Hello Codeforces!
I am quite excited to invite you to my first ever round, Codeforces Round 730 (Div. 2) which will start on Jul/07/2021 17:35 (Moscow time). The round will be rated for participants of Division 2 (having rating strictly less than 2100). As always, Division 1 participants are welcome to participate in the round but it will be unrated for them.
The problems of this round will be themed on the 2005 video game Need For Speed: Most Wanted. You will be given 5 problems and 2 hours 15 minutes to solve them. One of the problems will be interactive. So, it is recommended to read the guide on interactive problems before the round.
I would like to thank -
- Aleks5d for coordinating the round and not rejecting any of my problems :)
- KAN for reviewing all the problems and providing invaluable suggestions throughout the round preparation.
- towrist and yash_daga for listening to my problem ideas even before I proposed them, providing invaluable suggestions and testing the round.
- Akulyat, andr0901, Mazaalai, talibmohd, nnv-nick, robert9524, DimmyT, kassutta, 4qqqq, wxhtzdy, karavaiev and asrinivasan007 for testing the round.
- amgfrthsp for translating the statements.
- MikeMirzayanov for the amazing Codeforces and Polygon platform.
- All the participants for participating.
I tried my best to create interesting problems with clear statements, strong pretests and useful sample test cases. Make sure to read all of them. Good Luck and Have Fun. See you on the Blacklist ranklist.
The score distribution is here: $$$500-750-1500-(1000+1250)-3000$$$.
UPD: Editorial
UPD: Congratulations to the winners
Div. 1 + Div. 2 -
Div. 2 -
First Solves -
A: Geothermal
B: Geothermal
C: olmrgcsi
D1: h0up1ngze
D2: Golovanov399
E: tfg
As a true JaySharma1048576 stalker, I know he loves car racing games and there must be some non-binary search interactive problems in this round.
burn out the wheels and tighten your seat belts, I know this round would be amazing
You can't be more accurate than that.
The scoring distribution definitely highlights the Need For Speed on the first few problems.
I am afraid of getting my comment deleted but still I can't resist my hand. As difficulty of C is more than easy version of D . It is likely that rankings mostly depend on problem you solve :)
As a tester, this round made me feel dumb. Also JaySharma1048576 and his love for interactive problems continues Context
tnks for the round :)
As a participant, this round made me feel so dumb.
As a tester, I like problems!
Early Score Distribution. Nice!
How come this blog was not on the home page.
Insane TLs because of NFS? :)
As a Java-main, I hope not! xP
R.I.P python xd
C++ go brrrr
assembly go bam bam
While that may be true, you can't submit assembly code on codeforces
Then inline that assembly into C/C++.
Nothing much, just wishing You +ve∆
Note the usual start time of the round
Never solved an interactive problem in contest .i hope i solve one this time.
It's time to regain BMW M3 GTR from Razor !! :p
I think C is going to be that one problem :)
who is Razor? tourist?)
I don't know who is Razor, but I am Sergeant Cross.
Also may I ask, who is Mia ?
she is into a whole different sport now!
She quit racing long ago.
Linkedin please
As a tester, I need your support :)
Hope I can do my best to become Specialist and break my curse in this rank.
Don't put me into the Blacklist please. :<
Edit: I made it guys, thanks :3
all the best !
Scoring distribution is giving me courage to participate in this round. I just have one confusion. Does the scoring distribution necessarily indicate that problem D is easier than problem C?
Thanks for the contest!
nope
No.
Time to revise kinematics.
[Karavaiev][user:mcdx9524][user:Mazaalai][user:talibmohd][user:nnv-nick][user:DimmyT][user:andr0901][user:4qqqq][user:asrinivasan007][user:kassutta] Hey guys could you please tell me how do you get the opportunity to test a round as I also want to do so.I have participated in 80+ contests and my max rating is 1800+
See here
Pink slips instead of delta?
Was this an attempt to leak Problem C?
