Hello again, Codeforces!
I am glad to invite you to Codeforces Round 1004 (Div. 1), Codeforces Round 1004 (Div. 2), at Feb/11/2025 17:35 (Moscow time).
In Division 1 you will be offered $$$6$$$ problems. In Division 2 you will be offered $$$7$$$ problems. One of the problems in Div.1 will be divided into 2 subtasks. Round duration is set to be 2 hours.
Also, both divisions contain at least one interactive problem(s), so be prepared for those! Guide for interactive problems
I would like to thank,
FairyWinx for dedicated coordination, a lot of problemset discussion, and suggesting one of the problems idea.
Our testers: Dart-Xeyter, I_love_kirill22, RP-1, larush, Wansur, furt1ve, Iura_Shch, SomethingNew, Gabbasov, glllll, domovonok, Error_Yuan, Monogon, triple__a, evjeny_23, Blinov_Artemii, LeoPro, antontrygubO_o, A_G, N_z__, mainyutin for testing and providing feedback.
MikeMirzayanov, KAN for Polygon and Codeforces platforms.
As always, we hope you will like the problems. Have fun!
Score Distribution:
Div. 1: $$$750$$$ — $$$750$$$ — $$$1250$$$ — ($$$750$$$ + $$$1250$$$) — $$$2000$$$ — $$$3000$$$
Div. 2: $$$500$$$ — $$$1000$$$ — $$$1250$$$ — $$$1750$$$ — $$$1750$$$ — $$$2250$$$ — $$$3000$$$
UPD: Editorial
UPD2: We sincerely regret to inform you that we have discovered a bug in the interactor. A series of tests is currently underway to assess the full impact of this issue. Once we have the results, we will provide a detailed update. We deeply apologize for this incident and any inconvenience it may have caused.
UPD3: After the analysis, it was determined that this problem affected a small number of participants. There are no submissions that get AC with the correct interactor and erroneously received a non-AC verdict earlier. Therefore, the following decision was made:
If your solution worked with the old interactor, but does not work with the correct one and your rating has decreased, then the round will be unrated for you.
UPD4: Congratulations to the winners!
Div.1:
Also, special thanks and congratulations to rainboy for being one and only one solving problem F in division 1!
Div.2:
As a tester, I must say, FairyWinx is so orz.
.
hmm unusual score distribution D and E are same
I am guessing one of them is interactive
As a tester, I have a proof that upvoting this comment will lead to positive delta. But the proof is too long to fit the margin.
As a tester, ... are you really a tester?
yeah bro
I can confirm, yes.
Good luuuck Everyone
Hello, could someone please tell me when Codeforces typically runs the plagiarism check? For example, how many days after a contest does it occur? Also, for round 1002 div 2 , have the plagiarism checks (and the process of skipping certain solutions) been completed, or are they still pending?
they run them randomly ig. and yes they're pending since a long time
Authors should know that many aren't participating in rounds that have interactive problems, it's not good for you.
I do not think so. Always avoiding interective problems is not good for your coding ability.
Just write interactor in ~5 minuts and will be happy)
What's wrong with an interactive problem? It's not like they are that different from regular ones, all that really matters is that they force you to answeer queries online, opening up a lot of interesting possibilities
Bring Reporting Option in Codeforces MikeMirzayanov , so that the community can help cleaning away cheaters.
I saw some YouTube streamers providing solutions during the contest today.
Every problem starts with Hello-World but ends with infinite possibilities
As a tester, FairyWinx orz
As a tester, I should say, setters made one of the best interactive problem I've ever seen
As a coordinator, instead of disliking the post, disliking my comment!
my aim of this contest is to get rating 100+
same and by the way all the best
Div.1 A=Div2. C or E?
Usually, div 1 A = div 2 C
But if div 1 A = div 2 C, Div.1 D should be Div2.F, but Div.2 F isn't divided into 2 subtasks.
Maybe they didn't just copy and shift the problem set, or used only the easy version of div 1 D in div 2.
I think this time Div 1A=Div 2D
As a tester, problems are exciting. I recommend participating, and wish you good luck and have fun.
Finally, interactive problems are coming...
hello
Wish to reach Master!
why there are different score distributions in div1 and div2 ?
Hi! coders.
