The contest is over; I hope at least some of you enjoyed it :-) The editorial is published here.

The 7th April Fools Day Contest will take place on Monday April 1st. This is a joke competition in which solving the problem is often easier than figuring out what the actual task is.

In this round you'll be given several weird problems and 2 hours to solve them. The contest will use ACM ICPC rules (no hacks, the standings are decided by the number of solved problems and penalty time earned on them), and it will be unrated. You can submit solutions in any language allowed by Codeforces. To get an idea of what the contest will look like, you can check out the contests of the past years: 2012, 2013, 2014, 2016, 2017, 2018.

As usual, to enjoy competing in this round you'll need a sense of humor compatible with mine. Good luck, and have fun!

surprisingly tough ....XD

Good one

It's humor and

Acceptedthat create fun.The best contest of the year :D (not that I perform well in it)

I take back my comment.

before contest: so excited !!! after contest: whatever :)))))

Trying to solve those yes/no problems gets boring quickly when every hypothesis you make turns out to be wrong.

ravioli ravioli, what is in my 1st april contestoly?

I have been waiting for this contest

Will this year have April Fool's contests on other platforms? Last year there were other 2 big ones at least, but I couldn't find any this time.

Is it rated?

No

Yes. It's April 1st so yes.

.

boring guy lol :P

Unrated:Let's solve the problems just for fun!I hope everybody have fun.XD

No problem statements plz

I am tempted to follow this request, since for one of the problems writing the statement (in two languages) took strictly longer than writing all other pieces of the problem, and it certainly will be the case for another problem... But alas, I aim for variety of jokes above all, so some of the problems have to have a statement :-)

"When the contest begins" Surprise! There is no contest :)

"There is contest after 30 minutes, let's open cf" Boom!

Nobody:

Me:

It works XD.

A classified source just confirmed that this round will be rated :P :P :P

It's April Fools, so during the contest suddenly it will become rated =)

Why unrated? :(

I think it will be rated and open hack. because today is fools day

It's April 1st, the contest may suddenly starts 2 hour earlier XD

really?

2012,2013,2014-2016,2017,2018 Mathematically, u shouldnt be doing an april-fools contest this year o.O

I hope, in this year, this will not happen with me ...

image uploading site

It was normal that you got wrong answer. You had to make a program that makes 2 good bets in roulette. The results were made randomly (like in normal roulette). The best strategy was to bet on red or black (even or odd), because you have 48.6% that your bet will be good. That means 23.6% chance to pass main tests (only 2 test cases). While the results were random, the Answer section were made by the code of author that got accepted, so that can't be always good. In your case the result was 10 (black), so you got WA.

so, can I ask what is the contest id of the problem?

952D - I'm Feeling Lucky!

Thanks

To all the other teams which are attending ICPC WF, let's join this contest together!

SpoilerHA CYKE I AM NOT ATTENDING ICPC WF HAPPY APRILS FOOLS

SpoilerHa... Haha....

Exciting！

This is my first April Fool's Day contest. exciting!!!

Fools Day happy~~ : )

This contest won't get rated afterwards, right?

No one knows. It is April's fools contest, anything can happen :D

I hope I won't become a fool after this contest :3

`The 7th April Fools Day Contest will take place on Monday April 1st.`

Nickolas It's 32nd march, fix it.

Is it rated??

No.

real fun

chutiyaap

(https://drive.google.com/open?id=1BEDXr5N2ZR5Hf5Q4jHsbAxPjX69aO-Pu)

Spoilerand then you got WA because you print 4 but you should print "any number"

:)

Goodluck and have fun everybody!

It is should be April fools night contest for me :3

The image in the Russian version of 1145C - Mystery Circuit is broken for some reason, here is what it should be:

Should've supported Q# submission, missed opportunity. Tssk tssk.

It's the first time I get rank1.

I'm so happy.

XD

Wonder how the editorial will look like...

D has great speiling

I warned you, we didn't proofread the statement at all :-)

Did you proofread the sample testcases?

Yes, those are numbers, much easier to check :-)

ans is just Xor ;)

Don't tell me it was joke

yes it was a joke XD

you got April fooled

guessed Xor using D'Or

Due to the unethical and ugly behavior of some members of the community, this round will be rated. Let's solve problems just for fun!

Please post the editorial as soon as possible. Eagerly waiting (to submit :P).

http://codeforces.com/blog/entry/66327

The solutions for other two problems coming up soon...

Too late friend Nickolas, Round is over now. :D

How are so many people solving D can anyone post the answer :D :P

Toughest Constest everCan't even go through the problem statements.

I have seen some previous years April Fools Contest problems but this is the lamest April Fools Contest of all time. No offence

Tutorial for problem B please :)

I made a bet with my senior ray-asuka that if I could not become a candidate master before my birthday（4.2）, I would invite him and my teammates to dinner. This round was my last chance, and he got his wish.

best wishes

as the old saying in China, the most rich man never pubilsh his opinion with someone

Paper

No interactive AI on CF confirmed. :/

Felling like a real fool :(

two plus xor of third and min element?

