### Nanako's blog

By Nanako, 6 weeks ago,

"It's been a long time since I came here, and I've really been through a lot." Koxia muses as she chatted idly with Mahiru.

Now that the Winter Festival is approaching. Among the flash of fireworks, in the chimes of the New Year, what is waiting for them to encounter?

Armed with girlish courage, they stepped onwards.

Hello Codeforces!

We (Nanako, m_99, huangxiaohua, SteamTurbine, triple__a, Nezzar) are very pleasured to invite you to take part in Good Bye 2022: 2023 is NEAR, which will take place in Dec/30/2022 17:35 (Moscow time)!

This round consists of 8 tasks waiting for you to solve in 150 minutes, and will be rated for everyone!

On behalf of the author team, please allow me to express our sincere thanks to:

This round is supported by NEAR. The participants in the top 2047 places will receive prizes as follows:

• Ⓝ 1024 for the first place
• Ⓝ 512 for the 2-3 places
• Ⓝ 256 for the 4-7 places
• ...
• Ⓝ 1 for the 1024-2047 places

Score distribution will be announced soon.

Besides the regular editorial, materials to be public after the round will also include the Chinese statement and the Chinese editorial.

We hope you enjoy our problems and say Goodbye to your 2022 happily!

UPD1: Score distribution is 500 — 750 — 1250 — 1500 — 2000 — 2500 — 3250 — 4000.

UPD2: Congratulations to the winners!

Thanks all for joining!

UPD3: The editorial is available.

UPD4: The Chinese statement and the Chinese editorial have been added into the contest attachments.