This might be a round where tourist will get his badge back ||
+160??? awwww
It is not realistic even if he gets Rank 1 !! They don't get so much positive delta at that level..
Also, both divisions contain at least one interactive problem(s), so be prepared for those!
After a long time, yeahhhhhh
I'm a bit confused on how to efficiently write code for the interactive problems
From the linked post: "Input/output in interactive problems works much slower than in usual problems — try to use scanf/printf instead of cin/cout in С++, BufferedReader/PrintWriter in Java and etc."
Does this mean in Java, we should use BufferedReader/PrintWriter? Or that instead of using Bf/Pw we should use scanner?
And if the former, I guess all I need to do is write pw.flush() at the end of my code and it should behave somewhat benign?
You should flush after every output you make so that your query is not buffered and the interactive checker processes it instantly. Also, I have never really had problems with the performance on interactive problems, cin/cout were enough for me
From the score distribution, I would guess the 2 Div. 1 750s are the same as the 2 Div. 2 1750s.
Then the Div. 1 1250 = Div. 2 2250
And then Div. 1 2000(not the (750+1250) one) = Div. 2 3000
Now I'm thinking why they did not put the (750+1250) one in Div. 2.
There are already 7 scoring opportunities in Div2.
I'm guessing they didn't want to increase it even more to 8 so put the hard version only in Div2.
We considered D2 and E in Div.1 similar in terms of difficulty (as can be seen form score distribution). And since D had subtask it made sense to place it earlier in Div1. However, for "dead last" position in Div.2, we thought Div.1E is more appropriate, mainly because otherwise both Div.2F and Div2G would be some dp/combinatorics, and with Div1E we achieved some sort of topic balance in Div2.
I very like problems like "C+K+S".I want contest writer make more this problems in contest.
facts:
:(
I can't wait to take part in this anticipated contest!
This is my first (Div. 1, Div. 2) contest. I hope the qualities of the problems will be satisfactory.
As a participant,I am not a tester.
It would be cool if everyone got their nickname as a title
That would spoil the fun of competing in my opinion..
As a positive delta, wish everyone a participant!
nvm
my target pupil.
it was an observatory problem
Sorry for writing this in between the contest (I solved what i can and cant do the next one)
curiously I locked my problem and went to see room (i have never hacked a problem before)
i saw a solution (ofc i couldn't hack it) and went back
will this count as an "Unsuccessful hack" ?
No. Only when you submit your data or generator and the code doesn't have any errors, will it be considered as an unsuccessful hack.
Thanks <3
Unacceptedforces
Div2C is some cancer.
this contest consisted of problems that I'm just weak at... especially C.
I looked at C and thought it would be a nightmare since I'm pretty weak at number theory. I tried brute force and it worked first try
Is this contest an attempt at not having GPT-able problems? I enjoyed ABCD even though they kicked my a*s. Didn't really get to try E or F.
Hi, what's your thinking approach for B, I literally didn't got any smart idea to solve it?
after struggling a lot to phrase it in a nice way, I realised that you should try to push along as much as possible from the lowest ones to the bigger ones, of course without getting to the case where you have only one (because then you can't have it in both arrays). This approach is actually O(mlog(m)), where m is the maximum value of the elements in the array.
You can do it in O(m) by keeping a frequency array (arr[0] = # of 0s and so on). You can also do it in nlogn (n = number of elements), by using a sorted dictionary with a bbst implementation.
My approach was also similar, but I was confused about intermediate values, like what if newly converted values will disturb parity of intermediate values 1,1,1,1,1,2,2,4 Although for TC shown below, I was able to came up with formula 1 1 1 1 1 1 1 4 (4-1)*2+1>=index(4) Like intermediate elements should occurred twice, simultaneously they should form '4'. Above check should be performed for all elements with odd frequency But didn't worked for multiple cases :(
That is a decent point (about disturbing the parity). However, consider the following: If you started with an uneven amount at a low value you have to give at least 1 of the amount — 2 a higher value REGARDLESS, in hopes of finding another matching uneven in the higher values. If you started with an even, you will continue pushing an even and there is no problem there.
Hi sir, thanks for the explanation.
