Hello, Codeforces!
IzhitskiyTimofey, qualdoom and I are glad to invite everyone to participate in Codeforces Round 846 (Div. 2), which will take place in Jan/25/2023 17:35 (Moscow time).
This round will be rated for the participants with rating lower than 0x834 (i.e. 2100). Participants with a higher rating can take part in the round unofficially.
You will have 7 tasks and 2 hours to solve them.
One of the problems will be interactive. Make sure to read this blog and familiarize yourself with these types of problems before the round!
I want to sincerely thank everyone who provided invaluable help in preparing the round and made it many times better:
74TrAkToR for
rejection of 10+ tasksan idea for a problem, help in preparing the round and excellent coordination!KAN for upgrading the quality of statements.
055D, maomao90, maks_matiupatenko, loggerr, vdv09, alex.kudryashov, FynjyBath, alexey.shchepin, Alcabel, induk_v_tsiane, Siberian, polosatic, FedShat and tryharder for testing the round and useful feedback!
MikeMirzayanov for Codeforces and Polygon platforms!
You for participating in this round.
This is our first official round on Codeforces. We sincerely hope to your participation. We hope that you will like the proposed tasks!
The score will be announced closer to the start of the round.
We wish you good luck and have a good time! See you in the round!
UPD: Scoring distribution: $$$500-1000-1250-1500-1750-2000-2500$$$
UPD: Round is unrated. We're sorry — it's our fault.
UPD: Tutorial and comment about task C Once again, we apologize for the inconvenience caused.
It clashes with codechef starters 75 https://www.codechef.com/START75?itm_medium=hpbanner_1&itm_campaign=START75. Is it possible to change the time ? Thanks
You should be knowing that Codeforces >>>>>> Codechef.
Codechef tasks can be pretty good too, at least the ones at the end. Well, clashes are inevitable so just upsolve the contest you decide to skip.
their plagiarism checker is extremely bad... their community support is really bad...
I was plagiarised for one contest, in which I solved zero problem, and tried to solve one problem 16 times,,, and still I was plagiarised...
How can you plagiarise someone, who solved zero problems and tried to solve the problem 16 times !!!!!!...
if you copy code from someone, why wouldn't u get it right ???
https://discuss.codechef.com/t/successful-plagiarism/104943/2
UPDATE : I received response from codechef moderator regarding the plagiarism. According to them, I had solved 2 problems in contest and got plagiarised on 3rd problem.
Yup, it always have been Sir. Looking forward to it. will upsolve last 3 questions of codechef (they are worth it though).
Oh no that is an outdated statement. These days codechef problems are really really good.
Well, Codechef postponed it, but now they shouldn't have.
You mean Bits-chef??
As a tester, the tasks are quite interesting and the statements are clear.
What a joke is this round ? Unrated....
Why C cant be solved in given constraints? What's the problem??
Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains
Maybe there is, u will get a lot of money if you find one
my code is giving 12 answer for this what should be the answer ?
the answer is 13.
input is wrong n = you have given 9 times 1 , i got it now
Sorry about my shit testing(((
Git gud LOL, no harm intended!
It feels weird how this issue wasn't caught by so many people during the setting and testing phase.
i am happy because if this round was rated i would get minus :D
same, i barely made it to 1300 last round
how many questions you solved on this round for now
one completely, but i got stuck on some edge case on second one, not even trying now.
Me too, I took a really long time to solve A and B, and I have no hope of solving the other ones.
i did A in 33minutes :))
if this contest rated,@huanghaoxiang will get 1400+ rating!
HOw u had only solved A
Probably also c before it got deleted
As a clown**
Tester is me
Tester is me
Participant is me!
best testers
As a tester I can say that I am a tester
Young talent
Yeah
What did you test?
clashing with codechef, it would be great if timing is changed
Maybe codechef can adjust the timing too
I hope you know its codeforces who's helding on unusual time(saturday,thursday, etc), codechef's schedule is fixed
I hope now everybody got the answer Unrated round!! Thanks to the downvotes btw https://mirror.codeforces.com/blog/entry/111828?#comment-995528
BRUH MOMENT
Come from Real ID :eyes:
😢
On a side note tomorrow's problem are great and well tested :D We hope you participate
Yeah sure 😄 😄
*Clashing with codeforces . You mean right ?? Codechef should change the timing.
I Think it's Bitmask Round , I hope it is a misconception
I hope it is not (I like those type of problems)
almost OrangeForces lol
Codeforces round is not clashing with codechef round. Codechef is clashing with codeforces.
absolutely right.
Haha. Light Hearted Neme. Loved it Sir.
Love from India Sir.
I didn't get the joke the first time, sorry.
I take my words back ;(
DISAPPOINTED
omg orange round
Hoping to solve till D in this round.
wish every contester good luck and happy rating++ !
"One of the problems will be interactive."I think it will be "D".
Masters' Round!
Will the rating update of this round before educational #142?
Update: Now the rating has been updated.
What should I do To become Specialist
Practice ..
yes I will Thanks :)
hope i can solve problem C，so that i can change a color 。 i dont like green
this didn't age well
Fortunately, I solved C in the last thirty minutes, but there was some wrong with C。IS i solve C？（cry）
IS THAT A JOJO REFERENCE??????!!!!!!!!!!11!1!1!1!1!1!1!
No, you're watching too much anime.
bruv, the whole world is a jojo reference
Unrated?
yea
I too rushed to see the blog after the announcement.
unranted* they even made a mistake in the spelling TT. no offense
Why will this round be unrated?
The solution of the Problem C which was solved by the author is wrong.
I was off to a great start, and then they make the round unrated :)))))
so C is unsolvable or what?
Idk man, I just used a simple greedy which did pass the protests, but greedy algorithms are hard to prove
Greedy is wrong.
Consider following example: 102 people like dish 1, 104 people like dish 2.