I guess my name was not mentioned in the testers' list :(
Lol, yes. I double checked that to confirm.
As a tester of this round, I can say problems are very interesting and new. But make sure to read all the problem statements carefully.
As a Master, I am disappointed in my testing performance. GL to everyone
So it's going to be overly difficult and unbalanced for Div 2? Thanks for the warning I guess :/
Not really, testers traditionally brick rounds pretty hard (speaking from experience).
Yeah but so do I :)
Now that the round is over, will you concede that I was right?
Thank you for the ride. I knew you weren't blacklist material. But hey, where's your punk money now?
Last two contest were different level. Hoping that this round will give some confidence.
Two "orange" testers have already said the round was very difficult for them, so don't get your hopes up...
it's rather unusual for testers to actually hint at the difficulty. Most of them are usually like it's a nice round, the problems are nice, or would like to have everyone participate.
I interpreted this to mean this round will be especially difficult.
yes I meant these comments too.
Nothing objectively indicates anything about the difficulty but even so such comments are rare to come from the testers :P
Yeah I think it'll be a massacre but we'll see. I predict a few people who manage to solve 4+ problems will win big this round.
Pink slip from tourist
.
I also NEED some SPEED for a better rating.
hope that won't be another MATHFORCES contest
JaySharma1048576 the number $$$1048576$$$ is perfect and equal to $$$2^{20}$$$. Maybe we can expect some bit manipulation problems (or answer a binary search interactive in 20 queries), Its been a while since the last cool bit problem came :)
220 IQ? somewhat like 300iq.
after seeing the score distribution ,the theme for the contest Need For Speed: Most Wanted seems as a warning
Good luck and have fun to everyone!
see you on the blocklist. what a nice joke along with warning.
Proud moment for JaySharma1048576
The pun, see you on the Blacklist, XD lmao
lol
Wishing you all high rating!
Seriously?? Based on the theme Need for Speed!! Am so hyped up now.
Let me take on Razor once again before participating in the contest.
A B will surely be speedsolving.C will decide everything
Don't forget D part1
Now we know the answer.
Now it's time to increase my rating...
I am getting nervous. I only want that +19. :(
I am getting nervous. I only want that +29. :(
you better not think about it and just enjoy the problems when I became expert i wasn't tying to be expert I was just trying my best
I am only +34 away from specialist.I am little nervous.But once I started from 0.I should remember that.Good luck everyone!!!
Just remember that there were people with +1 away from their desired rank. Just a reminder. :)
Yeah.That's why Let's just enjoy the contest.Nothing is above than that
if you don't think about the rating, that's when you really grow.
Somewhere there is a child in every man, that doesn't grow.
still +14 away
You'll get there. Don't lose hope.
This round is going to be Speedforces
Need for SPEED.
I want some jerking acceleration for the first few problems, Hope not to get crashed or thrashed-->.
Even tourist is participating in this contest
Need for speed got many give a wrong attempt on A. :-)
C so tough that ranking is really based on Speed of A and B :-(
Am I the only one who was banging my head against a wall trying to understand the fourth problem? Great contest!
Me as well. How did >2.7k people solve it :/ Lets wait for the editorials!
It's just somewhat well known that a ^ x = y also means a ^ y = x, or any other rearrangement. Alternatively you can blame it on cheaters I guess.
I just wanna know, why ???????
Why god why ?
Am I the only one who is having hard time to understand problem C
I should've skipped this contest
Problem Statements seem too much messy.
segment tree is very indian
welp, late registration, and failed to solve anything. I will hold the L for now and practice more xd.
Not able to understood anything after solving problem A and B . C and D are bit mess
Speedy MathForces,as expected!
C is totally a mess!!
You could convert all the doubles/long doubles into long long int/int and then do a dp after that
Props to all the contestants who can imagine a hypercube in their brain.