Understood your point 1, that if we keep passing uneven amount of values then it will match at some point as per below example Example-1 1 1 1 2 2 3 4 4
1 1 2 2 2 3 4 4
1 1 2 2 3 3 4 4
Also understood your point-2, that we can keep passing even amount as well, we would still end up achieving, even parity of elements Example-2
1 1 1 1 2 3
1 1 1 2 2 3
1 1 2 2 2 3
1 1 2 2 3 3
Just one more clarification?
Can we say we can push element only when at least 3 identical element available in an array?!
Yes.
how did u solve B, I used the idea that If I want to form k ns then I need k+2 number of (n-1)s. sub Id:305692767
there were no intentions to make contest "anti-GPT".
Bro, it was an awesome contest! We want more like this, ones that aren't solvable by ChatGPT
Thanks for this one :)
trash contest
D seemed like such an interesting problem, I wish I had more time to solve it. B was super tricky for me and I'm surprised brute force worked on C
Thank you FairyWinx for delivering me to my coveted Grandmaster title in my two consecutive rounds.
^^
who else ask for "? 1 n" instead of "? i j" such that x[i] == 1 && x[j] == n in div2 D/div1 A xD
I miss the days when problem C used to be DP..
Report a cheater: byteplay_studios solved A~F and published them on https://www.youtube.com/watch?v=8249xj-tyS0
Now I get a surprise because I saw too many newbies solved D, E, F very fast......
In B, why the seventh test case is 'No' ? You can make the bags 244 and 244.
pass a 2, pass two times 4, increase two times one of the 2's in the first bag and done. What i'm not seeing? im dumb?
You need to have 3 in second bag to increase 2 second time.
you're right. thanks
i had same idea, but it is incorrect
when you increase 2 to 3 .. you can't increase it again.. because you can only increase a number which is in first bag.. the new number 3 is not in the bag now.
hope this helps
Once you increase the 2 by 1, it is no longer in the second bag.
Is this contest actually harder than usual or I am just very bad?
it was not just you
it was not just u
I feel the same way. At this rate, I should change my name to MaybeSpecialist (T_T).
Was this contest AI-proof or human-proof? (Not blaming anyone, I'm just here to vent...)
Problem D looks easy and inviting but I bet I am missing something.
But this is WA so what am I missing?
x = [1,1,2,2] y = [2,x,x,1] //hidden treat this as a graph y = [4,x,x,4] // hidden treat this as manhattan
both queries return 1, in both cases.
your idea is on the right path. However you can have this case, where you have a cycle from u -> v -> u, where u->v and v->u have the same length
What do you mean by
y = [2,x,x,1]? Could you give a concrete example withxandyvalues?I interpreted it as the 2nd and 3rd values in
yare the same. I think you meant they can be any valid values.yes, any values. it doesn't matter for our queries.
Sorting the array messes up the indices. You wanna query the index of 1 and index of n.
Thanks tiojuan and bramar2 I misread the problem about query values and if its arr[i] the index or the value. Pretty easy now that I read the question careful got AC. Thanks I had to read the comment again to rethink about my queries.
Can anyone confirm why my hack on Div1A/Div2D is unsuccessful? Am I hallucinating? 305640270
My test is
Cannot distinguish A and B by querying (1, 3) and (3, 1); since the shortest paths are 1->2->3 and 3->2->1; both are lengths of 2, and manhattan distance is |1-3| + |2-2| = 2
But why query that? I just query 4 1 and see that there is no path leaving 4
Haven't you tried hack once?
I didn't read the question properly. Tested it locally and it seems to answer B in both cases.
I have tried it by using test:
And the program's output is B. I want to know why as well.
I also had an unsuccessful hack against 305657749 with a similar test case, it seems like there's something wrong with the interactor.
Thanks for confirming! Now I think I can sleep well for my work tomorrow...
Also, this submission 305637703 should pass test 1, as it is correct when tested locally. However, the interactor returns WA on test 1.
Yes, we find bug in interactor, but this submission fails on tests
This submission is wrong. I'm complaining because I looked for format error for about 20 min due to this submission and the lack of interacting process for the sample.
I'm really sorry about this, and now we're thinking about what to do and how many people it affected
Funny thing is, after the rejudge my two WA 1 become WA 2, got another -100 and I would probably lose more rating than I already did.
Retesting is currently underway(
Btw, your solution is correct
I mean originally the two WA 1 don't count as resubmit but now they do. The change is minor but this whole thing is just hilarious.