Announcement of Good Bye 2022: 2023 is NEAR

• +1318

 » 6 weeks ago, # | ← Rev. 2 →   +28 EXcited to participate the ending of 2022 contest :") Good luck everyone..hoping for positive deltas :)
•  » » 6 weeks ago, # ^ |   +11 Thanks. I think this round will be great and I hope everyone will have a great result. Let's hope that only good memories remain from the year 2022
 » 6 weeks ago, # |   +114 @tester(done) https://codeforces.com/blog/entry/110638 you may start the contribution farm
•  » » 6 weeks ago, # ^ |   +67 As a tester, I spammed my message here so as to not clutter
•  » » » 5 weeks ago, # ^ |   +4 As a tester, I give you respect
•  » » 6 weeks ago, # ^ |   +51 As a tester, thanks for reminding!
•  » » 6 weeks ago, # ^ |   +33 As a tester, the problems are very interesting!
•  » » 6 weeks ago, # ^ |   +36 As a tester, I wish you a happy new year! :D
•  » » 6 weeks ago, # ^ |   +41 As a tester, I tested the round.From my perspective, the problems are really interesting and I strongly recommend everyone to take a chance to participate!
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   +13 As a non-tester, hope i get a chance to be tester someday
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +10 As a tester, I can say it will be a very good round for you~Goodbye 2022 and happy new year~
•  » » 5 weeks ago, # ^ |   0 As a tester, mark. Happy new year :)
•  » » 5 weeks ago, # ^ |   +2 I guess everyone who upvoted the blog regrets now.
 » 6 weeks ago, # | ← Rev. 2 →   +153 Receiving 1 near nowadays is the same as getting 0
 » 6 weeks ago, # | ← Rev. 2 →   +52 As a writer, I think the little cartoon girls in the picture are cute.
•  » » 6 weeks ago, # ^ |   +22 As a participant, I think the little cartoon girl in your profile picture is cute.
•  » » 5 weeks ago, # ^ |   -16 nice profile picture
 » 6 weeks ago, # |   +34 I would participate instead of testing if I knew there was a prize LOL , anyways the quality of the contest is really high and good luck to all of the participants :D
 » 6 weeks ago, # |   -36 The girls are so cute (,,>﹏<,,) I wish I could participate ૮₍ ˃ ⤙ ˂ ₎ა
•  » » 6 weeks ago, # ^ |   +25 Wrong account
•  » » » 6 weeks ago, # ^ |   -117 I graduated from a famous school in mainland Mehedinti, at least I am not stupid. My account is not high because I don’t have much time to train. If I spend a lot of time training, I should be able to train to Master, but I really don’t have that time.
•  » » » » 6 weeks ago, # ^ |   -6 I think you misunderstood the joke
•  » » » » » 6 weeks ago, # ^ |   +67 No, I did not.Marinush
•  » » » » » » 5 weeks ago, # ^ |   0 Marinush is epic
 » 6 weeks ago, # |   +16 As a returning geezer participant, I ask you young uns to take it easy on us.
 » 6 weeks ago, # |   0 Goodbye,2022!
 » 6 weeks ago, # |   +1 As a tester, I can say it will be a very good round for you~
 » 6 weeks ago, # |   +5 Beautiful BG, another step to the Animeforces :3
 » 6 weeks ago, # |   +57 Me and my bois calculating the monetary value of 1 NEAR.
•  » » 6 weeks ago, # ^ | ← Rev. 3 →   +26 1 NEAR Protocol~=1.3 USD
•  » » » 5 weeks ago, # ^ |   +19 How to receive these gifts?
•  » » » » 5 weeks ago, # ^ |   +8 Waiting for an answer,too
 » 6 weeks ago, # | ← Rev. 2 →   -12 Happy new year!
 » 6 weeks ago, # |   -28 Hope I can become candidate master!
 » 6 weeks ago, # |   0 I want to know the difficulty of this game, div1 or div2 or else？
•  » » 6 weeks ago, # ^ |   +1 Div.1 + Div.2
•  » » » 6 weeks ago, # ^ |   0 Thanks for your response（
 » 6 weeks ago, # |   +7 As a tester, the problems are very good! I hope you will have fun in this round!
•  » » 5 weeks ago, # ^ | ← Rev. 3 →   0 Every tester says the same thing. It means something is there. I'm waiting for tomorrow. It's time to turn the table
 » 6 weeks ago, # |   +17 I would like to see Santa Claus or a Christmas tree on the announcement
 » 6 weeks ago, # | ← Rev. 2 →   -32 OK I surrender no more downvote plz
 » 6 weeks ago, # |   +2 all high rating in 2023!!!
 » 6 weeks ago, # |   +31 I wonder if instead of "Accepted" we will see "Happy New Year"
•  » » 6 weeks ago, # ^ |   +1 Only that instead of Wrong Answer was not Happy New Year :)
•  » » » 5 weeks ago, # ^ |   0 we will see *Bye Bye 2022".
 » 6 weeks ago, # |   0 Am I only one who feels that the girl on right is staring at us.
•  » » 5 weeks ago, # ^ |   0 You are too tall. She is probably staring at everyone except you.
 » 6 weeks ago, # | ← Rev. 2 →   0 This year was awesome, wish everyone all the best!!! :)
 » 6 weeks ago, # |   -7 excited for my 21st round.. will try to be more consistent in codeforces rounds in 2023
 » 6 weeks ago, # |   -22 lol
 » 6 weeks ago, # |   +1 Everyone Best of Luck Good Bye 2022 Happy New Year Everyone
 » 6 weeks ago, # | ← Rev. 2 →   +4 This contest will be in my Heart. Good bye 2022. Welcome 2023. Wishing the new year will be good one for everyone.
 » 6 weeks ago, # |   +10 I hope I can get a good result this time and get off to a good start for the New Year!
 » 6 weeks ago, # |   -14 Will this be div1 or div2?
•  » » 6 weeks ago, # ^ |   +22 It's a combined round like Div.1 + Div.2.
 » 6 weeks ago, # |   0 I wish you all the best, more ratings to everyone! :)
 » 6 weeks ago, # |   0 Let's go !
 » 6 weeks ago, # |   0 All the best everyone for the last contest of 2022 and Advance Happy New Year.
 » 6 weeks ago, # |   0 mister is too ORZ
 » 6 weeks ago, # |   0 Thank you for your work!
 » 6 weeks ago, # |   +25 Delete this immediately.Marinush
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   -32 [Deleted] :(
 » 6 weeks ago, # |   0 Really excited for participating in last contest of this year . Started journey on CF this year and learnt alot :)
 » 6 weeks ago, # |   0
•  » » 6 weeks ago, # ^ |   0 solved too much easy problems.
 » 6 weeks ago, # |   0 I'm Nanako_fan!
•  » » 6 weeks ago, # ^ |   -8 :blobheart:
 » 6 weeks ago, # |   +33 One NEAR Ⓝ is about \$1.3. Hope I can get NEAR in this contest.
 » 6 weeks ago, # |   +24 This is my last chance to become Master before New Year's Eve. Wish myself the best of luck!
 » 6 weeks ago, # |   +8 As a participant, I wish all participants have a great time and get positive deltas! Good bye 2022!
 » 6 weeks ago, # |   0 Koxia and Mahiru >>>>