Tables are: 51, 51, 26, 26, 26, 26
Pretests must be weak as shit lmao
I think even setters must have thought greedy was right
It's not that the pretests are weak (the round wouldn't have been made unrated if that was the reason), but the problem setters probably didn't realize that the greedy solution is actually incorrect.
is the answer not 179?
The answer should be 206, right? I think that's what my solution would give?
If your strategy is to assign biggest group of people to biggest table then this approach fails on this test:
1 9 3 1 1 1 1 1 2 2 2 2 4 3 2
answer is 8? edit. got it, 9
It is 9.
Assign 1's to the tables of size 3 and 2, 2's to table of size 4.
What answer should it be for this case? Edit: sorry, should have refreshed the page.
My strategy was assigning the smallest that fits the table. That would still work?
1
11 3
1 1 1 1 1 1 2 2 2 2 2
5 3 3
11
Explain your algorithm more deeply since I have misunderstood smth.
Maintain a (multi)set with the group sizes. For each table (sorted in descending order), find the smallest group that has at least that size, and put on the table, and adjust the group size in the set. If none exist, put the largest among the set.
EDIT: I didn't prove correctness, but I tried to anticipate the problem with the direct greedy approach
your solution fails on this test case: 1 10 4 1 1 1 1 1 1 2 2 2 2 3 3 2 2
answer should be 10
1
11 4
1 1 1 1 1 1 2 2 2 2 2
5 3 2 1
Ans: 11
Yes, you're right; I get 9
My code worked too. It gave me result 11
Hello sir, can u tell me ur greedy algorithm?
How about no?
Why not? Round is unrated.
Because its incorrect
Same
Unrated??.. First time solved 3 questions in 40mins
"Unranted"
what an absolute bruh moment
Oh, I think there will be plenty of rants actually...
quite ironic indeed
Solved A+B+C in 26mins , thought I would finally become a specialist :")
But turns out the round is unrated. Sad :(
Anyways, Nice problems , thanks to the authors <3 !
Hello sir can u tell me ur solution for C? I am curious and ur help will be greatly appreciated. Thank you.
It's greedy approach. But will fail on certain testcases. so yeah..... No solution exits in given constraints.
Sad that round will be unrated. Anyway, I enjoyed solving problems, especially D
I was getting some positive delta (110+) after a long time. and now it's unrated. was it really necessary to make it unrated?
Edit:- ohh c is not solvable that's why it became unrated!
Yes, since problem C is unsolvable. The problem setters thought that a greedy approach would solve C but it turns out that greedy doesn't always work. During the contest they realized that C is actually unsolvable within the constraints.
And no, it wouldn't be enough to just not count C towards the ratings, because different people spent different amounts of time on C and it just wouldn't be fair.
What is "can't be solved under given constraints"?? Last I saw, 2752+ correct submissions are there on C
same question?!
Yeah I'm kinda confused since I thought C was kind of easy. Maybe the test cases are weak?
Hello sir, can u pls tell me your solution for C? It will be greatly appreciated. Thank you.
Though the round is unrated it would be great if this is discussed after contest, ig?
did you use greedy approach ??
can you solve for this,,, lets say..
25 people wants to eat dish 1. 15 people wants to eat dish 2.
tables are 15 , 13 , 12 .
greedy wont work here... I was stuck here... also, I got stuck in B somehow... got 3 wrong subs..
is the answer not 38? table 15 dish 1 -> 15 satisfied table 13 dish 2 -> 13 satisfied table 12 dish 1 -> 10 satisfied
Answer is 40.
Everyone who wants to eat dish 2 sits at table 1.
The rest of the people (people who like dish 1) split themselves between table 2 and table 3. Therefore, there are no dissatisfied customers.
why the greedy wont work here? isn't the answer 15 + 13 + 10 = 38?
Answer is 40 ...
we will make 25 dish-1 people sit on 12 + 13 table...
and 15 people from dish-2 on table 15 ...
idk man, maybe pretests are well below the constraints.
Maybe they mean that the mistake cannot be solved within the time constraints of this competition, as the mistake would need be corrected in just a few minutes.
This might make some sense
Nevermind, this comment explains why the greedy (which I at least did) doesn't work.
I suppose they mean the actual testcases, not the pretests (which are not comprehensive).
I skipped C, solved D, and after 5 minutes round became unrated. Not cool.
same
The way it was going was almost sure of becoming CM today and it became unrated
+1
+1
+1
+1
kinda sus
almost 3k people solved a unsolvable problem :/ how? misread? :/
Maybe they used greedy approach, but it was actually wrong.
the round is unranted, not unrated guys.
What does that mean?
that means round is unranted
I hope I get positive ranting delta
nothing here
The problem maker have their faults, but you're not expected to be so rude.
how is problem C not solvable
because it cant be solved using a greedy algorithm. If you have used greedy algo, try this : 7 people want dish "1", and 5 people want dish "2" and we are given 3 tables with accommodation 5, 4 and 3
SPOILER : the solution is 12 guests can be made happy and not 10
Garam krke thanda kr diya -_- .
Lol
Here We Go, After Solving ABC Under 30 mins, the round is unrated. WoW.
I feel you, this was my best performance in a long time and the round becomes unrated
The only reason u solved c is because the problem is wrong
Can u tell me solution for C? It will be greatly appreciated. Thank you.
N=17 M=5 Guests : 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 Seats : 4 4 3 3 3
16
I wanted to know what the greedy solution actually was, cos the greedy solution that I came up with was something I already knew didn't work. So i was wondering what greedy solution others came up with and thought worked but actually didn't
maintain a priority queue and sort b in reverse, and then greedily pop the maximum element from the pq and and assign them to the maximum table currently available, if not all of them fit in that table then add num_of_people — size_of_table to the priority queue and repeat the process. But this fails on so many test cases so yeah
Oh right. That was actually the first greedy solution that I thought of as well. Kinda weird that so many ppl just assumed that it would work when it doesn't.