Any faster alternative to flush operations besides switching to c++ (for java users?) Had no idea why I was tleing until I switched to c++ real quick
Mathforces and Messforces
I tried to wrote a DP solution for C but failed.
I think we can solve this with tracking values(recursion type).
That's what I did. Link You just have to be careful when converting the doubles/long doubles into integers I think.
Good
I was thinking of recursion while implementing my dp solution but i was too late.
Is it just me or was A harder than B and D1?
It's just you
How to think for problem C and D1? How to get that idea? What's the approach?
lower bound on v is 0.1, so at every step at least 0.05 probability is transferred to P so 2^20 recursive calls at max as a very very loose upper bound, alternatively set like 1e-12 epsilon since only 1e-6 is needed to pass
E(c,m,p,v) = 1 + c*E(c-v,m+v/2,p+v/2,v) + m*E(c+v/2,m-v,p+v/2,v) with a little adjustments for when c or m is 0
It's not very nice to edit your comment's meaning after posting, now you make all the people who replied about problem C look silly lol.
For D, we have $$$n$$$ queries and $$$n$$$ possible candidates for the password, so we're able to check every single candidate, which is the maximum amount of information we could hope for on this problem. To deal with the password changing, we just need to apply the same transformation to the new queries. Let's say the original password was $$$x$$$, and now it is $$$f(x)$$$. Then if we want to ask if $$$x = 1$$$, we should instead ask $$$f(1)$$$. For D1, the $$$f$$$ function is just xor of all previously asked queries, because the inverse of xor base 2 is itself.
Sorry, I thought it was problem C as I was solving it after A,B, I immediately tried to edit it.
First of all you need to know that if (a xor b)= c then (a xor c) =b.
Now you can run a loop for all the possible initial passwords ie [0,n-1] and check if the password was initially x then what would be its value right now considering all your previous guesses. Which means xor this x with all previous guesses which u can store in a variable and keep updating it on each step.
In C , can someone explain me , why only adding setprecision(9) gets the same code accepted , where as without setprecision same answer is treated as wrong answer. Is there any rule for how many digits is shown after decimals if I simply print a double variable without setprecision. Also , is this feature compiler dependent?
I am pretty sure the default precision is upto about 6 decimal places, so in problems like this you have to make sure you manually set up your precision. Had suffered a lot of WA because of that once :(
Exactly , thats what I use to think till now. And thats why I didn't mind writting it explicitly in the code bcz the error was asked to be kept less than 10^-6. But guess what , I think it is not upto 6 digits , because today I suffered atleast 10 times without setprecision and same code passed now.
because if you dont c++ can decide not to output 9 digits of precision. im not sure what are the guidelines for this but if you do the test cases there is a good chance when you actually print the numbers there will be less than 6 decimal digits, and for the answer you need at least 6 as the minimal error allowed is 10^-6
Yeah , correct. I missed this today and realised it so late. In the last input of Sample Test Cases , without setprecision , the compiler prints 4.26016 and misses the 6th digit. I guess , its definitely not 6 decimal points by default.
C was a good problem but can anyone explain how this gets WA and this gets AC.
Could anyone help me know why am I getting WA on case 3 of pretest 1? Is it an implementation error or is it some floating-point precision trick that I am missing?
https://codeforces.me/contest/1543/submission/121640564
Edit: instead of if(c>v) use if(c-v>1e-6). RIP.
What was that C? I spent a long time trying to understand that statement, please provide more clear statements next time,and also how to solve it? I didn't like any of the problems of this contest.
I spent more than half an hour to understand the problem C !!
This round is even worse than the last one.
you are right
Damn, I knew this round was going to be garbage based on the tester comments but I decided to try it anyway. RIP my rating.
How to solve C if
0 < v < 1? Is there only math solution?Btw, my C was incorrect due to big eps, it forced me to use
__float128.Why not using esp to eliminate precision error?
Me entire round:
when author wants you to guess the solution
Those are rookie numbers :
_
You gotta pump those numbers up
WA on test 1 because "The initial password chosen was 3", in the example test case was 2.