•  » » 6 weeks ago, # ^ |   -10 ;_;
 » 6 weeks ago, # |   -7 the picture is nice!:)
 » 6 weeks ago, # |   +3 EXcited to participate the ending of 2022 contest :") Good luck everyone..hoping for positive deltas :)
 » 6 weeks ago, # |   +24 Chinese Editorial! Good job!
•  » » 6 weeks ago, # ^ |   +14 Thanks >_<
 » 6 weeks ago, # |   -8 As a tester, thanks for reminding!
 » 6 weeks ago, # |   -11 Do anyone know which country is 'Ⓝ' belong to?And what's the value?Thanks.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +11 Virtual currency offered by sponsor NEARhttps://codeforces.com/blog/entry/110638?#comment-986079
•  » » » 6 weeks ago, # ^ |   0 So is it possible to exchange NEAR for cash or is it only a voucher?
•  » » » » 6 weeks ago, # ^ |   +8 I guess exchanging into cash means many procedures but I'm not really sure ;_; maybe you could ask for participants of Goodbye 2021.
•  » » » » 5 weeks ago, # ^ | ← Rev. 2 →   +16 it is possible using crypto exchanges like Binance and Huobi
 » 6 weeks ago, # |   +3 As a participant, i wish you a happy new year! :)
 » 6 weeks ago, # |   0 As a tester, I may ask whether the test is reted or not shinzanmono
•  » » 6 weeks ago, # ^ |   +10 Oops, you are a participant instead of a tester. As for rating stuff plz read the announcement carefully.
 » 6 weeks ago, # |   0 hope to become blue from orange!
 » 6 weeks ago, # |   0 Hoping to enjoy a great contest at the end of 2022 :)
 » 5 weeks ago, # |   +5 I will be ranked #1 in this contest. I am feeling it
•  » » 5 weeks ago, # ^ |   0 Feeling it too please post an update after the contest!
•  » » » 5 weeks ago, # ^ |   0 Sure, Good Luck!!
•  » » » 5 weeks ago, # ^ |   +3 Went as planned. Just 8198 ranks away from what I was feeling xD
•  » » » » 5 weeks ago, # ^ |   +1 Counts as #1 to me, amazing job!
 » 5 weeks ago, # |   -24 girlish power ??? what the **** is that ?
 » 5 weeks ago, # |   +24 Authors, please, read this: https://codeforces.com/blog/entry/56119
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   +32 Oops, thanks for the reminder. Although this round is themed, I'm an opponent of long statements so they will be literally very short and with only necessary pictures.As for the announcement, I have checked with the coordinator and the admin before posting, so I guess it's in a reasonable range. If you don't like it, I personally feel sorry for you, but don't blame other authors cuz it's my personal decision to add it. ;_;
•  » » » 5 weeks ago, # ^ |   0 No, that's kinda cute, that was just a reminder that someones internet connection might be not that fast :)
 » 5 weeks ago, # |   +42 Is it rated?
•  » » 5 weeks ago, # ^ |   +39 Read the announcement carefully and it's a yes.
•  » » » 5 weeks ago, # ^ |   0 sorry if the answer has appeared before, but why this contest currently unrated?
•  » » » » 5 weeks ago, # ^ |   0 It looks like a temporary rollback for excluding cheaters or something.
•  » » » » » 5 weeks ago, # ^ |   0 okay, thanks for the information
 » 5 weeks ago, # |   0 i am so excited
 » 5 weeks ago, # |   0 OMG! a "Good Bye" contest.
 » 5 weeks ago, # |   0 Why so many pfps of animated girls here?
 » 5 weeks ago, # | ← Rev. 2 →   +6 This year I will focus more on learning and skills than on rating
•  » » 5 weeks ago, # ^ |   -10 I thought rating reflects these. Thanks for enlighten us all that rating is calculated from out of blue
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 Ah I meant just blindly giving contests instead of upsolving.It is true that even rating will decrease in that case
 » 5 weeks ago, # |   0 i hope it will be a great contest
 » 5 weeks ago, # |   +11 Last chance to see myself as CANDIDATE MASTER Adarsh8409 on NEW YEAR EVE.
 » 5 weeks ago, # |   -8 can we move the contest to tmr? i would like to participate from 31/12 straight to 01/01
 » 5 weeks ago, # | ← Rev. 2 →   0 As a participant, i wish you a happy new year! and hope all everybody high rating in 2023 :>Good luck to everyone
 » 5 weeks ago, # |   0 have fun!
 » 5 weeks ago, # |   0 As a Desi_Boy from village. Today I'll do my best. I hope one day I'll become good problem solver
 » 5 weeks ago, # |   0 Excited for this!
 » 5 weeks ago, # |   0 Very excited to participate!
 » 5 weeks ago, # | ← Rev. 2 →   -25
 » 5 weeks ago, # |   -38 The comment is not hidden because of too positive feedback, click here to view it
 » 5 weeks ago, # |   0 Happy New Year to all coder in the codeforces.
 » 5 weeks ago, # |   0 Hope I’ll become a pupil this round
•  » » 5 weeks ago, # ^ |   +13 You can, even right now
•  » » » 5 weeks ago, # ^ |   0 Nice!
•  » » » 5 weeks ago, # ^ |   0 haha, good joke
 » 5 weeks ago, # |   0 Hope for good results!
 » 5 weeks ago, # |   -8 I will give back my 1Ⓝ for 1 more picture
 » 5 weeks ago, # |   0 How many problems I should solve to reach 1900 rating?
•  » » 5 weeks ago, # ^ |   0 Just try your best
 » 5 weeks ago, # |   0 Good luck everyone..hoping for positive deltas :)
 » 5 weeks ago, # |   +10 Leaving behind my Christman celebrations. Leaving behind my institute's Cultural fest. Leaving behind a dinner party.Just won't miss 2022's last contest ^o^
 » 5 weeks ago, # | ← Rev. 2 →   +1 To all the participants, good luck in the final contest of 2022, and happy new year! may 2023 be a year of great achievements for all of us! :)
 » 5 weeks ago, # | ← Rev. 5 →   0 Am very sad Because due to travelling i am not able to attend it *_*
 » 5 weeks ago, # |   0 Hope to perform better
 » 5 weeks ago, # | ← Rev. 2 →   0 As a participant, I wish you a very Happy New Year(:
 » 5 weeks ago, # |   +3 As a member of ABC squad, I was surprised m_99 is on the writer's list. I surely succeed in the game!
 » 5 weeks ago, # |   +5 Skibididop dop dop oh yes yes yes Shibididip dip Shkibidop ShkibidiW W W.cout<<(("Good luck")?"DOP":"DIP");
 » 5 weeks ago, # |   0 Hoping for positive rating deltas!!! may everyone a happy new year
 » 5 weeks ago, # |   0 I don't want to miss out the last contest of 2022 more than I don't want to lose rating
 » 5 weeks ago, # |   +13 Last chance for me to be an expert ( + 154 ) before 2023 [ fingers crossed ].
•  » » 5 weeks ago, # ^ |   +16 oops. forget about expert, became pupil again :(