How is problem C unsolvable ?
Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains
Oh! I understand now. My code (greedy) is giving output for this as 12.
But, ideally, it should be 13.
Can anyone explain why Problem C can not be solved?
I think the intended solution was a greedy algo, but it appears, that there are some tests, where it doesn't work
Maybe the intended solution of C is not fully correct and maybe exist some counter case of this solution which makes problem more complicated than supposed to.
Unrated. Thank you so much.
Why even this single comment can receive downvote I can't understand
Why unrated? Sad. I think constraints are ok
Codechef round was postponed for an unrated round.
Who would have thought the sequel would be as good as the original https://mirror.codeforces.com/blog/entry/103170 https://mirror.codeforces.com/blog/entry/103108
18 coders could not find this before and what were the testers doing. what a waste of time:(
...
... Trust me its not.
I was on the verge of becoming a specialist after so long and now the round in unrated..sigh :-)
same here :P
Is it ranted?
*unranted
I first time passed pretests in D problem (hopefully AC) in contest , At least there is something to be happy.
same with me first time i did a d problem in contest
What does "unranted" mean?Unrated?
it means whatever happens please don't rant about it
How did so many people get AC in C?
Only pretests are run during the contest and the solutions probably would've failed when run against the proper testcases.
how did the testers code pass then
The judge's code probably used the same greedy algorithm everyone else used. They didn't realize that it is actually incorrect before the contest.
The test data is too weak and the testers used the wrong method to produce the test data.
can we make it the first contest then where editorial is updated before the contest ends.
+165 Delta and it's all gone. Thanks for the great round !
It say's unranted, what does that mean?!
No rate updates for the official participants. You can find your rate history in the graph in your profile.
Can someone explain what they mean by that problem C is unsolvable in given constraints?
my rank would've increased this round :(
This is just sad.
Bye Bye +125. Top 500 performance.
what extension are you using?
https://chrome.google.com/webstore/detail/carrot/gakohpplicjdhhfllilcjpfildodfnnn?hl=en
Congratulations for not being a guessforces Andy and thanks for adding me to the friends list lol
Thanks. :)
Me too :(
muda muda muda muda muda muda muda muda
bye bye +100 (real)
I was wondering that how come C be 1250 worth of points but seem impossible to me. I am dumb but not that that dumb(hopefully).
It had an easy-to-think-of greed solution, but now it seems to be a wrong one. The correct solution may be the Knapsack problem, but it cannot achieve the required time complexity.
How did so many people falsely solve C? I stared at it for like 20mins and had no idea, but seeing that many people solving it so fast, I started to doubt myself. I tried some stupid greedy ideas but all failed on paper.
Easy hacking greedy passed the pretests.
Given it was only 1250 points, proof by AC is easier to try than a real proof or looking for a counter example :P
Can you please explain what is the issue with the constraints?
exactly what I thought lol
Speedcoding just does that to you
I guess this really says something about how many people "solve" problems by simply guessing a reasonable-looking greedy.
B was guessable too
I am still not sure, how to solve problem B optimally...
It's ideal to split the array into 2 subarrays.
darn it... I am so dumb :(
I mean maybe, but the point is that C is literally unsolvable — so anyone who actually proves their solutions wouldn't have solved it.
I think the problem is that everybody writing was trying to get AC — and the greedy prrof is somewhat easy to think up really fast. None of the testers really struggled on the task, except me.
The issue should have been caught by me — when we were "testing", I did not manage to solve C (it was B then) in contest, and I submitted like 7 wrong (all greedy) solutions to it. Then, when I was asking the author about the solution, I was told that it is "a simple greedy". Then, I decided to believe him and did not upsolve that task. I should have caught it.
So I think that the CF system of less points with more time will always incentivize this sort of "half-done" proofs.
I proved B before solving it. But when I came to C I just guessed some unproved greedy approach and got WA for silly mistake then I started doubting this approach and couldn't come with another one.
My problem solving skill is slowly goes from random guesses to prove-first approach or even partially-proved approach after watching many streams from tourist, um_nik, and many other legends.
Is this a common thing among highly experienced users when it comes to simpler problems (say div 1 A-C)?
If you go to the "status" page and look at all C AC's, the fact that most of them are 15ms should point at a sub-quadratic solution
I see many people solved C !!
why c is unsolvable!?
why the round unrated?◑﹏◐
Seems that greedy solution of C is not correct
Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains
+114 ...and then its unrated
Testcase for problem c that dosen't work whit the greedy metohd 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5
Great!
don't do it unrated pls
Not Happy with the contest making today :(
This contest is a disgrace to the Joestar Bloodline!
What was the problem with C?
did you use greedy approach ??
can you solve for this,,, lets say..
25 people wants to eat dish 1. 15 people wants to eat dish 2.
tables are 15 , 13 , 12 .
greedy approch will fail,, for 25 dish guests we can pick 12 + 13 = 25 , and 15 for rest.
I see. How did I not see that lol? Speedcoding I guess. What would be a good dp formulation for this assuming bounds are low enough?
after solving A , I tried solving C... couldn't solve that...
it is basically knapsack problem with 'K' sacks given to us...
where 'K' is number of distinct elements given in array.
Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains
Is the answer 12 ? I mean its working fine on my local environment. can someone hack my solution please I am feeling to much smart for getting my solution accepted.
u put the 9 ones at the three 3 ppl tables and the 4 guys at the table of 5 and the answer is 13
What the fuck?
The solution is $$$8$$$, not $$$9$$$?
It is 9, author's solution is wrong.
Maybe the standard code also used the wrong greedy method and fails on this.
