### wuhudsm's blog

By wuhudsm, 11 months ago,

Hello, Codeforces!

We are happy to invite you to TheForces Round #22 (Interesting-Forces), which will take place on Aug/27/2023 17:35 (Moscow time)

What is TheForces Round?

Editorial

You will have 135 minutes to solve 6 problems.

We strongly recommend you to read all problems.

The round is TheForces rated!, after the round you can find your rating changes here.

Discord Server (1000+ people)

Contests' archive

• +57

 » 11 months ago, # |   +7 As a tester, I can confirm that the problems are very interesting. And very enjoyable to solve.
 » 11 months ago, # |   +15 Hopefully, it would be a INTERESTING ROUND for you! We've made a lot of adjustments to make the problems better. Wish you all the best!(plus: plz give me contribution, thx!)
•  » » 11 months ago, # ^ |   +3 Upvoted!
 » 11 months ago, # |   +17 As an author, I'd like to thank everyone who contributed to this round, especially Yawn for testing :DThis is my first time ever to be an author :DCouldn't have been done without my G EyadBT, special thanks to him :)
•  » » 11 months ago, # ^ |   +6 No u u
•  » » » 11 months ago, # ^ |   +6 :pleading_face:
 » 11 months ago, # |   +9 thank you for the round and I hope the problems are interesting
 » 11 months ago, # |   0 This round is very challenging. As a tester, I enjoyed problem solving. All the Best to everyone!
 » 11 months ago, # |   +8 The registration is open!
 » 11 months ago, # | ← Rev. 2 →   0 Is there any simple solution to B?Or is it just casework like I did? if n=1, then miller rabin test if k even, then 0 check 111...222 check 5...evens check with 3 (probably had some issues here ig) check 1...222...(even)
•  » » 11 months ago, # ^ |   +3 when n=1 the concatination of k with its self will never be prime if k>1,but if k=1 it will. thats beacuse when you concatenate k with its self it will be writen like k times 10 to some power puls k and this expression is divisable by k which is greater than 1
•  » » 11 months ago, # ^ |   0