•  » » » 5 weeks ago, # ^ |   0 I can feel you bruh
 » 5 weeks ago, # |   +11 As a participant i am wondering if there are XOR problems
 » 5 weeks ago, # |   -9 is it rated?@2021_yes
 » 5 weeks ago, # |   0 Good luck to you in the last contest of the year 2022 and happy new year!
 » 5 weeks ago, # |   0 Good luck everyone! Happy New Year everyone:)
 » 5 weeks ago, # |   0 Damn..Just realized I wasn't registered and to wait 10 extra minutes for registration pushed me to don't join and wait for virtual, sad.
 » 5 weeks ago, # |   +1 I can't solve the 1st problem (⓿_⓿) Every solution gives me WA on pretest 2 XDXD
•  » » 5 weeks ago, # ^ |   0 first time, huh
•  » » 5 weeks ago, # ^ |   0 I got 5 WA (crying in corner)
•  » » » 5 weeks ago, # ^ |   0 Me too , I got 5 wrongs but still can't solve it!!(-_-)
•  » » 5 weeks ago, # ^ |   0 Think about it... which number should you replace with B_j ? The smallest number, of course. Now just implement this with a set<> or priority_queue. I missed this easy solution and went for the guessing ride.
•  » » » 5 weeks ago, # ^ |   +4 You don't even need to use a set or priority queue actually. Since the constraints are very loose $(n, m \le 100)$, one can simply sort the array $a$ after every replacement of $a$'s smallest number, i.e., a[0] with b[j]
•  » » » » 5 weeks ago, # ^ |   0 I did that but wrong answer on pretest 2 , maybe I missed something so I sm gonna write the code again ✍(◔◡◔)
•  » » » » » 5 weeks ago, # ^ |   0 I see that you have sorted the array $b$ too. You can't really do that because the order of elements in $b$ is important. The operations are happening in sequential order.
•  » » » » » » 5 weeks ago, # ^ |   0 My bad!◑﹏◐ Thanks for your help ❤
•  » » » » » » 5 weeks ago, # ^ |   0 why can not we sort array b. Is it mentioned in the question
•  » » » » » » » 5 weeks ago, # ^ |   0 In the $j$-th operation, we are changing a number on the whiteboard to $b_j$. This implies that the ordering of the array $b$ is significant and plays a role in the problem. Hence, you cannot change the order.
•  » » » 5 weeks ago, # ^ |   0 only set might give you wrong answer. you should use multiset in that case
 » 5 weeks ago, # |   +27 Kinda ruined my birthday ＞︿＜
•  » » 5 weeks ago, # ^ |   +1 Happy birthday, Nayra!❤❤
•  » » » 5 weeks ago, # ^ |   0 Thanks Sweetie, wishing all the best to u :)❤️
•  » » 5 weeks ago, # ^ |   0 Happy birthday, Nayra!!(❁´◡❁)
•  » » » 5 weeks ago, # ^ |   0 Aww Thank u ,may u reach all ur beloved goals ❤️
 » 5 weeks ago, # |   +21 Is problem a harder than usual or am just that bad?
•  » » 5 weeks ago, # ^ |   -29 it's just a very bad question(
•  » » 5 weeks ago, # ^ |   +5 Nope , You are right ! It is a very hard round ◑﹏◐
 » 5 weeks ago, # |   +22 Why such a bad round in New Year ?
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   +6 Why such a bad comment in New Year? To be more precise, what's bad with this contest? I disagree that the contest is bad.
•  » » » 5 weeks ago, # ^ |   0 Problem C was ugly. Did you look at scoreboard? In a Div.1+2 round, only 3k managed to solve C, while there were 12k participants. Does this problem seem appropriate for position "C"?
•  » » » » 5 weeks ago, # ^ |   0 YES.
•  » » » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 Of course. For higher rated participants the order/difficulty of easier problems doesn't matter, right?. They're gonna solve it anyways, why would they even care ?Except that, solution for C was uploaded on youtube during contest. For most of Div.2 people, this will cause something called, Rating Inflation.
•  » » » » » » 5 weeks ago, # ^ | ← Rev. 3 →   0 The scoreboard and predictor says the rating of C would be about *1600/*1700. This difficulty is really not strange and there are a lot of problem C with this rating.About solution leakage, unfortunately I didn't heard about that, but even if that is true, what does that have to do with the competition being bad? Just that behavior is bad, right?
•  » » » » » » » 5 weeks ago, # ^ | ← Rev. 4 →   0 For 1700 rated problem, that amount of ACs is low.Behavior is bad. But the contest Will be Bad too. Its more like, who googles first, that wins. No?
•  » » » » » » » » 5 weeks ago, # ^ |   0 Well, there were some contests which suffered such behavior. But were these contests regarded as essentially bad contest?
•  » » » » » » » » » 5 weeks ago, # ^ |   0 No they weren't. But this C is not 1700 rated (harder than that). Which means this contest is not suited for most Div.2 people. But it depends on how you solved it. By guessing? [Hell yeah, free 1000 points. Such a good problem!]
•  » » » » » » » » » 3 weeks ago, # ^ |   0 Now it got *1700 rating tag.
 » 5 weeks ago, # |   +4 I thought that last contest of 2022 will be easy but It says me goodbye from pupil rating
 » 5 weeks ago, # | ← Rev. 2 →   -30 Man, problems are kinda difficult. Please include problems of Graphs, DP and Trees in A,B,C and D's'. I am only able to solve these but these are usually adhoc and there are no concepts of DSA used in these problems. It doesnt gives the satisifaction to solve these adhoc problems.
•  » » 5 weeks ago, # ^ |   +1 Yeah but you can go for weekly and biweekly on LC
•  » » » 5 weeks ago, # ^ |   0 Yeah, I give them but I want to enjoy cf rounds too. Dont get me wrong, they might be perfect for other users but not for me as I am not that good.
•  » » » 5 weeks ago, # ^ |   0 I think what he meant, problems which boil down to those problems(not specifically to DP, Graphs, Trees etc..) may be an array,Binarysearch,2 pointers or which involve a bit of thinking and observation Because in recent contest there is a lot being packed with pure math, Pruning, Brutally Brute Force problems in A,B,C. If you see my problem topics you will find math being solved more inspite of me being not targeting them at all  Bring some level of math is fine but Bringing whole algebra, number theory like Mobius Inversion, Chinese Remainder Theorem, Some Extended Euclidian Theorem Question which was asked in Problem B and Taking some math problem and solving it and giving it to Arabic Writers. I think they even will create calculus, differential problems and say to print it out as cout<<"My Math Answer"<<"\n"; 