C looked so hard to me with given constraints. Looked like a multiple knapsack problem. The knapsacks are your guests that like dish $$$i$$$ and the items are the tables. In this version you can keep feeding a full knapsack but gain no score. I tried greedy strategy on papers, they all had edge cases. Couldn't find a dp. Best algo I found was like $$$O(m^{\sqrt{n}})$$$. I really wonder what happened there :)
the dp i came up with was something like dp[i][j]->max satisfied customers considering till ith type and till j seats. so dp[i+1][j]=max(dp[i+1][j],dp[i][k=0 to j] + min(count[i+1],summation till k)) ps:i did sort the tables though
I'm not sure what you mean with "count[i+1]" and "summation till k". Have you tried your solution with the counterexemples to many greedy approaches that were given in the comment ?
Only if the contest makers had a stand for stopping time like ZA WARUDO
Notice that it is unranted instead of unrated. Does it mean that this round still needs to be rated?
UPD:Now this round has become unrated. It is really a frustrating round.
In an interactive problem, TLE means i am taking more operation than the available operations ?
Read the interaction instructions carefully: "If your program performs more than 30 operations for one test case, subtracts a number x greater than n, or makes an incorrect request, then response to the request will be -1, after receiving such response, your program must exit immediately to receive the Wrong Answer verdict. Otherwise, you can get any other verdict."
why this round is unrated??
sorry to hear that it's unrated..
and a wrong example for many solution including mine:
the answer is 7 instead of 6.
What is that extension bro?
Do you mean why the answer should be 7?
let the room be $$$c_1, c_2, c_3$$$ , then we can match $$$c_1 \rightarrow 2*3$$$, $$$c_2 \rightarrow 1*2$$$ and $$$c_3$$$ the same.
If you use a greedy like me, the match would be $$$c_1 \rightarrow 1*3$$$, which is obviously wrong ;_;
Carrot
sorry, i misunderstood...
it's Carrot, also CF predictor is another good choice.
I'm curious how these testers test this round?
lmao
Bye bye my hopes and chances to become pupil...
Funny that a large portion of the top 100 participants didn't even attempt C because they knew it to be impossible
Trash Russian Round.
There has been 2 times in the past year when CF rounds and CC starters rounds are planned to clash with each other. One of it was postponed both times, and in both occasions, the round that is not postponed became unrated [Lol]
Why the announcement says
Unfortunately, the round will be unranted. We apologize, we made a mistake in problem C and it cannot be solved within the given constraints.
instead of "unrated"?
Unranted is not unrated, so it's still rated.
Well, Unra'n'ted.
So, does the std solve this problem with greedy algorithm? lmao.
answer:
7Could have attended the Codechef contest instead. Whatever ...
That actually has been shifted to tomorrow ,so can be attended .
Good news, thanks!
How to prove B?
If g divides a and b, it divides a + b. That implies [there was typo mistake] gcd(a + b, c) >= gcd(a, b, c), which means that if you have some partition you wouldn't get worse solution if you delete some intervals.
Let's say your optimal answer has k partitions, whose gcd is 'x'. We can merge the first k-1 partitions and the gcd will either increase or stay the same.
Suppose that k >= 3 in optimal answer. Let d be the answer for testcase. Then you can combine some two neighboring segments in one segment and get a solution no worse than the previous one. It's because if d | a and d | b then d | (a+b) and you got answer for k — 1. So in optimal answer k = 2 and you get your solution
you only need to split array into 2, If you get some gcd 'x' by splitting more than once, then you can club all untill there are 2 subarrays because each subarray is multiple of 'x' and sum of them will also be multiple of 'x'.
Why not just remove problem C and extend the round by 15-30 mins rather than declare it unrated?
because some people have spent time on this problem, and some people haven't
Guys go easy on the Downvote Button
Mistakes were made
To all the people that solved C, How did you fakesolved it? I'm interested in the "expected solution".
Greedy. Didn't even realize it was wrong lol
As a contestant it is highly expected to come up fast with a wrong solution by intuition especially with greedy. But as a round tester a plenty of time is there to test the problems thoroughly. Such kind of mistakes wastes time of others.
I completely agree.
Such a frustrating moment solved 3 problems in 30 mins and then what??UNRATED. Good Bye +170
considering solution of B is cringe and everyone typed it without any proof i guess its fair
The proof is simple though. If $$$gcd(a,b,c) = k$$$ then $$$gcd(a + b, c) \geq k$$$ since $$$ k \mid (a + b)$$$.
Hi @AndreyPavlov & fellow setters and testers
Questions are great, mistakes happen! I really enjoyed the questions and logic used. Doesnt matter if round is unrated but I really enjoyed the questions. Thanks for the contest. Much love <3
Indeed. The problems(except for C) are very interesting.
For C,
1 7 3 1 1 1 2 2 2 2 2 2 3Actual answer is 7 with optimal selectiontype 1 table 3type 2 table 1, 2But Greedy gives 6type 2 table 3type 1 table 1, 2one person with type 2 is not satisfiedVolveré y seré millones
respect. GOA T
Those who solved C what was your approach?
Greedy
I wonder why coordinator and many testers didn't even realize this problem.
Imagine Masters not recognizing NP-Hard problem when they see one.
when masters see NP-Hard problems they say: no problem.
Is C unsolvable? We could just add stronger tests and re-judge all submissions. It is solvable surely because the first AC solutions were by GMs. Would like to know more about the reason behind taking such a big decision, skipping other alternatives like re-judging solutions.
It can be proved to be unsolvable in polynomial time by reduction to 3-partition. The grandmasters who solved it were wrong.
topg for a reason
Can you please elaborate a little?
to prove that a problem is NP-hard you can consider another problem known to be NP-hard, 3-partition in the comment by Everule, and then show a relation such that if this problem is solvable then 3-partition is solving.
this means that this problems is atleast as hard as 3-partition which we do not believe to be solvable in polynomial time.
everule is topG because he never guesses and anyone who never guesses is topG
I see thank you. :)
In C, sum of all numbers is bounded so you can't really make a reduction.
3-partition is only NP-hard in number of elements, not when the sum if bounded
Prolly, I didn’t look into the reduction, was just explaining what the process is
You mean the other way around, reducing 3-partition to this problem?
