TCO20 Round 2A and Parallel Round of TCO20 Algorithm Round 2A are scheduled to be held on Thursday, July 9 at 07:00 UTC -4.

Both the rounds will be _rated_

Please note that you must register for this round in the Arena. Registration is now open for the round in the Arena or Applet and will close 5 minutes before the match begins, so make sure that you are all ready to go. Click here to what time it starts in your area.

**Members eligible to compete in Round 2A include:**

- Members who won an Automatic Berth* a.k.a Bye* for Round 1

- Advancers to Round 2 from Round 1A or Round 1B

**Members eligible to compete in the Parallel Round include:**

- Members who did not qualify for Round 2

- Members who didn't compete in Online Round 1

- Qualifiers for Round 4 or TCO20 Algorithm Finals from Online Stages 1,2 and 3

**Don’t know how to compete in Topcoder Algorithm Competitions?**

Check out this guide to successfully compete in an algorithm match.

You can compete using:

- Topcoder Java Applet - You can refer to this guide here to set up the applet. (Note that those who have Java 8 installed on their machine will see a security issue — You will have to add Topcoder in security exceptions in Java Control Panel. Please refer to the details in the guide.)
- Topcoder Web Arena(Beta) - Please watch this video for step by step guide

Best of luck to you in the Arena!

Gentle Reminder: The rounds begin in 4 hours.

Why is this contest not held on a weekend :(

It becomes really hard for people who are working to participate in the contest.

Agree! But a lot of factors come into consideration while scheduling, however, next year we will try and keep TCO Rounds on weekends preferably. However, there's a Round 2B on Saturday, July 18 12:00 UTC -4. Looking forward to seeing you compete in one of these :)

Does the Parallel Round have same problems? Also, are these Div1+2 combined?

Yes, parallel round has the same problems and there are 3 problems, open to both Div I and Div II members

I don't seem to be able to register (although I got a bye, according to https://tco20.topcoder.com/competition-overview/algorithm/algorithm-byes). Anyone else having the same problem?

Checking and sorting it out!

You are already qualified by Stage 2 SRM points, to TCO20 Round 4 directly. Congratulations!

Oh, wow, didn't know that. Well, good for me then :)

Haha! I was checking everywhere — missed checking you for Round 4 :D Thanks square1001

Maybe not good, round 1, 2, 3 are opportunities to increase your rating. Competing in parallel is harder :)

How to find the string for l=7 in problem B?

Try to find any period with digits 1,3,7,9 Shortest length of the period is >7 which is why some many failures

Any Intuition behind this, cause I thought it was trivial to solve for a number which is prime for 6 out of 7 possible cycles. And then print the number in such a way so that the 7th cycle is non-prime.

It's because if we take any digit of 2,4,6,8 it will be divisible by 2 when that digit comes at last place. Same with digit 5(divisible by 5).

I was trying using only 1,3,7 :(

Not really a fan of B which involves pattern finding via program and then hardcode the pattern :( (ALL submissions in our room has a pattern string in their code)

I hardcoded the solution for B as well. During the challenge phase I found dreamoon_love_AA 's solution and was awestruck by how beautiful it was. In my room, dreamoon_love_AA, me and IH19980412. He ruined all of our dreams by hacking all the B's solution in the room.

After long time to do experiment, I got to know "repeating L-digit prime won't work when L = 7" and luckily this findings help me gain 3 challenges :)

Well, the key point of this problem is "strongly believing that it is possible to gain perfect score" and I think problems of this type tend to have low success rate in SRM.

I just googled something like "primes rotation" and found wikipedia article about Circular primes where I read the same thing which you discovered after long experiments :)

Hovewer, it didn't help me to solve the problem in time, but also gave some additional points on challenges ^_^

Hope Topcoder has the function that we can remove our submission. Before the contest end, I found my solution is wrong but I cannot withdraw it...

Not sure about others. But for me, both 300 and 400 were finding patterns. Too much work for 75 mins.

I think 300 was fine to some extent. But 400 was boring.

Actually, 300 is not too much work. It is actually solvable in 180 seconds...

(Some coders actually gained more than 297/300 points for this problem)

Editorials: https://www.topcoder.com/2020-tco-algorithm-round-2a-editorials/

The URL doesn't work.

Sorry — try now!

It works, thanks.

what is the cutoff for being eligible in next round?

Once the results are official, top 200 will advance. See the diagram from TCO rules

Exercise: What happens for L=8?I can die without knowing it.

If my computing are correct, then there is no string of length $$$n > 32$$$, where each substring of length $$$8$$$ is prime.

Thanks, my life is changed after this information :)

I do not know how did you validate your answer, I implemented dp solution in compelxity $$$O(4^l \cdot n)$$$ based on observation that answer must contain only digits (1, 3,7 and 9) — I think such idea can be used for bigger values of $$$l$$$ in case there is solution for each prime substring.

I had bug during contest, but I can rewrite it fast if you have different way of calculation.

Well, I just generated all $$$8$$$-digits primes, and for each of them calculated all valid transitions (if we have some prime P = p1p2...p8 then I add directed edge to any prime Q = p2p3...p8x). And now if there is cycle in graph, then there exist string each substring of which of length $$$8$$$ is prime. Otherwise, we are interested in the longest path (for $$$L = 8$$$ it's length is $$$25$$$).

Upd. Ah, now I see that this is exactly what was written in the editorial...

Now I'm interested in what happens in $$$L = 9, 10, 11,\dots$$$ and so on.

However, the cycle

existsif $$$L = 19, 23, 317, 1031, 49081, 86453, 109297, 270343$$$. That's simply because, repunit $$$11111\cdots111$$$ will be a valid number.