What was the bug in the interactor?
We skip edge x_n -> y_n
Wait... If we ignore an edge, the graph will not exist any cycles. Does this mean that many incorrect solutions that do not check for cycles could pass this problem?
Wait, can't you still have smaller cycles anyways?
This is false, 1->3 2->1, 3->2, 4->1
Hack correct :)
this is not fair to the contestant since he wasn't hacked in contest, i think you should keep the contestant's points (or in general take a bigger look at what is fair, such as giving an option to unrate people whose verdicts might change).
It's not the final decision or anything. We are still investigating how much impact this issue dealt to a contest.
I guess we certainly will not reject people who had AC in contest, only because of interactor bug, since it is similar to "weak tests" situation.UPD: no,sorry:( .An announcement will be made eventually, when investigation is complete.I think that the interactor have some problems which make the system has given the wrong shortest paths. But I trust this test can really make the submission give the wrong answer.
I don't know if the problem's author can upload the interactor of this problem and it may give the actual reason.
And the interactor hiding the interacting process makes this mistake totally unbearable.
Edit: To clarify, if the interaction process had been visible, the bug in the interactor could have been found and fixed immediately. The bugged interactor incorrectly calculates the shortest path in the sample, and many contestants could have noticed the bug right away.
However, due to the lack of interacting process, these contestants gets WA 1 with misleading information from the interactor. As a result, they wasted time debugging the sample for a non-existing problem, making this mistake somewhat unbearable.
In my opinion, contestants whose submission initially got WA 1 during the contest and later changed to WA 2 after the bug was fixed were also affected by the bugged interactor, not just those who gets WA after fixing the bug. To the best of my knowledge, there are several such cases.
Edit2: Softened my tone in Edit section a little bit ^w^ (original version in Rev. 4)
I used BFS for C and kept getting a Time Limit Exceeded (TLE) error. Can anyone suggest the correct way to solve this task? here is one of my sub: https://codeforces.net/contest/2067/submission/305699504
tysm btw ^-^
Check out my submission in the contest, just brute force
tysm, I'm just dum^2
How do I even begin to solve problems like this one's B? One hour of thinking only to give up because I'm totally clueless.
Same
What helps me solve questions like B is going through the testcases. We have an operation and we need to figure out when should we use that operation.
One question I got while going through testcases was: Suppose you have 4 1's: 1 1 1 1, should I leave it as is, as it is balanced or should I increase two of 1's to 2's: 1 1 2 2?
The following sample testcase made me realize it is always best to increase whenever possible as the incremented nos. might help in the future.
8
1 1 1 1 1 1 1 4
Couldn't even solve B, have to go back to the drawing board now. How's B becoming a pain to me is beyond my understanding.
welcome pupil
I felt terrible after the contest for solving only A, but am starting to feel better now after seeing others having the same struggle
If I solve only A I got -160 delta lol
Not able to solve problems in contest is way painful to me than loosing some rating. Rating means nothing to me.
youre right you can gain the rating you lost if you study
C is one of the worst problems ever, why would you put a bruteforce problem as div2C? Problem B seemed weird too, I think this is what they meant when they were talking about anti-GPT problems. On the other hand I really liked problem D, unfortunately I didn't get further than problem E because I spent too much time on B and C.
Solutions in ~7 lines)
:( 7 nested for loops
guessforces
This is a contest that you thought was inexplicable before knowing the solutions, but felt extremely foolish after knowing them.
Wtf Kevin114514? Explain yourself please.
So, what's wrong?
My submission failed pretest 1, even though it passes on my local machine.
https://codeforces.net/contest/2067/submission/305693896
Any idea why this happens?
i must be strictly less than 10 as you only have 10 elements in the array, but you were iterating up to 10, which means you accessed an 11th number that's a garbage value.
Problem setters took AI a little bit seriously :)
I've managed to properly think only about ABC.
Personally, it was a very great experience to not back down after such a failure at B (huge bad luck + small skill issue), keeping emotioins at bay, and then figuring out C (good luck). Enduring such a variance in one contest is amazing if you don't give up.
what is observation in C . please tell
If you add nines then you only add fixed number of nines. Then just bruteforce.
oh thank you for sharing your idea.