And even then it doesn't imply unsolvability in polynomial time (probably)
I may have messed up the direction of reduction, because I don't have experience in theoretical CS, but solving this problem in polynomial time breaks the fact that 3-partition is strongly NP-complete. i.e., That even if the integers are bounded by polynomial in $$$n$$$, which it is, the problem is still not solvable in polynomial time.
Let us pick a set $$$S$$$ of with sum and size divisible by $$$3$$$ to 3-partition. We bound every element in $$$S$$$ to $$$sum(S) \times \frac{3}{4n}$$$ and $$$sum(S) \times \frac{3}{2n}$$$. This is also strongly NP-complete. Now we make $$$n/3$$$ groups of people, each consisting of $$$sum(S) \times \frac{3}{n}$$$ people, and tables consisting of the sizes of $$$S$$$. The optimal solution to this is a 3-partition if it exists. Notice that any solution with all must be a 3-partition due to the bounds on set sizes of $$$S$$$.
I just say that if P = NP then you can solve NP-complete problems in polynomial time
That's not polynomial solution, that's pseudo polynomial solution, as the total sum, T, is bounded.
I believe 3-partition can be solved in $$$O(NT^3)$$$ (which is pseudo polynomial) with a 3D DP, similar to double knapsack.
3-partition is strongly np-hard, so probably you can't even solve it in poly-time when the input is given in unary (base 1).
it looks like you are confusing dividing a set into 3 subsets with dividing an array into triplets
edit: oops, looks like it was everule who confused them. you're right that the reduction everule posted is solvable in pseudopolynomial time.
Yeah I was wrong, splitting an array into 3 sets with the same sum isn't the same as splitting it into triplets...
I wonder if it's a first proven NP-hard problem on Codeforces
Yeah, I've fixed the reduction, I messed up while looking for it.
3-partition can't actually be solved in pseudopolynomial time (assuming P != NP). And it's probably a different problem then you're thinking.
The problem is: given an integer B, and a set of $$$3n$$$ integers, partition the set into n groups, such that sum in each group is equal to $$$B$$$.
This problem is strongly NP-complete, so even if the numerical values in the input are encoded in unary (so a value of $$$A$$$, takes $$$A$$$ bits to encode), it is still NP-complete. The reduction from 3-partition to problem C is as follows:
Given arbitrary instance of 3-partition, where numbers are encoded in unary. Then make $$$n \times B$$$ guests, divided into n groups of equal $$$a_i$$$, each of size $$$B$$$. The set of $$$3n$$$ integers S, just make that into $$$c_i$$$'s. Now you can show that answer to problem C for this input is $$$n \times B$$$ iff the instance of 3-partition was solvable, Almost... One problem we face is that inside one bucket there can be more or less than exactly 3 elements. To fix this, we can add a sufficiently big enough number to each number in the set S, and add three times that number to B. This enforces that not too many numbers can end up in one bucket.
And it's a polynomial time reduction, because $$$n \times B$$$ is polynomial in the input size, (and even if you increase by that sufficiently big number, you can still choose that as polynomially as big as the input).
Jeroen there is another cool result, that makes the constraint on triplets quite easy in the wikipedia article
I find it so funny to see many people complaining about the round going unrated saying: "I solved three problems" when they literally fakesolved one of them.
Some of them solved A, B, D. So their complain is genuine.
"Rant" means "roar".
Then "Unrant" means "No roar"
Then writers means that "The contest is without roar."
Well...
For problem D,
my idea was to keep subtracting by 2^input — 1 until it eventually hit 0.
Is this wrong? I think it's probably wrong, but looking at the #of submissions is making me think that smth like this might work. Can anyone tell me if this is the intended sol? Please no other spoilers about any intended solution btw.
try with 35
Thanks, I see the problem now. I'll try smth else and pray it works lol
good idea, but i think there is an arbitrarily large number with weird binary sequence that can counter that, because the number could've just decided to troll you lol (A.K.A the input is always small, in which the queries would exceed 30). better way is to subtract it in a really interesting way to the point that its final form would be 2^input-1.
Yeah, I found a solution that will work. I'm such an idiot lol. Will code it up after school. Edit: Got AC
I tried the same.
It does not work. Offline I tried for 10**9 and it needed more queries than the limit.
I would've become red today, if only it was rated :( /s
author, tester and contestant all guessed solutions
Contest would have been fine if C was just removed during testing or something
People wasted time on it , that's why this is not fair
"If C was removed during testing"
Yes !
some people got pretests passed fast and others took time
this time affected other problems
so it is not fair(although I was really good in the contest)
No, you don't understand. By testing I mean (and I suppose the original commenter also meant) the phase where testers try the problems before the contest.
If you mean this then I fully agree
Yeah I meant that only
Nah , some people who knew that greedy solution is wrong may have had wasted time on C which they could have used to solve other problems.
ABCD in 45 minutes and really near of reaching CM
If the author did a mistake it's ok ,but what were the testers doing !!!
I think the main testers job is to find things like this
There is no need to be harsh on tester guys! Everyone makes mistakes . I hope fellow testers learn from this and try not to do such mistakes in future!
Does anyone figure out the correct answer to Problem C even if it may get a TLE or MLE?
You can just brute force all the possible combinations. There is probably also some dp solution, but it has non-polynomial time complexity.
I've found an easier version of problem C, how to solve it?
Try to break your limits and make a solution
If I'm not mistaken, this task is also NP-hard and you just need to write some heuristic to solve it.
I would get about + 120 and the round was unrated ToT. Anw still a good contest
Is the rating change just a guess or is there anywhere to calculate this estimate?
There are extensions like CF Predictor or Carrot that let you see the rating change you are going to get while the contest is still running.
These update live during the contest depenging on how you're doing compared to others but they are not always exactly correct (typically the actual rating change is within +/- 5 points compared to the predictor).