I want to now why the third part of the sample in problem C is always a little different ?
maybe you didn't control the condition like if(c>eps)
During the contest I solved D1. Then I tried to solve D2 but my solution got TLE. I changed my solution and tried again, but accidentally I sent my solution to D1 instead of D2. And unfortunately the system considered that as a resubmission. :(
For C, i got the correct answer but relative error is more than 10^-6. How to solve this issue?
I solved that by assuming $$$0 = 10^{-6}$$$
Can you explain why that works?
I was using
long doubleand because of floating point exception even if a number was supposed to be $$$0$$$, it can be something like $$$10^{-16}$$$. So I compared it with $$$10^{-6}$$$ instead of $$$0$$$.same
You can use the function of precision:
cout.setf(ios::fixed | ios::showpoint);
cout.precision(x);
Here in the brackets instead of "x" you can write what decimal you want.
All excitement for interactive went in vain.
how to solve D1 ?PLZ EXPLAIN IN DETAIL
dont worry buddy :)
Since we can guess n times let's try guessing for each of the n possible first passwords.
Obs 1: if the password is p and we're guessing the password q then the newpassword is actually p^q (^ in this case is bitwise xor)
Obs 2: we can mantain all of the changes we did to the starting password before a certain point. If we guessed the numbers 1 and 2 before and the starting password is x, then the starting password was x, then it was x^1, then it became x^1^2, so we can store 1^2^.... as we go on
What we are actually going to do is let xr be the changes we did to the starting password, then make a loop from 0 to n-1 and for each iteration we will guess xr^i. If the feedback is equal to 1, then we got it right. Otherwise, since we guessed xr, the new xr will be xr^xr^i = i. We can do that in O(N) with no problems (i assume the solution is the same for d2 but i didnt have time to debug my code for it)
Never do P(a) > v .. while dealing with doubles ! NEVER </3
For problem C, to get a AC, I should have been imprecise :))
can you elaborate
Okay. I was using c!=0 , m!=0 and p!=0 at various places.
This was causing issues because, at the end my answer varied from the expected answer by 10^-4.
I should have instead used c>=1e-6, m>=1e-6, p>=1e-6.
The former fails, but the latter passes :))
wow, this is terrible. now im wondering if the original test cases were actually imprecise, and they thought the program would just run for a really long time if they didnt do this
edit: it's most likely some floating point nonsense
How can I solve for the fractional error in the C problem (I'm having trouble with the third in sample 1, which is 5.00573 all the time)? Even with long double, the precision was not enough...
Yeah, i got the same issue, can we request author to consider these kind of solutions?
The whole "point" of the problem is to use some stupid trick to avoid floating point error. Apart from that it's just trivial implementation.
The thing is, I guess most people who used the trick are just using "cargo cult programming" and would not be able to prove the correctness of their solution.
So basically it's a problem that only stupid people who know the trick, or geniuses can solve.
I just assumed a good error would be 1e-8 so instead of checking if it's over 0 I checked if it's > 1e-8... And then prayed a lot that I don't FST because I would have needed 1e-7 maybe.
Exactly, you fall into the category of people who know the trick but don't know why it works.
I mean, double has a particular precision. That's because doubles are represented internally as exponent and base as binary, with a set number of bits. Obviously you can't write any number and what can be an "finite" number (with a finite number of decimals) in decimal might be an "infinite" in binary and vice versa. That means that there is an error in representation of a number past a certain precision so you might say 1.6565*10^(-7) but actually it gets approximated as 1.656565...*10^(-7). Double guarantees just some decimals to be correct. But yeah, I wouldn't know how to pull the exact number for the error. Adding up numbers can have some effect on the error while multiplying them can have some other effects on the error.
That's much more in depth than cargo cults...
Writing some code without understanding how/why it works is the definition of cargo cult programming.