I did something similar but couldn't prove anything .. fingers crossed for system test
Can you tell an approach about a i tried making some pattern like x should be equal to y-1 and when x is divisible by 9 check for 1 and something like that but it failed can you tell me your approach?
I am sorry I don't have a concrete proof but this is what I did
I was hoping that imagine one of the digit 'x' in the input will be converted to 7 ... so we just put 9 for that digit ... meaning if we are targetting 3rd digit.. we can keep adding 999 till 3rd digit become 7...
and I did this for every digit
so first keep adding 9.. then keep adding 99, 999, 9999 , till you cover all digits .. and find the minimum among all these digits
B and C felt very tricky to me .. hoping to pass system test because I didn't prove them much rigorously and just went with intuition
I cannot even prove correctness of my solution to A
ha ha LMAO.
I google plot of sum of digits graph and was able to make an observation from there.
Am I the only one who couldn't solve A even tho solved B-C?
whoa thats crazy, did you struggle with the case where after 9 the sum gets decreased ?
I did a little mistake.. My code :
I didn't check that x have to be smaller then y when check abs(x — y) == 1.
-_-dump..
oh no, I am sorry to hear that.
wishing you all the best for upcoming rounds
I only solved A, Brute from 1 to 1e9 and found pattern :)
actually I googled for sum of digit graph and I got idea from their.. ie I got the pattern observation from there.
me too couldn't solve A :(
lol me too
i tried doing c by adding 9's and incrementing ans , whats going wrong in this? Can anyone explain.
May not be optimal. Did you just try with 9s or even tried 99, 999, ... which can go TLE but not wrong?
I just kept adding 9s only and for each addition I check for the sevens
The contest was a little too hard for me- Gotta grind more I guess...
I want to report a cheater... pasit used AI in many contests... Including this contest (Div2) (I think he uses AI in problem E and F. He used AI to generate the code, then change the variable name.)
Agree.
I agree, AI should only be used for debugging :(
fyi this is not allowed during live contests. see to this post for more clarification
Oh okay, thanks for letting me know!
I agree as well. AI should only be used to ask for syntax.
I can not believe C is brute force. I spent the whole contest trying to come up with some crazy rules and they kept failing every time.
yeah, I used BFS and failed hard lol
same here..
painful.....
I have no idea how I passed B and C lol. I just guessed something, and it passed.
yeah me too, I feel like a fluke :(
I want to know c problem is Once you've chosen the 9 form, can't you change it anymore?
In C, the maximum answer which can we obtain is 7 operation, so you write (10!*6) and it worked. It took 1800ms. But it passed.
I am sorry if this is dumb but how is the brute-force solution of problem C faster than the BFS solution. Cant people just make a very big testcase for which bruteforce solution reaches max time.
Smart Brute force will not fail in any case !! People just need to figure out that minimum operations would never exceed 10 (I guess it should be 7, but doesn't have a formal proof) and even if they go for all 9999....999 till 1e18 that would take 18 iterations and for all the iterations at most (say 10) operations :- in total 180 operations for a single testcase !! 180 * 1e4 = 2e6 (won't fail even if you add log factor for digits) !!
and in BFS solution which I did and got TLE; we are checking for all 18 numbers or whatever at each possible number which is not a good choice.. It's better to use a same number (999...999) and keep on performing operations until it contains 7 !!
Why is pretest 1 failing on CF but not on my laptop?!
It passes on my local machine. But the output shown on CF is totally different, causing pretest 1 to fail. Please let me know why this happens. Some other test might fail, but how is pretest 1 failing?!!
https://codeforces.net/contest/2067/submission/305693896
Your vector v only contains 10 elements, but you are accessing v[i] with i up to 10, so it causes undefined behavior.
It worked on my local machine. I dont know how! :)
try to add
#pragma GCC optimize("Ofast")before all your code in local and delete it before you submit your code.Tried, it still works! This is so weird. It works on the left extension bar as well as in the terminal.
Here's the link, incase the image doesn't load: https://drive.google.com/file/d/1ngtq0oBwQ_-N-cGHZxonyJ1i9qY75VJ4/view?usp=sharing
its undefined behavior. The result depends on the compiler being used, classic c++ moment
My guess is that the 8 bytes in memory after
von your machine were all 0, so it didn't affect your answer, while on CF they were some big number with a 7 in it, so the answer is always 0 or 1 because you can just add this big number to get a number with a 7.On making this change, the all the tests pass. I got AC. But it is still weird how the original solution passed on my local machine.