F
Actually today's situation remind me of a contest held many years ago, whose div1.b was also a greedy hacked by a talented coder Um_nik
Funny you should mention me. Screencast, timestamp on when I send the clarification. Another good moment
I think the testers/setters should be required to submit the bruteforce solution (for small but as large as possible cases) along with the intended solution.
Maybe C was a blunder, but after all last 3 problems are quite a challenge, so the setters are not as bad, as you think
Are you sure?
I'm curious. You didn't even prepare a brute force solution to stress-test problem C? I thought stress-testing is a "must" when you prepare a problem?
agree... completely agree with you...
Most normal contest in Ohio 💀
Thanks for the problems. D was a good problem btw but it seems more suited as a replacement for C.
gcd round :(
No, we say gcdforces.
NPforces
And this round
its still nice problemset mens, shit happens, but i like this round anyway
Use me as a dislike button
so sad it got unrated, even though D is a fun question to do (as a math nerd and interactive problem fans lol :p). also, how do you dp problem C?
Problem C is NP-complete meaning that there is no known solution that can solve it fast.
i thought about using dp but with the priority guest group being different for each dp (A.K.A dp brute force because better than brute force^2 i guess :p), but idk how to do that. Guess my concept is kinda similar to that.
You can solve it as a greedy
It can be fake-solved using greedy
I want divide something, is it OK?
Me after the announcement:
for the sake of interest, it's still worth doing strong system tests against the greedy approach. my solution has been undergoing stress testing for more than half an hour. sorry for poor english
Problem F seems intresting. How to solve it?
Sort the numbers, find the divisors for each number and then use mobius function with prefix sums and some tricks.
Regardless of problem C and the round being unrated, I really enjoyed the problems and it could have been my master round.
Thanks for making an amazing contest :D
Agree if C was removed with the problems starting from D being shifted, it could've been one the best rounds made recently :)
Yes, the author's solution in C was incorrect. We haven't noticed it in 9 months. We stressed this solution (as it turned out, we stressed it terribly), I even wrote an analysis with a proof of this greed. Now an exercise for the reader, find the error in this tutorial:
First, let's remember for each dish how many other \bf{unplanted} people like it, let for dish $$$i$$$ this value will be equal to $$$b_i$$$.
The following greedy solution works: let's try to take a dish $$$j$$$ such that $$$b_j$$$ is the maximum of the array $$$b$$$, consider another free table with a maximum capacity of $$$k$$$. Then you can put $$$\min(b_j, c_k)$$$ guests at table $$$k$$$, and change $$$b_j := b_j - \min(b_j, c_k)$$$. After that, you need to continue the algorithm, if there are still free tables and unsettled guests. If there are unseated guests at the end, we will seat them at any tables — they will not be satisfied, that is, we need to minimize the number of free seats.
Let's prove this solution:
1) Obviously, it is not profitable to put dish $$$i$$$ on the table and at the same time not seat the maximum possible number of people who have dish $$$i$$$ as their favorite.
2) It is profitable to take the maximum $$$b_i$$$ and seat $$$k$$$ at the table with the maximum $$$c_k$$$. Let's assume that we seated at table $$$k$$$ not $$$b_i$$$, but another $$$b_j \le b_i$$$. Then if:
$$$b_i, b_j \le c_k$$$; then it is obvious that it is better to seat $$$b_i$$$, because there will be $$$c_k - b_i \le c_k - b_j$$$ free seats.
$$$b_j \le c_k \le b_i$$$; then there will be $$$c_k - b_j$$$ empty seats at table $$$k$$$, but if $$$b_i$$$ were seated, there would be $$$0$$$, so this option is not better.
$$$c_k \le b_i, b_j$$$; then it would be no worse to plant $$$b_i$$$, since $$$b_j - c_k \le b_i - c_k$$$. That is, at the next iterations of the algorithm, it is better for us to have a maximum as large as possible; when choosing $$$b_i$$$, the maximum will be no less than when choosing $$$b_j$$$.
That is, taking the maximum $$$b_i$$$ is no worse than taking another $$$b_j \le b_i$$$, so we are sure that at each iteration of the algorithm we seat the maximum possible number of guests.
1
15 3
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
7 4 4
Courtsey — NemanjaSo2005
I think he means to find which part of the tutorial is wrong not to find a stress test.
In the second case, if $$$b_i$$$ are seated, then there are $$$0$$$ empty seats, but there are also $$$b_i-c_k$$$ extra people, compared to $$$c_k-b_j$$$ empty seats if $$$b_j$$$ are seated. It's not obvious that seating $$$b_i$$$ is better.
In the third case, it's not necessarily true that having a larger maximum is better.
It happens :) I am happy you did not let me lose my sanity over 1 hours while trying to solve C :P
It's kinda hard to find a particular error in a text that doesn't make any sense. For some reason you are minimizing empty seats, how is it connected to the actual problem — I don't know. Then there are just random phrases like "it is better for us to have a maximum as large as possible". Why is it better?
This is how I understand it: for each table they choose people from the same group — these are going to be the "satisfied" people for that table. The number of free seats is then some constant minus total number of satisfied people, so we try to minimize it. What they do in the "proof" is they prove that the solution is greedy, i.e. at each step we are indeed minimizing the number of free seats for the current table. The error is that they never prove that greedy actually works.
"stressed this decision" -- solution, not decision
It's incorrect translation from Russian, I guess
Upd: fixed
The main problem in the proof is that you never took into account that your choice for the current table might affect the other tables. If you were assigning people to only one table, the greedy is perfectly valid. Additionally, if all tables had the same capacity, your greedy is completely valid. Bullet point 3 is the one about which I am talking. You said that it is better for us to have maximum as large as possible, without providing proof for why. As it turns, that statement is incorrect as can be seen in the example above. I know that the proof looks quite logical and it seems to intuitively make sense. However, intuition can be really misleading sometimes.