•  » » » » 5 weeks ago, # ^ | ← Rev. 2 →   -15 I think Extended Euclid theorem is kinda standard basic stuff. They can be ok for Div2 ABC
•  » » » » » 5 weeks ago, # ^ |   0 Agreed
•  » » 5 weeks ago, # ^ |   +20 Read this blog. My thoughts are roughly the same.Also funny that you mention D, because D is actually a graph problem after you get some observations.
•  » » » 5 weeks ago, # ^ |   0 Actually, I didnt read D for this contest as I was stuck in C. I was not particularly pointing to this contest but all cf rounds in general.
•  » » 5 weeks ago, # ^ |   0 Adhoc = Math ?
 » 5 weeks ago, # |   +28 As i've expected, Good Bye Expert: Newbie is NEAR! Kinda thrown this one
 » 5 weeks ago, # |   +7 Problem A could've been framed better imo
•  » » 5 weeks ago, # ^ |   0 how did u solved it
•  » » » 5 weeks ago, # ^ |   0 Solutionyou just need to replace the smallest element of array A one after another with elements of array B and at the end the total sum is the ans
•  » » » » 5 weeks ago, # ^ |   0 did the same thing but getting wrong answer on pretest 2PS: i did not sort the second array B
•  » » » » » 5 weeks ago, # ^ |   0 did you first sort A? like without any operations first sort A and then m times swap and sort?Or you can use priority queue too
•  » » » » » » 5 weeks ago, # ^ |   0 i first sort A then just swapped m times if m>=n then basically take all the greater elements of B else we will have to take all the elements of B + remaining largest n-m elements of AI dis this...
•  » » » » » » » 5 weeks ago, # ^ |   0 You'll actually have to swap the smallest element everytime. Let's say A = 1, 3, 7, 8 and B = 3,5,2 Then after 1st swap A should like: 3, 3,7,8 2nd swap 3,5,7,8 3rd swap 2,5,7,8 So it is optimal to pick the lowest element from A everytime and swap it
•  » » » » » » » » 5 weeks ago, # ^ |   +1 oh i forgot that after swapping from B that element itself may be smaller than those in A so I must keep on sortingThanks got it
•  » » » » » » 5 weeks ago, # ^ |   0 well this isn't optimal way cause let's suppose there are my first array is 3 4 and second array is 1 8 6 now there can be three operation's done so if I only take maximum values of second array and swap them with first array values then after three operations my first array will be 4 8 and second array will be 1 3 6 however if we pass on 1 first then we can make our first array equal to 8 and 6.
•  » » » » » » » 5 weeks ago, # ^ |   +1 The thing is that, you need to swap values from array B one after another, 1st element then 2nd and so on That is why I mentioned it could've been framed better
•  » » » 5 weeks ago, # ^ |   0 just used a priority_queue
 » 5 weeks ago, # |   +95 Today's contest was a perfect embodiment of my 2022. Disappointing and making me doubt my very existence. I hope 2023 is better :)
 » 5 weeks ago, # |   +20 OK finally I got it,from Master to Expert.
 » 5 weeks ago, # |   +37 I found problem D amazingly beautiful. what a problem and process of solving it was really good
 » 5 weeks ago, # |   +4 How to solve problem E?
 » 5 weeks ago, # |   +28 Who else made at least one WA on A because they sorted the array b (or equivalent) before performing the operations ?
•  » » 5 weeks ago, # ^ |   0 us
•  » » 5 weeks ago, # ^ |   0 Wow.. That was the reason I couldn't get AC. I thought I could choose the order of operations.
•  » » » 5 weeks ago, # ^ |   0 same...ACed at 2:16
•  » » 5 weeks ago, # ^ |   0 why sorting leads to WA ?
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 If you read the problem statements carefully, you will see that you cannot take any element from array b arbitrarily; you should always maintain the sequence. That's why you shouldn't sort the array. Also, you always need to replace the smallest element of array '**a**' with '**b[j]**'.
 » 5 weeks ago, # |   0 Was problem C related to check if the solution to the congruence equations created by each pair exists or not?
•  » » 5 weeks ago, # ^ |   +5 just to give a hint. if you have 2 odd numbers and 2 even numbers the answer is NO ! now think about 3 an so on
•  » » » 5 weeks ago, # ^ |   0 i tried this method like if we have both (odd,odd) & (even,even) pair then ans is NO otherwise YES.this is what i did but its failing testcase-4.187368333
•  » » » » 5 weeks ago, # ^ |   0 that is what i meant by so on ! now what if you have 2 numbers that have reminder of 1 with divided by 3 and 2 numbers that have reminder of 2 when divided by 3 and 2 numbers that have reminder 0 ?
•  » » » » » 5 weeks ago, # ^ |   0 can u elaborate on C?This was my approach https://codeforces.com/contest/1770/submission/187361827
•  » » » 5 weeks ago, # ^ |   0 can you share your code please
•  » » » » 5 weeks ago, # ^ |   0 check my submissions
•  » » » 5 weeks ago, # ^ |   0 Is there a theory or mathematical prove for this solution ??
•  » » 5 weeks ago, # ^ |   0 i tried it using brute force which i think should have workedby adding i=1 to i=100 and then checking th gcd
•  » » 5 weeks ago, # ^ |   0 I too have the same doubt, thinking the same
 » 5 weeks ago, # |   -9 A disaster -_-
 » 5 weeks ago, # | ← Rev. 2 →   +35 Are these accounts bots? They sent correct solutions all at the same time. There are more of them with the same timing.
•  » » 5 weeks ago, # ^ |   0 I guess somebody wants to steal some NEAR tokens...
 » 5 weeks ago, # |   -6 Definitely not the ending I hoped for
 » 5 weeks ago, # |   +5 I... don't even have the will to make the obvious "anime girl on internet == trap" joke. Happy new year, everyone...!
 » 5 weeks ago, # |   +4 Misread A for 20 minutes and skipped the x is positive part(even it is highlighted) in C. Feels bad.
 » 5 weeks ago, # |   +14 I think the record of difficulty of A in div 2 is broken today
 » 5 weeks ago, # |   +47 Problem E was a great problem. (Hint for people who haven’t solved E: the tree edge e_i divides the tree into two disjoint sets)
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   0 I've noticed it but I don't know how to calculate number of successful moves along each edges
 » 5 weeks ago, # |   0 Problem A is such a nightmare.
•  » » 5 weeks ago, # ^ |   0 for me also bro , 6 wrong submissions, can anyone explain problem A