YES!

how did you even come up with that solution?

How did people guess this on the first attempt, with only one sample case?

It's the wrong letters, they make up sentence.

true... nice logic man !

exactly !! how ??

Xor seems intuitive... but what about min^3rd + 2 ...

Its impossible without extra hints

Note that there are typos in the statement. Write down all the letters that should be replaced by some other letter for the correct spelling. This is the message you get.

My solution for E: if $$$i$$$ gives remainder 1, 4 or 5 mod 6, then yes. Sadly, it's a wrong solution, but it perfectly fits.

ye, I also trusted oeis :(

`((min(id,25)+id)%(2+id%3))>0`

I found it after 1min of the contest...

THAT'S AWESOME!

UPD: Problem EAnd where did this magical condition come from?

It comes from the Fourier transformation of the image 1 to 20. (my poor English :D

Well played :)

That's neat!

Well, I managed to brute force B after 50 attempts...

i do the same for problem C :)

I made the same for both :P

How do you bruteforce C when you need to guess a number? Seems like it would take a lot more attempts :o

Tests turned out to be ordered by numbers in their input)

I made it with supposing that the answer is also between 0 and 15. I wasn't sure on it, but I tried it. Later I noticed that each number has different answer.

Input has 4 bit. Circuit looks like some kind of bit shuffling / adding. So we can guess that output is in range [0, 15] and has some kind of pattern.

--> basically just submit some educated guess. You can look at my submissions to see how I guessed it.

curious about how to brute force on YES/NO problems

Make sure it answers correctly for first 4 tests. Then try to get 5th test with binary search — answer no to everything (else), then yes to 50..99, etc. Eventually I realized the test cases are in increasing order which made things a lot easier — just need to find the next value to "flip" the answer on.

That's pointless on any serious contest because the number of attempts it takes will give you around 0 points.

You: TWOPLUSXOROFTHIRDANDMINELEMENT

Me, an intellectual: LTABRACCANEPGZEFBOHPNOSIFASOMC

An intelleqtual

I knew this fact quickly because i use google translator... kk

And wtf is that "mine element" :|? Got this message pretty quickly but had no idea what to do with it.

Notice how there is only 1 e, not 2 e(minelement isn't mineelement).

fuck

I guess it is "min" element. btw which message are you talking about?

By message I meant hidden statement "two plus (...)"

fece is a correct word, so MIN element ...

... but it took me half an hour to figure it out :D Also had MINE there at first ...

SBEVE

1% of time to read statements, 99% of time to guess and attempt, lol

You missed 1% (I wrote it before he edited his comment).

Why can't I view others' solutions? System Checks?

Only one task... Ahahahah, the best contest)) I want more!

my wrong solution for C: A circle with cross indicates 4(because of cross); A dot indicates 1(because it's a dot); Note that the input number is between 0 and 15, so make it become binaries. for the example: 3->0011->the last line(4+1+1+1) + the third line(4+1+1) =13;

I think this is reasonable but got WA , sad...

It's actually just some quantum logic gates of 4 bits.

QUANTUM LOGIC GATES?

Apparently ,this isn't a NEAT WORD.

Use the most foolish way to spend April Fools' Day lol！！！

I am so happy to be the second person to solve problem C!!

All because I listened carefully in the [Quantum Computing] class when COIWC2019(Chinese Olympiad in Informatics — Winter Camp)!!

lol Are you taking quantum computing classes for the Olympiad?

I don't know XXDD! Isn't that crazy!!??

And wtf is that problem about?

It's just a Controlled Not gate. https://en.wikipedia.org/wiki/Controlled_NOT_gate

4 bits 1-2-4-8

The symbols represents NOT, CNOT, TOFFOLI gates etc.

[https://en.wikipedia.org/wiki/Quantum_logic_gate#Controlled_(cX_cY_cZ)_gates](https://en.wikipedia.org/wiki/Quantum_logic_gate#Controlled_(cX_cY_cZ)_gates)

Don't ask me why is this used in April's Fool instead of Q# winterOMG! I thought they are XOR gates because they were like xor symbol.

Why the problem B wants to ban KAN? Just a joke

in D i used grammarly to get the grammatically correct sentence to make it readable. rip

Solution of A:N/4

really? i brute forced the whole solution :P

In the first test case n was 4 but the answer was 4 not 1

Now I regret because I voted that I love this contest before it starts..

Please help in problem A.

In each step, delete either the right half of the matrix or the left In order to get a part of the numbers ascending sorted Print the longest length of a sorted array you can obtain

Problem A is straightforward — no sense of humor required to solve it.

Will the links to contestant solutions be available?

You should be able to see the solutions now.

congrats to applese for winning the 2019 fools contest!

We are used to look for the editorial in the end of the post, not in the beginning.