You didn't even solve C. Please give me the mathematical proof for which it would be a particular number for the threshold. Because that's literally the only thing I don't know for sure.
Nevermind I think it's anywhere between 1e-5 (the probabilities are given with 1e-4) and 1e-14... since double is 1e-15 and we would have about 10 additions at most so 10 times the error.
Assuming that precision isn't the problem (and I can't see your submission yet), you may need to do
cout.precision(12)before outputting. You can also docout << setprecision(12)but that needs <iomanip> (and not everyone uses bits/stdc++.h)Since comparing doubles is messed up, try something like
(c-v)<=1e-6instead ofc<=vyou should not compare doubles as integers a < b || a — b <= 0.000001 instead of a <= b should work
I changed eps from 1e-20 to 1e-9 and 5.00572993744602889486 turned into 5.00505077652118574591. Didn't submit C because of this. Isn't my solution more precise?
exactly
1e-20 is ridiculously low. Doubles only have precision for 1e-15.
I got the same issue too. I think the third set of examples in question C has accuracy problems If we think that a number is Invalid, we set it to -2.0 and think that a number less than -1.0 is Invalid, we will get the answer 5.005729937446. I think this is more accurate than directly judging whether the number is greater than 1e-6.
Solution for D:
Test $$$0$$$.
for $$$1 \leq i \lt n$$$, try $$$2^{\textrm{ctz}(n)+1}-1$$$ (take the number of trailing zeros, add one to it, and test the number with that many ones in binary)
How I came up with it:
I tried small cases by hand, found that $$$"0\ 1\ 3"$$$ worked for $$$n = 3$$$, $$$"0\ 1\ 3\ 1\ 7"$$$ worked for $$$n = 5$$$, and then generalized the pattern.
Well, the main idea is "if the answer was 1 and we asked query 0 how will our answer change?" and based on that we make this query. In D1 the x^z=y and x^z^(y^z)= y^(y^z) and x^y=z so we can simply calculate the next number based on the query we just asked. I found 2 approaches to this problem. -The first one is to get the prefix xor of asked queries and change the checking number based on that number. -Second one is changing queries based on the last query. if we asked if the answer was X last turn, this turn we will check if the answer is X+1 and the query is basically the change between X and X+1 which is X^(X+1).
And on problem D2 I upgraded the second approach and made it work in base K. To solve D2 first we have to be able to do xor operation on base K. And see that after an odd number of queries the query we have to ask is inverted. for example, base=5, and the number we have to ask in query is 12341 in base 5 we have to ask query 32103 instead of 12341.
the solution is here: https://pastebin.com/embed_js/eDqDVUWQ
Sorry for my English, it might be bad. (It's not my first language)
Is it even possible to get AC on C with bitmasking? The precision error never left me. And all the top position in standing Reds seem to have used recursion.
https://codeforces.me/contest/1543/submission/121654469 Yes. Sadly, I didn't submit it because I had a precision error on the third sample test.
using double in C: relative error is like 10^3 using long double in C: relative error is like 10^4
fuck me i guess
C was the worst question ever asked on Codeforces. 1e-6 is never equal to 0. I don even understand how this question was even allowed for a rated codeforces round.
What do you mean by 1e-6 is never equal to 0?
Why can't we take 1e-18 to be equal to 0. It's not a lesson for me, it's the stupidity of setters.
You should know that double numbers are dangerous. If you don't, this is a good lesson for you.
Maybe this proves something code here
3 things I learned after the round:
Use
a > 1e-9instead ofa > 0.Use
a > 1e-9instead ofa > 0.Use
a > 1e-9instead ofa > 0.bruh, literally this changes answer from 5.0301143 to 5.0050777 ?? xD thats like 0.025 difference, yo?
anyway goodbye candidate master
Same error! Good lesson for comparison of real numbers XD.
Does greedy colouring not work on E? Otherwise what is TC 4?
