Hey All!
Topcoder SRM 773 is scheduled to start at 12:00 UTC -5, Dec 21, 2019. Registration is now open in the Web Arena or Applet and will close 5 minutes before the match begins.
Thanks to misof for writing the problems and coordinating the round.
This is the last SRM of Stage 1 of TCO20 Algorithm Tournament.
Stage 1 TCO20 Points Leaderboard | Topcoder Java Applet | Next SRM 774 — Jan 11, 12:00 UTC-5
Some Important Links
| Match Results | (To access match results, rating changes, challenges, failure test cases) |
| Problem Archive | (Top access previous problems with their categories and success rates) |
| Problem Writing | (To access everything related to problem writing at Topcoder) |
| Algorithm Rankings | (To access Algorithm Rankings) |
| Editorials | (To access recent Editorials) |
Good luck to everyone!
Reminder: 2 hours to start! Good luck and enjoy the last SRM of 2019!
Haven't been able to get past the "Completing login... One moment please..." screen in the web arena for the last fifteen minutes.
Would recommend you to use the applet. Web arena has known issues :(
http://topcodr.co/SRMGuide
Nice problems, but 900 wasn't hard enough for the last spot. My solution is to replace all tests by the maximal one, compute LCM and try adding divisors of the LCM that increase GCD(sum, LCM) until LCM divides the sum. Simple and effective.
Funnily enough, adding divisors of LCM(1..22) greedily in descending order till the total sum is LCM(1..22) also works.
So the task gets more interesting if you can't add elements with value less than $$$23$$$
Nope, you can find the answer with simple greedy even if you will forbid adding numbers that are at most $$$22$$$. You can look at my submission.
At least it wouldn't let me make an unhackable solution trivially.
Div1 500 was the unlucky problem publishment for misof. The very similar problem has appeared on AtCoder Japanese Student Championship Qual Problem D — Classified. Just this time sheyasutaka was few months earlier.
However, in Div1 500, it was not specified to "minimize" the number of airlines. If "minimize", the thinking step will be much trivial and I can come up with solution in few seconds. But this time, since it was not specified, it took about 4 minutes to come up with it. So I emphasize that it's somewhat different.
It's $$$\lceil{log_2(N)}\rceil$$$ the minimum number of airlines required for the Div1 500 problem ? If it's true how to prove that is a lower bound ?
If $$$2^k < n$$$, then you can find two vertices which are always in one part. So there is no edge between them.
Hahaha, wtf is that 500pt. It should be like B on some educational round. I've seen this like million times, it's so super standard trick.