Topcoder SRM 821 is scheduled to start at 12:00 UTC-5 on Jan 8, 2022.
Registration is now open for the SRM in the Arena or Applet and closes at 11:55 UTC-5 .The coding phase will start at 12:05 UTC-5, so make sure that you are all ready to go. Click here to see what time it starts in your area.
Please take a look at our How to Compete guide to understand Topcoder Algorithm rounds better.
Some Important Links:: Match Results (match results, rating changes, challenges, individual test case results), Problem Archive, Problem Writing, Algorithm Rankings, Editorials and Older Editorials(SRM 710 and before),
Best of luck!
- The Topcoder Team
There was some momentary bug preventing <100 and >100 character titles being submitted, so we created an exactly 100-character title for this particular round.
The stage leaderboard is pretty misleading. It displays and sorts by the total score over all SRMs in the stage, but your rules claim that the tiebreaker is the total score over the SRMs all tied competitors took part in.
Stage leaderboard: https://tco22.topcoder.com/competition/algorithm?tracks[algo-tco22]=1&tracks[algorithm-tabs]=3
Rules: https://tco22.topcoder.com/competition/algorithm?tracks[algo-tco22]=1&tracks[algorithm-tabs]=2
I was only checking the standings on clist before, and I accepted that I failed to qual. When I finally glanced here, it gave me false hope :(
Actually, I also have a question. Do people qualified from previous stages count in the tiebreaker rule (regardless SRMs to be part of together)? Seems like it affects something here.
Noted and taken forward, thank you. There were no current ties for the qualifying position, but in such a case the decision would be made based on the rules themselves. Please let us know any other questions that may arise, and we'll do our best to clarify.
Thanks for consideration.
I'm still not exactly sure who's qualified from stage 2 based just on the rules. Can you maybe list the people, please?
tourist from stage 1, Petr from stage 2.
Sorry, I was more curious about those who advance to round 4.
Gotcha. We will publish official advancer lists as soon as possible and make sure to also comment on this thread with it as well.
Sure, thanks.
Also one more important question -
Let's say -
A participated in SRM X (div. 2) and SRM Y (div. 1).
B participated in SRM X (div. 1) and SRM Y (div. 1).
Also, let's suppose A and B are tied. Would points of SRM X & Y both be taken into account for breaking ties? Or points of just SRM Y would be taken into account?
Div2's 1000 pt problem is often as difficult as Div1 250 pt. It's very likely that A would have much higher points than B if we take into account SRM X.
Topcoder_Updates Topcoder java applet isn't working for me, "Failed to validate certificate, the application will not be executed"
It was working fine till last srm, I already have exceptions added in java, I am using windows 10, please help.
EDIT: Ok, it's weird, I downloaded applet again from here it worked, but earlier one from here didn't.
For the first one, I am getting
Caused by: java.lang.NoClassDefFoundError: com/topcoder/client/contestant/CoderFor second,
Cannot grant permissions to unsigned jars. Application requested security permissions, but jars are not signed.So you too aren't able to open the applet?
Yes on Ubuntu. For some reason it works on my Mac.
Try this this or this
This worked for me, thanks!
Have you fixed the error with unsigned jars?
No I have tried both solutions mentioned here https://itectec.com/unixlinux/running-unsigned-javaws-code/. But doesn't work.
Is there any solution to the Divison 1 250 / Division 2 600 which does not involve the (row+column)%3 idea? I can think of several adhoc solutions but can't prove to myself that they work, and obviously a huge number of people have been successfully challenged. I studied over a dozen passed solutions and they all have (row+column)%3. Just curious, thanks!
(row-column)%3
Jokes aside, I don't think there's a non-heuristic (i.e. proved) way to do it other than splitting into 3 sets such that any 3 consecutive cells cover them all. This is just the simplest way. It's a well-known trick, you'd do well to learn it.
What if it is modified to a minimization problems and ask what is the least number of bullets are fired?
There were only 2 working solutions in Div2 for B, less than for A.
One tends to underestimate it, because it is "only B". Patterns wouldn't work. I calculated the ranks for each free cell, at first how many free direct neighbours does it have. That wasn't enough even for pretests. But with neighbours of neighbours (in same direction) it was. Of course failed system tests. Maybe it would work in a similar way. Just moving through all possible 1x3 and 3x1 windows, and ranking free cells — in how many free windows does the cell participate. Then mark the highest ranked cell, and repeat from begin.
I started with that approach and it is particularly good for sparse battlefields (with many obstacles) but it doesn't work well on some battlefields with all/many free cells and "bad" shape (like 8x5 etc).