Me while reading C
https://codeforces.me/contest/1543/submission/121580369
can anyone explain this coding style.....hope everyone get what I am trying to convey
ant++,wasp---
I think when real hackers hack the codes like this they try hard to find the main code.
haha++ hacker_work++
These are just ways to bypass plagiarism. I guess by now, its a well known fact that there are telegram groups and youtube channels which leak the solutions during the contest and people like this (author of this code) just copy and do this stupidity to not get banned.
Me after solving A and B
Every one is talking about problem C , is there any body like me who couldn't solve the problem A
me :D
mathforces let's go!
For me A was much harder than B. Solved it, but wasted a lot of time and also felt really nervous waiting for system tests.
And I did fail solving problem A in one of the older contests too. That's nothing to be proud about. Just shows that more training is necessary for these particular types of problems.
I swear I'll never give any writer's first contest.
Who can tell me why java8 made D1 overtime.My algorithm is the same as others.
Try reading bytes directly from
System.ininstead of wrapping it withBufferedReaderandInputStreamReader.directly?by Scanner? I submitted the same code with C，ac .= =!!!(f**k)
I think
Scanneris slow, you can try with myFastReaderclass: https://codeforces.me/contest/1543/submission/121654156For non-interactive problems I wrap
System.inwithBufferedInputStream.wow！thank
The problems were as bad as getting tle on test case 47
TLE on 47 is ok if you are tourist
We want old codeforces back. It was used to be so much fun.
(⋟﹏⋞)(╥_╥)
I read Problem C at least 10 times.but I couldn't understand what type of probability was that?
Can anyone who passed pretests on C tell the number of pretests on that problem?
I believe problems A, B, C, and D1 had 3 pretests each.
A, B, C and D1 had 3 pretests each, D2 had 7 pretests and E had 16 pretests.
can anyone explain me what was problem C(problem statement) in human language, pls ?
you have a chance to get one of 3 things a,b and c, each with probability pa, pb, pc. if you get a or b, you keep getting more things, but when you get c you stop
in addition, if you get a or b, the probability of it decreaces by v, and that ammount is distributed to the other things. for example, if pa = 0.2, pb = 0.2, pc = 0.6 and v = 0.1, if you get a the new probabilities will be pa = 0.2 — 0.1 = 0.1, pb = 0.2 + 0.1/2 = 0.25, pc = 0.6 + 0.1/2 = 0.65.
i think i have a solution if you want but floating point nonsense screwed me really hard in the contest
Thanks man, I got the solution also. Understand the problem statement seems hard than actual solution for me.
I just don't understand why my code for C outputs correctly on my computer but wrong on codeforces,
I don't understand why 0.6*10000 on my computer is 6000 but on codeforces 5999
i think this is because trying to parse 0.6 into binary gives an infinite decimal, so depending on how the implementation of the multiplication is in your computer and on codeforces computers you can get either answer
JaySharma1048576 : I tried my best to create interesting problems with clear statements
clear statements??? you sure?
What was unclear?
Statements
https://codeforces.me/contest/1543/submission/121632020
why this is wrong?
long long, while n and n-x can be regular integers, nothing is bounding n*(n-x) to be smaller than 1e9
I have defined long long as int only
my god... my dear god...
1- stop using defines, it's for your own good
2- if you're going to use defines, define it without conficting with existing typenames
not only defines do strange behavior because it's litterally copypasting what it's defined in the code but there are infinitely more predictable alternatives like typedef for types, functions for things like your "TotalSum" and const int for values.
Do that and your code will work, there is some evil black magic fuckery going on with it.
edit: it turns out the problem was not what i said but i still stand by it. If all of the evil stuff wasn't going on, it would be obvious what to pinpoint what was actually wrong with your code because the only thing that could go wrong is the function you didn't write, accumulate
The type of the returning value of accumulate() is the type of the third parameter. In your code you used 0, which is by default an int. In this case you should use 0LL.