I honestly feel quite bad for you and the rest of the authors as this was your first contest and it went the way it did. I cannot imagine how bad you all are feeling, especially with all the hateful comments from some individuals in codeforces community. But I do not suggest you quit either CP or problemsetting because of that. (Your choice, I am just giving a recommendation) Everyone makes mistakes and it can be a great opportunity to learn and improve.
Local best solution doesn't equal to global best solution I think. For example, let's seat 4*1 and 3*2 into 3 2 2, and surely placing 1 or 2 into 3 can fill this single table, but the cases after that become very different. Just like why we can't use DP sometimes.
I think one problem is that you are assuming that "at the next iterations of the algorithm, it is better for us to have a maximum as large as possible", which is incorrect. If we have c = {3, 2}, b = {3, 2} gives a score 5 while b = {4, 1} gives only 4 although it has a larger maximum.
what you are trying to convince in the last section of 2nd point is that for a fixed c array the lexicographically bigger b1 is always better compared to a smaller b2 (both having same sum and b1,b2 are sorted),now,what if c is exactly same as b2?
Could you please update rating only for those, who has positive delta?
UPD: Nevermind, just found out about C a little bit late
Could you please update rating only for those, who have 0 delta?
I know the complex progress of contest preparation. It needs a lot of effort and attention, but there are many testers I am shocked by this annoying situation. The reason for my participation in Codeforces contests is not ranking. I am joining contests for having fun and learning some new things. But I thought my solution will get the wrong answer verdict while submitting but when it got accepted I got shocked. I hope testers and problem setters of this round will be more careful in the future.
why "locale=ru" in the tutorial link
Oh it's fixed now.
A: If there's no odd numbers, or n==3 and there's 2 odd numbers, no solution. Otherwise we can find 3 odd numbers or 1 odd number and 2 even numbers.
B: Note that common divisor of (a,b,c) would also be common divisor of (a,b+c), it's always optimal to divide the array into 2 segments.
D: First let q=1 and query(q). If the bitcount is reduced, the last bit of n is 1 and you've just cancelled it, then you need to try for q=q<<1 (which means, left-shift q to the next unknown bit). Else the last bit of n is 0, and by count the difference of bitcount before and after the query, you'll know how many zero bits was on the tail of n, and you've just fliped them to 1 (and a 1-bit to 0). Then you can cancel these 1-bits in one query, and left-shift q to the next unknown bit. In the first situation, you've got the information of the least significant bit by one query, and in the second situation, you've got the information of at least 2 bits by 2 queries. Thus by repeating this process we can get the answer.
E: Sqrt decomposition. Consider for gcd(u,v) in [1,ceil(sqrt(L))-1], [ceil(sqrt(L)),L-1] and [L,R/2] respectively.
PS: It's regrettable that the contest was ruined by an incorrect problem, and many great contestants lost their positive delta. It's a pity that problem D,E are pretty well-designed.
So what happens to C about it being it problemset. I think C must be removed from problemset too.
Brief solutions(hints) besides G,I'm weak in string problems.
100 or 111
Dividing the sequence into 2 subsequences must be best.
NP complete.Can't be solved.
Solve it from low digit to high.
If you ask $$$2^i$$$ and the $$$cnt$$$ decreases $$$1$$$,add $$$2^i$$$ to the answer,and $$$i=i+1$$$.
Else $$$cnt$$$ increases $$$d(d \gt 0)$$$,we add $$$2^{i+d+1}$$$ to the answer,and $$$i=i+d+2$$$.
Repeat this until all the $$$1$$$ digits are found.
For the value of gcd $$$g$$$,we only need to check if the second minimum multiple of $$$g$$$ which $$$\ge l$$$ is $$$\le r$$$.
It can be expressed as $$$ \lfloor \frac{l-1}{g} \rfloor +2 \le \lfloor \frac{r}{g} \rfloor$$$.
Notice that $$$l\le10^9$$$,use math blocking skills to solve it(Calculate all the $$$g$$$ which has same $$$\lfloor \frac{l-1}{g} \rfloor$$$).
The number of total schemes is $$$n \choose 3$$$.
Sort $$$a$$$ first.
And then if $$$i \lt j$$$ and $$$\gcd(a_i,a_j) \gt 1$$$,reduce $$$j-i-1$$$ to the answer.
Use
Mobius inversionorInclusion And exclusion.Unfortunately, G is just a suffix automaton exercise :(
I haven't learned SAM yet :)
Solve count for C during contest shows why cf is just guessforces nowadays
rip my >100 positive delta ;-;
Anyways it’s still a fun round with other problems being interesting
How did you calc your delta?
Carrot extension
Is it possible without extensions?
Yes, but you have to do a few thousands of calculations
hahaha so funny mr obvious
what answer did you expect?) lol
I dont know what answer should i expect...thats why i am asking
First time I solve 5 problems in div2. 140 positive delta. I would've been div1. Big bruh moment.
first time got approach of D but unrated , its OK :)
E could have been written as "Constructing a graph with vertices from $$$l$$$ to $$$r$$$" instead of "deleting all other vertices and all other edges", would be much easier to read and understand lol
E could have been written without any graphs at all. You just assign weights and right after that throw them in the set to count the number of unique values. There is nothing graph-y about this problem.
this is my first time that I solve the last problem and rk110- in div.2,although it is unrated,I think it still be a good contest
change to rated -_-
Meanwhile 1000 cyans solved it
This is a really good joke!
lol
AndreyPavlov Please remove problem C from codeforces as it can be harmful for people who try to solve it later during practice or something.
What ever happen, I still love codeforces
what's wrong with problem c? this is the quickest i solved a,b,c and this round gonna be unrated fek
You have solved C incorrectly
It's so terrible to do system testing C with the wrong test case.
I am kind of curious: how were there so many AC on C, even tho it could not be solved? Did people just submit a greedy solution?