No, it's not weird, you can read some C++ documentation about undefined behavior. Basically when it happens, you cannot predict what your program will behave, it will give you any kind of unexpected results, even different results on the same machine.
Some compiler or environment sets default value as zero for debugging mode. maybe this can impact to your case. Uninitialized variable may dependent to your machine and can behavior different with CF server
Thanks to all for replying. Is there any way to detect such overflows in my local machine? I have tried
#pragma GCC optimize("Ofast"), but it doesn't solve the problem.I tought like every time A was see and do and wasted minutes.....
NOW I FELL LIKE I AM EVERGREEN.......(hope less).
Don't give up bro, You'll finally get it!
I choose to continue to believe Codeforces Round Div.2 after participating this exciting and interesting round. Thank all problemset designers & testers.
Nice contest. Enjoyed the problems despite being a complete brick. Dont get the point of D1 though (mine trivially solves D2)
For me D1 was easier than A (and maybe B), and my solution to D2 approaches the problem from a different direction (for D1 I went from higher to lower, and for D2 from lower to higher and had to check a lot more things).
i guess i got lucky that my approach generalizes easily then. About difficulties, it seems all of us have very different experiences (after talking with others). Personally I found C < D1 ~ D2 < A < B which is quite funny
Screencast
I always like your problems. Totally went insane on C, but that's obviously my issue, not the issue of the problem.
E was very easy, spent all my time on C. leaving would have been a good decision
Can someone please explain why my code fails for problem B
Finally, specialist!
Can u help me with the second problem I don't know why its giving wrong answer please
My observation- If freq(x) >=3 then we can always convert freq(x)-2 x to x+1 ex- consider (3 3 3 3 3 5) () (3 3 3 3 5) (3) (3 4 4 4 5) (3) (3 4 4 5) (3 4) (3 4 5 5) (3 4) (3 4 5) (3 4 5) Thus its partition can be done This is my code-
PS- I know I'm really bad at explanation
I did something similar What I did is I started Iterating from the largest number possible and I said If that number appears odd number of times I need to have 3 x-1 numbers right ? I maintained a Required array and checked whether actual freq and required are same or less or greater But, if req> freq then I require rer-freq+2 times x-1 . I thought its correct why is it failing I can't get it
can you share ur solution link?
https://codeforces.net/contest/2067/submission/305715816
Your logic is wrong! try dry running it in this example (3 3 3 3 3 5) the same one I did in prev reply. If I go by ur logic since freq(5)=1 and freq(4)=0 then your code will result in "No" even though the answer is "Yes" for this problem.
it got accepted I changed some lines in the else condition , ya u are right
FairyWinx round = be ready for WA2
Nice problems, Div1B is too nice that I couldn't solve for hours.
NICE contest and Problem DIV2E just nice
...
We sincerely regret to inform you that we have discovered a bug in the interactor. A series of tests is currently underway to assess the full impact of this issue. Once we have the results, we will provide a detailed update. We deeply apologize for this incident and any inconvenience it may have caused.
Also comment for dislike
Upvoted
Its fine :) it happens. Take it easy. Nice contest nevertheless
oh hell nah
Mistakes do happen, and I respect your decision. Can you tell me a bit of information about the following things?:
Also, regarding weak data:
Uphacks in 24 hours are happening over and over for the contest for Educational Codeforces and Div. 3 contest. In that case, some AC submissions that passed the pretest during the contest finally fails in system testing.
The final decision seems somewhat "These submissions finally fail on system testing, and if your solution got WA on the system test, you can make your round unrated. (So only positive deltas are applied)." Why should it be the only case of rounds with uphack? I know good things are good, and asking this kind of question to the problem-setters who made small mistakes seems harsh or blunt in some sense, but I want to know the thoughts behind the decision; it's fine if the thoughts are "good things are good." Because as I wrote in the first place, mistakes do happen.
~140 participants received an incorrect answer after correcting the interactor
There are were ~5 hacks in total, and apparently they were very unlucky.
~100 maybe
don't understand question(
After all, the mistake is on our side. The format of educational rounds and regular rounds is different.