•  » » 5 weeks ago, # ^ |   0 The solution is simple: iterate over array b and switch element in array b with the smallest element in array a. do this for every element in array b.
 » 5 weeks ago, # |   0 Before this contest I thought after this contest I might become candidate master and have a very happy New Year. After this giving contest I am depressed and sad on the new year. Hopefully the next year will be better.
•  » » 5 weeks ago, # ^ |   +1 Master next year, sir!
 » 5 weeks ago, # |   +35 There were gaps between BC and EF but the problems are very interesting! Thanks for the amazing round in the last of 2022! One regret thing is I can't proof my solution for C during the contest.
•  » » 5 weeks ago, # ^ |   0 Thank you for your endorsement! The proof of C is given in the editorial.
 » 5 weeks ago, # |   0 How to solve E? It's sufficient to calculate numbers of successful moves of each directions on each edges, but how can we calculate it
•  » » 5 weeks ago, # ^ |   0 For each edge u--v, we cut this edge to get 2 subtrees, and let b(u)=number of butterflies start on the subtree which contains u, similarly b(v)Then if m1= number of successful moves from u to v, m2= number of successful moves from v to u, then the contribution of edge u--v is m1(b(u)-1)(b(v)+1)+m2(b(v)-1)(b(u)+1)+(2^(n-1)-m1-m2)b(u)b(v)So how to calculate m1 and m2?
•  » » 5 weeks ago, # ^ |   +8 Consider the dp(u, i), the probability of after i iterations, the vertex u is covered with a butterfly. Clearly, dp(u, i)= dp(u, i-1) if the other edge doesn’t include u. And the probability of u and v, where e_i = (u, v) should be calculated fairly simply, so we can calculate dp(u, n — 1) for all u in O(N) as all values except two changes between iteration. So using the probability array, we can calculate the probability when there is a movement between the two disjoint trees divided by the ith edge, and thus the contribution of the ith edge to the answer.
•  » » » 5 weeks ago, # ^ | ← Rev. 3 →   +8 But what if probabilities of u and v are not independentUpdate: Now I know probabilities of u and v are independent. Because each edge would be moved only once, and before being moved, moves of u-subtree and v-subtree don't intervene each other.
 » 5 weeks ago, # |   +2 goodbyes are harsh
 » 5 weeks ago, # |   +20 Ending the year with -100 ig :)
•  » » 5 weeks ago, # ^ |   0 Oh sh here we go again
•  » » 5 weeks ago, # ^ |   -8 same
•  » » » 5 weeks ago, # ^ |   +5 Oh man, you lost CM after so very work, much harsh!
•  » » » » 5 weeks ago, # ^ |   +6 thank you for understanding)) I'm actually not that upset, since it's only a title and nothing more. Maybe it's still a little bit too early for me, yes, but at some point i'll get there. So i try to be positive and think of it as a useful experience!
•  » » 5 weeks ago, # ^ |   +5 same x)
 » 5 weeks ago, # |   +1 can someone pls explain logic for problem c?Thank you.
•  » » 5 weeks ago, # ^ | ← Rev. 6 →   +3 lets look at prime number p and every pair of 2 elements in the array (a[i] and a[j]). if a[i]%p!=a[j]%p, we can choose every x and independent on x%p gcd(a[i]+x, a[j]+x)%p!=0. if a[i]%p==a[j]%p we can't choose x%p==p-a[i]%p. considering n<=100, we should look at small primes* (and for every prime we have to decide — is there a "good" x%p — if not, there is no solution) *because for big primes there are not so many pairs
 » 5 weeks ago, # |   0 Oh no!How to do D?My idea is to convert it into a forest, and then find the rings. If it is the self ring, answer times n, otherwise the answer is multiplied by 2. But I got the WA on test 2.my submissionAnd how to do C...... I am not good at gcd.
•  » » 5 weeks ago, # ^ |   0 Literally me. Did the same thing in D, got 998244353 WAs on pretest 2
•  » » » 5 weeks ago, # ^ |   +16 Hold up, doesn't that mean you got 0 WAs?
•  » » 5 weeks ago, # ^ |   0 I had the same idea, but in the end i realized that there might not be cycles: 5 1 1 1 1 1 1 2 3 4 5C is (good imo) an observation problem.
•  » » 5 weeks ago, # ^ |   0 your idea was correct for D. however there are some twists. think about these 1 : 2 self rings in a component (or any two rings in a component tbh) 2 : a tree in the forestboth of these can be handled with a simple if ! if the number of edges in a component is not equal to the number of nodes then simply print 0.
•  » » 5 weeks ago, # ^ |   +3 Your solution is correct when the answer is not $0$. For every connected component, the number of vertices and the number of edges must be the same, otherwise there will be same numbers appearing in $d$.
 » 5 weeks ago, # |   -26 I suggest to make the round unrated because the problems are not New Year's)
 » 5 weeks ago, # |   +33 I had a sol for E for a version where the butterflies could go to another node even if there was already another butterfly. Spent 1 hour coding that and noticed it was wrong after finishing.Certainly one of the ways to end the year.
•  » » 5 weeks ago, # ^ |   -24 Please explain C hope u did
•  » » 5 weeks ago, # ^ |   0 Oh no. Haven't noticed this before reading your comment.
 » 5 weeks ago, # |   +1 Good bye, ratings. :(
 » 5 weeks ago, # |   -8 Deadliest Contest Goodbye 2022 :\
 » 5 weeks ago, # |   -15 Why is this code showing error for 2nd pretest in 1st q? Please help me #include using namespace std; int main() { int t, n, m, i, j, k; long long s; cin >> t; long long a[t][100], b[t][100]; for (i = 0; i < t; i++) { s = 0; cin >> n >> m; for (j = 0; j < n; j++) { cin >> a[i][j]; } for (k = 0; k < m; k++) { cin >> b[i][k]; } sort(a[i], a[i] + n); sort(b[i], b[i] + m); a[i][0] = b[i][m - 1]; for (j = 1, k = m - 2; j < n && k >= 0;j++,k--) { if (a[i][j] <= b[i][k]) { a[i][j] = b[i][k]; } else { break; } } for (j = 0; j < n; j++) { s += a[i][j]; } cout << s << "\n"; } return 0; } 
•  » » 5 weeks ago, # ^ |   +8 You shouldn't sort the B array.
•  » » 5 weeks ago, # ^ |   0 You cannot sort array b. Have to perform operations in order. I took forever to fix my mistake on this too
•  » » » 5 weeks ago, # ^ |   0 Oh nooo ;( I wasted much time on this... and didnt do the other questions....
•  » » » 5 weeks ago, # ^ |   0 in the question it is given you have to perform operation but doesn't that in the order
•  » » 5 weeks ago, # ^ |   0 The last element of $b$ must be used.
