?
•  » » » 8 years ago, # ^ |   +3 nope! anyone else?
•  » » » 8 years ago, # ^ |   0 Received mine today. Anyone else?
 » 8 years ago, # |   0 I've received my shirt now as well. Thanks for this!