On detailed explanation of question: You can refer this blog post for detailed rating display. Suppose you are the second time participating codeforces Round and got 500 -> 800 (+300). The internal rating actually went down from 1400 -> 1350 (-50). What about this case?
About the unlucky ones — why aren’t hacks rejudged just like submissions?
Format of regular rounds — The point is that even in the regular rounds, you can get pretest passed and systest failed, and this is the current case.
LOL, I guess all the solutions have not been judged yet and the rating changes happened XD
An example
MikeMirzayanov is it a bug?
Read UPD2 of the blog.
what about ratings on a problem?!
can anyone please tell me why this fail on test case 4 i will be gratefull if someone can find the error for me SUBMISSION
In problem B of div2 the test case 6 2 2 2 4 4 4 we can make the first bag as 2 4 4 and second bag as 2 4 4 but why the answer is not "YES"
how you convert 2 to 4?
I got it . I forgot that 3 must also be there also bro i need to ask one thing if this problem really belong to 1200 range i saw on clist ?? And how there are more than 7k ac on this problem this is surely more than 1400 range
IT's not much hard
rejudge 305643299 plz, I think it is not a mistake
I rujudge your solutions in polygon thirds times and your solution getting IL every time
O , i find the bug , thx
will the ratings be recalculated after judging problem A again?
I write in the upd3 decision
Sorry, it's not explicitly mentioned and I'm not really sure I understand it correctly. Ratings will be recalculated after the rejudge of A, and then people who get WA on A and have negative delta will become unrated, right?
Current standing correct. And if in the current standing you have negative delta and your get non-AC after rejudge for you round will be unrated
So this is not relevant, correct?
Yes(((
My solution worked with the old interactor, but does not work with the correct one and my rating has decreased.But why the round still rated for me?
I am sure that my solution was accepted during the contest.
Same here, I'm now receiving RE on pretest 2 due to a wrong assertion in the code. The code I submitted during the contest (and that got accepted during the contest) should fail basically all reasonable test cases. If I had received the RE during the contest, I would have easily fixed the problem. I'm not sure whether the policy also holds for RE or only for WA, but in my opinion it would be fair if it also applied here. Maybe, they'll just make the round unrated for us later.
Finally, the situation is as follows:
Rejudge your submission. If it now receives WA instead of AC, it is given the WA verdict.
Now, with the updated contest scores, everyone's rating has been recalculated. If 1) affected you and your rating decreased, then the round will be unrated for you
This feels wrong to me. Why does 1) hold place? Currently, it is treated as if the 'good interactor' was only available during system tests, but not on pretests. Though it is factually not true – the test that makes the wrong codes fail is actually in pretests.
Also, if we still treat this whole situation as 'weak system tests', it still feels wrong: Ahmed_Salah7 tried to hack me during the contest using this test. The test should have successfully hacked my solution, but it didn't since the interactor was wrong.
So if we treat the 'good interactor' to be the part of system test, then in reality, some users like Ahmed_Salah7 should've received their +100 hack points instead of -50, and some others should've been hacked during the contest and should've gotten the time to change their solution so that it handles the wrong case.
I feel this is kinda unfair. I believe that we can't really do anything great in this situation, but maybe there's another solution: for example, if it affected only a handful of people, why don't we just leave it as it was? ;)
Yes, the situation turned out to be extremely ugly and we tried to find some way out (
Fun fact: In the Div.2 contest yesterday, there are 19 cheaters in the top 50 contestants.
This is especially unfair to me because I happened to be the rk50.
According to this blog.
Hey could you please tell me are only the last round's ratings rolled back, or are all pending contest ratings rolled back?
My solution worked with the old interactor, but does not work with the correct one and my rating has decreased.But why the round still rated for me?
Oh, the process of solving Div. 2 C for me was strange.
Firstly, I found the answer is small, and then I tried memoization search, but TLE.
Then I found $$$ans\le 7$$$. And I tried to use DFS with numbers $$$9\sim 9999999999$$$, but TLE again.
Then I shortened the DFS range to $$$9\sim 999999999$$$, getting AC.
This process wasted much time, which made me unable to think about the rest of the problems with a clear mind.
According to upd3 I should be unrated. Then why my rating decreased. :(
I meet the same problem.