 » 5 weeks ago, # |   0 All I can say after the contest, is the name of the contest itself ("Goodbye 2022" XD). Hope 2023 turns out to be better overall for everyone!
 » 5 weeks ago, # |   0 I did a weird solution in C that passed the pretests, could someone explain why is it correct or incorrect ? 187365787
•  » » 5 weeks ago, # ^ |   +1 It seems that everyone solved C like this.
•  » » » 5 weeks ago, # ^ |   0 The solution is posted on tg, during the contest.
•  » » 5 weeks ago, # ^ |   0 Can you elaborate your thought process for writing this solution? It would make understanding your code easier.
•  » » » 5 weeks ago, # ^ |   0 For any number k, if the array has two or more occurrences of all of the values from 0 to k-1 modulo k, then that means for any x we choose we can always have two number in the array giving gcd as 1. Let's say x%k=p then we already have two numbers in the array which give the value modulo k as (k-p) so k will always be the gcd in such situation.And also k has to be smaller than 50 if it exists. This is one situation, it can be concluded that if such k exists then there is always an x, but I cannot prove that if x exists then there is always such k.
•  » » » » 5 weeks ago, # ^ | ← Rev. 3 →   +1 Your solution is always correct. I have an intuitive understanding of why, but I can't give a formal proof.It's enough to check all primes $p \le 50$ since if the array is bad for some composite number, it is also bad for all of its prime factors. And now if it was good for every prime, there exists a set of congruencies you need to solve to get $x$:$x \equiv a\ (mod\ 2)$$x \equiv b\ (mod\ 3)$$x \equiv c\ (mod\ 5)$$\cdots$$x \equiv o\ (mod\ 47)$Becase all of the modulos are prime, they are also all coprime with each other, which means that there exists some $x$ which satisfies all of the congruences based on the chinese remainder theorem.UPD: I realised that the set of congruences up to $p \le 50$ aren't enough to gurantee that the specific $x$ works. We would need to add a lot more congruences for larger primes. But we know that there always must exist a valid congruence for each larger prime, which means we can still apply the Chinese remainder theorem to calculate $x$.
 » 5 weeks ago, # |   0 Hey, was the problem C really more difficult than standard Div2 C or was I panicking for nothing??
•  » » 5 weeks ago, # ^ |   +9 Nope, it was in fact harder. It was a bit more math intensive.
 » 5 weeks ago, # |   -8 Ending with still Pupil :(
 » 5 weeks ago, # |   +12 Fuck Bye 2022.
 » 5 weeks ago, # | ← Rev. 2 →   +15 The statement for problem A is quite unclear It doesn't clarify the fact that you have to use the numbers in array b in the same order they're given.It just says that you have to perform all m operations... Spent 90% of the contest's time trying to figure it out and lost a lot of points :(
•  » » 5 weeks ago, # ^ |   +1 i also thought the same and demotivate since it is A problem and not able to solve Question language was very unclear
•  » » 5 weeks ago, # ^ |   +1 The j-th operation is to choose one of the whiteboards and change the integer written on it to bj.I think you misread the problem. Happens with me too many times.
 » 5 weeks ago, # | ← Rev. 2 →   +1 How to solve Problem C. Could Someone Help
•  » » 5 weeks ago, # ^ |   +1 For any $i$ and $j$, if $a_i \equiv a_j \ \mathrm{(mod\ m)}$ for some $m > 1$, then $x$ must meet the condition: $x + a_i \not\equiv 0 \ \mathrm{(mod\ m)}$. Otherwise, the $\mathrm{gcd}(a_i + x, a_j + x)$ will be a multiplication of $m$.Generally, for any prime $p$, $x \not\equiv a_i \ \mathrm{(mod\ p)}$ for any $i$ and $j$ such that $a_i \equiv a_j \ \mathrm{(mod\ p)}$. Also, if there are two or more $a_i$ such that $a_i \equiv r \ \mathrm{(mod\ p)}$ for every $0 \le r < p$, we can't find $x$ since there will be a pair of $a_i$ and $a_j$ which are multiplication of $p$ for any $x$. And otherwise, $x$ always exists. (can be shown by Chinese reminder theorem)So the solution is to check whether $r$ exists such that there are at most one $a_i$ which meet $a_i \equiv r \ \mathrm{(mod\ p)}$ for any $p$. Here, we don't have to check for $p > n$ since $r$ will always exists such that no $a_i \equiv r \ \mathrm{(mod\ p)}$ due to the Pigeonhole principle.Iterate for every prime $p < 100$, and iterate over $i$ in each $p$. Count $a_i\ \mathrm{mod}\ p$ in $p$-length array and check if 0 or 1 exists. If not, print NO.
•  » » » 5 weeks ago, # ^ |   0 Actually, you only need to iterate over every prime $p \le 50$ since due to the pigeon hole principle for every $p > 50$ there must exist at least one $r$ such that $a_i \equiv r\ (mod\ p)$ appears at most once.
 » 5 weeks ago, # |   0 I think C>E>D.
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   +19 C was basically noticing that we only have to check primes under 100 as for primes above 100, by Pigeonhole principle, there would be always a free residue which x can take. I guess the hard part was thinking this as a CRT/prime problem
•  » » » 5 weeks ago, # ^ |   0 Omg shit shit... How it clicked man?
•  » » » » 5 weeks ago, # ^ |   +9 Solving a degenerate amount of number theory problems for MO
•  » » » 5 weeks ago, # ^ |   0 Can u tell it in more detail? I haven't gotten the idea, why can we always find x > 100 to fit that ?
•  » » » » 5 weeks ago, # ^ |   0 Because n is bounded to 100, the number of residue classes that are forbidden is strictly less than the total number of residence classes
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 As n <= 100, at max we can make 50 pairs right?, so isn't 51 enough. Just couldn't wrap my head around this.Edit: It got Accepted, now someone should confirm whether my assumption is True or Tests are weak
•  » » » » 5 weeks ago, # ^ |   0 Oh yeah for each prime, if there is a residue class with less than 2 elements, the prime is good
 » 5 weeks ago, # |   +7 goodbye to 2022 and goodbye to lots of my ratings :)
 » 5 weeks ago, # |   +12 This round was tough.
 » 5 weeks ago, # |   0 I was literally only seconds late from finishing C. Fuck this shit!!!
 » 5 weeks ago, # |   +73 At the last minute!
•  » » 5 weeks ago, # ^ |   0 can you explain your solution please??
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5 weeks ago, # ^ |
Rev. 2   0