Also curious: what would be the best possible runtime for a correct solution?
can problem B solved using binary search on answer?
Thank you for not making me specialist.
Can someone give a test case where this approach gives a wrong answer to C?
Calculate how many meals of each type and how many tables of each size there are.
Go through the meals, if there is a table with the same size as the count of that meal, then put those people to that table.
Put remaining meal and table counts to two priority queues. While there are still some empty tables or not all people has a spot: take the top of meals and put them at the biggest empty table, if some people who like that meal are still left, check if there is empty table with that amount of spots, if there is put them there, otherwise put that number back to priority queue.
1
15 4
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
5 4 4 2
The answer is 15, but your approach would give 14.
Thank you sir
hello sir, my approach works for some of these type of test cases, can you tell me where my approach can go wrong? Sort the tables array in descending order, and we keep the frequency of elements in a multi set, and for every table from largest to smallest we find the lower bound of the table size in the multiset, and we assign the table size to that many people of same type, and erase the people that are sitting in the table in the multiset and the extra people left will be put inside the multiset again, we do this if the lower bound exists, and if the lower bound does not exist, we just assign the maximum element of multiset to the corresponding table. In test case 16: the array was 1 1 1 1 1 1 2 2 2 2 2, and the table given was 3 3 5 And the answer to this test case according to my solution is 11, but the given answer is 10 which is wrong.
Your approach assigns people to the table without caring about what kind of effect it might have on other tables. That's the main reason why no greedy solution works.
Nice one. I didn't solve C during the contest and was wondering how 1000 people solved it. Shows how people don't understand their solutions and can't prove it.
I hope Codeforces will hold a good competition next time, instead of being as rubbish as this one
D is really an interesting problem
i didn't participant in the contest yesterday, and only see people are talking about the unsolvable problem C. but now actually i even can't see what exactly is the problem statement, where can i find it? thanks
The problem was something like this:
You are inviting people to a party. There are $$$n$$$ people, and the $$$i$$$ th person likes the dish $$$a_i$$$ $$$(1\leq i\leq n)$$$. You have $$$m$$$ tables, and the $$$j$$$ th table has a capacity of seating $$$c_j$$$ people $$$(1\leq j\leq m)$$$. Now, for every $$$1\leq j\leq m$$$ you have to select a dish $$$x_j$$$ to serve at the $$$j$$$ th table, and the people seated at that table who like that dish will become happy. You have to maximise the total number of happy people by selecting what dish to serve at each table.
Input: Arrays $$$a$$$ and $$$c$$$ were given in input
Output: Just print a single integer, i.e. the maximum number of happy people possible
Constraints: $$$1\leq n\leq 2\cdot10^3$$$, $$$1\leq m\leq 2\cdot10^3$$$, $$$1\leq a_i\leq n$$$, $$$1\leq c_j\leq 2\cdot10^3$$$ and $$$\sum_{j}c_j \geq n$$$ (i.e. total number of seats is more than number of people, so every person can be seated)
Hi guys! If you wanna know why problem C was wrong, and the editorials for A,B,C,D then you can check it here — https://www.youtube.com/@GrindCoding
Does this solution for C work?
Take the array of guests and put 1 extra element in it, let's call it breaker. Now go trough each permutation of that array. To process a permutation, you are going to only assign people before breaker to the correct table and you can ignore the rest. (ignore means to assign them to any table) How do you assign them? Consider each consecutive subarray [l,r] of equal elements such that a[l-1]!=a[l] and a[r+1]!=a[r]. (a is the permutation) You will add it's length (r-l+1) to the array b. Obviously, you are ignoring the elements after the breaker. Now that you have array b, you are going to assign each b[i] (apart from the last one) to exactly one table, such that the sum of capacities of tables that you assigned is minimal. That problem can be solved with greedy. Now, fill the remaining tables with the last element. If it is possible to do so, such assignment is valid. Now you just select the best of all valid arrangements and output it as a solution.
Why does greedy work? So, let's say that we have 2 arrays x and y who will describe any optimal arrangement of people. On table i, I will put x[i] people who like dish y[j]. Now, if we rearrange x and y in such a way that none y[i]=y[i+1] other than on some suffix of array y, we get a permutation. First x[1] elements are y[1], then x[2] elements are y[2] and so on. After that we can put the breaker and unused elements. Now, it is clear that all b[i] apart from the last one will be assigned to exactly 1 table. The optimal assignment of b[i] is so that capacity of unoccupied tables is maximized, so that last b[i] has enough tables to arrange the people from it.
Complexity of the solution is O((N+1)! * (N+M) * log (N+M)). Meaning that for N=10 and M=10, the solution would use around 4 Billion operations. Note that this is probably not the optimal solution and that it is likely that there exists a faster one.
Also, if you noticed any mistakes in my solution or proof, please let me know. Example:
15 4
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
5 4 4 2
One of the optimal permutations is:
2 2 2 2 2 1 1 1 1 2 2 1 1 1 1 B
It will result in an array b like this: 5 4 2 4. We will assign 5 4 and 2 to the tables, leaving one table with capacity 4. We can assign the last element of array b (4) to it.
Obviously you can do dp on subsets in $$$O(3^n m)$$$. Store the mask of people you didn't already sat somewhere, iterate over all tables, choose a mask of people to sit at this table. That's basically memoization of your bruteforce.
But then we only care about the multiset of sizes of groups with the same taste, and the sum of these numbers is at most $$$n$$$. So we can write dp with memoization on reachable states, there will be at most $$$O(m \sum_{i=0}^{n} p(i))$$$ states where $$$p(x)$$$ is the [partition number](https://en.wikipedia.org/wiki/Partition_(number_theory)). This should work reasonably fast for $$$n \le 50$$$ or something.
Problem F says "Kira is not very strong in computer science, so he asks you to count the number of ways to invide friends."
It should say "invite" not "invide" I think.