I meet the same problem. sevlll777 please check it. :(
I meet the same problem.
In Contest:
After System Test:
Nevertheless, I admit it's my fault not to check whether the graph form a cycle, but IMO I should be unrated according to the announcement.
I meet the same problem.
Me too ;(
Hello, yes, it must be unrated, apparently there was an error, I'll pass it to Mike
When will the rating be fixed. I mean it's already 2 days.
what's wrong in my div2 E/div1 B 305690629
please help
UPD : if(arr[i]==0){ zero--; if(zero!=0)continue; }
added this to get AC (skip all 0 other than first 0 )
Can someone try to hack my submission for Div1C/Div2F? Feels like it should overflow at
3 * dp[vec[i - 1]] % modbut it somehow passed all testsvector
dpis of type long longI think this contest should be unrated. It is extremely unfair to make the participants bear the responsibility for the problem setter's mistake in writing the interactive system.
According to the announcement your submission would be incorrect with both the wrong and fixed interactor, so how does it make a difference?
In fact, my solution was correct and should have passed the test cases. If it didn't pass during the contest, I could have spent more time fixing it. However, my rating was not corrected.
if your submission passed pretests with wrong interactor but failed with the new one then it should be unrated for you. They must have made a mistake...
Maybe rating hasn't been updated yet, because I think many people have encountered the same situation as me.
Yes, the situation is extremely ugly, but it affected quite a few people, so the following decision was made
Does it include those who got WA 1 during the contest but should be WA 2?
What about the "-50" points for unsuccessful hacks that should have been successful and should have earned +100 points in the contest (+150 points now)? Is this fair?
I also tried to clarify within the contest to check if there was an issue with the hacks in problem A, but no one seemed to care to look into it.
So, can you please FairyWinx either add +150 to my contest score (if that's possible) or make the contest unrated for me?
include
using namespace std;
void solve() { int x, y; cin >> x >> y;
if (x + 1 == y) cout << "YES" << endl; else if (y == 1 && (x % 9 == 0)) cout << "YES" << endl; else if (x - y == 8) cout << "YES" << endl; else cout << "NO" << endl;}
int main() { int t; cin >> t; while (t--) { solve(); } return 0; }
what is wrong in this? for Problem A)
x = 20 y = 3 Answer should be YES as 299 + 1 = 300
Wxssim Orz
Been following him since his Blog.
Way to go man! Road to red starts <3
sevlll777 My Div. 1 A passed pretest during contest but failed on pretest after contest, so please unrated the round for me.
I meet the same problem.
Same here
Rafi22 Sigma
Same
me too
Me too.
Thanks for taking away my first GM performance in div1. It took me 1 minute to repair my solution to A.
sto sto Wxssim orz orz
why there are different score distributions in div1 and div2 ?
Really simple solution for problem B // this is the code ~~~~~ void solve() { int n;cin>>n; vi a(1001,0); FOR(n){ int x;cin>>x;a[x]++; } int extra=0; FOR(1001){ extra+=a[i]; //if extra spend if(extra==1){ cout<<"No\n";return; } else{ if(extra>1){ extra-=2; } } } if(extra&1)cout<<"No\n"; else cout<<"Yes\n"; } ~~~~~
Excuse me, but why the round isn't unrated for me yet?
As you can see in the contest, I submitted problem A pretty early (after about 14 min), and the reason I only submitted 1 time is I passed the pretest with the old interactor (and also I got an Accepted verdict after the contest too). But my code wasn't correct, so I had a Wrong Answer on pretest 2 verdict after the bug was fixed.
I thought the contest would be unrated for me after a few days, but it has been almost a week already, so I'm curious if the round will be unrated for me in the future or not, and what's the reason for the long delay
Thanks for reading my comment
(Sorry for my bad English skill, I'm not a native English speaker)
(Edit: fixed some grammatical mistakes)
Same for me, I'm also waiting :(
Nevermind, see this comment: https://codeforces.net/blog/entry/139639?#comment-1247772 Seems like it's going to be fixed soon!
Sorry for the mistake in problem 2067B of Codeforces Round 1004 (Div.2). I was using ideone as well and wasn't aware that it had public or private mode. I regret the mistake and will take care not to repeat this in the future. I humbly request you not to block me or issue any penalty. Sorry for any inconvenience caused.