I was waiting for it to pass system tests. Now, as it passed, I can explain it

You can say NO only if you proved that all positive numbers till ∞ will cause a problem for at least two numbers.

### Now consider the divisor 3

suppose you have two numbers in the array that if you added 1 to them, they will be divisible by 3,

Now, if you added 1, the first two numbers will not work. If you added 2, the second two numbers will not work. If you added 3, the third two numbers will not work. And lastly if you added 4, it is the same thing as you adding 1 (by taking mod 3).

And the same will happen for 5 and 6 and so all other positiver numbers, so there is no solution.

This explanation is for 3 only, you need to check this also for all other numbers $≤ n/2$

 » 5 weeks ago, # |   0 Greedy came to mind when I looked at problem A, but it was quite a simple one. It seems like a trap to me.
 » 5 weeks ago, # |   +3 contest is hard
 » 5 weeks ago, # | ← Rev. 2 →   +6 Did only I struggle..Edit: (Mistake) Sorry I am soo exhausted rn.. It was in Problem A
 » 5 weeks ago, # |   +16 Goodbye 2022! Goodbye my rating!I think that problems are interesting but a little too difficult QAQ
 » 5 weeks ago, # |   -45 time to downvote
 » 5 weeks ago, # |   -12 2022 hasn't been a good year for me and this contest is just the confirmation , thanks
 » 5 weeks ago, # |   -39 I fucking hate this round
•  » » 5 weeks ago, # ^ |   +16 Clarificaton: because it was too good,
 » 5 weeks ago, # |   0 This Is Probably The Toughest Contest I've Ever Appeared On CF :(
 » 5 weeks ago, # | ← Rev. 2 →   +79 Goodbye 2022. Goodbye mom of authors :))
 » 5 weeks ago, # |   +64
•  » » 5 weeks ago, # ^ |   0 What does that mean, are they same like problem G in today's contest?
•  » » » 5 weeks ago, # ^ |   0 Not exactly the same, G is reduced to this problem, but the reduction is very simple.
•  » » » » 5 weeks ago, # ^ |   0 can you explain how to reduce it? I watched your screencast, where you coded the function solve by hand instead of copying from the old submission. Is it because the reduction is not exactly the same as the original one?
•  » » » » » 5 weeks ago, # ^ |   +27 I'm just stupid. The original code was written before I wrote fft library for myself, and I was afraid that it might have been too slow for this problem (the new code is also very slow smh), so I decided to write it in the new style.As in the editorial, we'll solve for ) and ( independently. Focus on (. Let's say that the minimum balance is $-m$, then we have to remove $m$ brackets before the first occurrence of the minimum balance, and we have to make sure that after removals all the balances are non-negative. It's the same as saying that we have to remove at least $k$ brackets before the first occurrence of balance $-k$. Let's build a Young diagram corresponding to these limitations. Each move will mean the next bracket, if we move up, we remove it, if we move right, we don't. So we are interested in the number of paths from the lower-left corner to the upper-right corner, but some cells are forbidden, as we know that we have to move up at least $k$ times before the first occurrence of balance $-k$. It is easy to see that these limitations describe a Young diagram.Example:$( \color{green}{)} \color{red}{)} \color{red}{)} ( ( \color{green}{)} ( \color{green}{)} \color{green}{)} \color{red}{)} ( \color{green}{)} \color{red}{)}$ ...... ...... .....# ..#### ..#### `
•  » » 5 weeks ago, # ^ |   0 Just curious, how do you keep track of similar problems like these? Or do you just remember bits of problem statement and use that to Google them.
•  » » » 5 weeks ago, # ^ |   +39 This one is just a very natural-sounding problem (count the number of paths in Young tableau) and I was very proud when I was able to come up with the solution in training, so I remembered the circumstances when I solved it (it was a training with my ICPC team, somewhere in 2019, the contest was in CF Gym), so I was able to search the trainings we participated in that period of time. The actual process from my screencast.Good problems stay in your memory, what can I say.
 » 5 weeks ago, # |   +36 Brain Limit Exceeded.
 » 5 weeks ago, # |   +14 Thanks for the round! Regardless of my performance, I think the problems are nice, especially D and E. This is my short comment for the problems.A. Normal problem. I used priority queue in D2A for the first time(although naive works) XDB. Constructive problem. It requires some observation(and maybe some intuition?), normal problem.C. It requires some intuition again :P, but good problem. I got WA on last pretest because I mistook the limit of n as 50. XDD. Nice problem. Graph modeling and following observations are good. I heard the implementation can be hard according to approach, but easy implementation exists.E. Nice problem. The idea is hard, but clear. How to create such problem?F. I read it, but currently has no idea.G, H. Not read.My performance in Good Bye 20xx contest has many ups and downs, this time my performance went up, maybe I can reach GM for the first time?
•  » » 5 weeks ago, # ^ |   0 can u explain C pls?This was my approach https://codeforces.com/contest/1770/submission/187361827
•  » » » 5 weeks ago, # ^ |   0 We have to see all prime numbers less than or equal to n/2. For some prime number p, get the remainder of each element divided by p, and if every number between 0 and p-1 appears twice or more, the answer is NO. If no such prime number exists, the answer is YES. Consider this case, #1 6 5 7 9 11 13 12whatever the value of x is, there should be always 2 multiple of 3, so the answer is NO. Your code print YES in this case.
•  » » » » 5 weeks ago, # ^ |