Note that the type always depends on the third parameter only. So if A is vector long long, accumulate(A.begin(), A.end(), 0) is still wrong.
It wasn't really good. The difference between B and C was very huge. It could be better
So fun contest! Cars, racing, hacking... Very thanks to the authors! It's was really cool.
Thanks for the round! For anyone interested, here's a link to my screencast.
Woke up half an hour before the contest , gave it , and now i think i should have slept instead.
Problem D1 : https://codeforces.me/contest/1543/submission/121622589
I didn't put the break condition if the solution is found but the main tests passed. How is that possible ?
the judge is adaptative, so it will try to "change" the initial password to see if you can guess all of them. so the judge is like "let's say the initial password is 0" and you get it right, then it says "let's say the initial password was 1 instead" until you get to the last one.
No matter you think this round is interesting or not, honestly, it's a different round.
Solved 4 Question after long time.. Expert Again!!!
When I implemented my solution for problem C, it printed
4.263965352for the subtest 4 of test case 1 but it got AC when I submitted. My submission is here.I tried to run that test on AtCoder as well and it gave the same result. I even tried to run some AC codes with that test and the results were also the same. The output of that test case on Codeforces was
4.260163674, so weird.Is there any problem with the compiler?
Actually, for problem B it's much easier to prove that the cars have to be as evenly as possible. You can assume that array a is sorted withouth losing the generality of the problem, so when you do the sum of modules you will get a2-a1+a3-a1+..+an-a1+...+an-an-1. It's very easy to see that a1 comes with the most negative signes in front, then comes a2, then a3, and so on. So in order to minimize the total sum you need to have a1 as big as you can,..., an as low as you can. But initally you assumed that a1<=a2<=a3<=...<=an. So it's now obvious that you must distribute the cars as evenly as possible.
poor english and bad google translation caused me fail to understand what pC was talking about
That was fun. Sweet and viscous masochistic fun. It's becoming a painful habit without Educational Rounds.
I think LordVoldebug is a Div1 contestant and you mean LordVoIdebug, winning the 4th place. Wow, they are the same person xD
Sorry. I misunderstood
linstead ofIin his username. Both of them look so similar.Thanks for a round, especially for problems D and E — surprisingly simple, yet challenging and mind-bending.
Couldn't even get A right... I suck
I am newbie class with low contest rating (891), but I scored 1812 in the contest. I would have expected some rank points. Does anybody know why I am not getting any?
The user ratings hasn't been updated yet for today's content. If you see the final standings, there should be rating update tab.
its been 10 hours and still rating is not been changed
I think the problems will be better if they have no or shorter story backgrounds.
Anyway, it is a good contest!
Is the content unrated?
"The round will be rated for participants of Division 2 (having rating strictly less than 2100)."
Then why my rating is not changed at all?
It will get updated soon. Don't worry.
Why my rating is not increase or decrease in Codeforces Round #730 (Div. 2) as I solve 3 out of 6 questions with a standing of 1863. What's the criteria of rating increment.
nobody got their rating changes yet, relax
You did not solve bro . You cheated . Your Submission to this problem clearly has a message "forwarded from" in the first line itself . Cheaters like you should be banned .
what it means by forwared from ? means copy paste?
his submission has this line "forwarded from" in the first line and he got a compilation error due to that . someone might have sent him the solution somewhere and he just copy pasted it and with the message he also copied and the forwarded thing that appears when someone forwards you a message in some chat apps . Hence , caught XD ...
Yep. Quite similar to this solution (taking out all the ants and wasps)
Pure comedy gold, this needs to be framed and kept in eternity
Hahaha cheater caught
You didn't think Cheating was gonna get caught?
sir that's the truth whether you believe or not I don't care
well, the number of skipped submissions in your submission history is the proof that u are a cheater. So please do not try to explain .
Why rating change take so long?
To me this round is a little bit hard compared to the previous rounds... I think
can anyone explain why at n=1 in problem B answer is 0?
Inconvenience is defined as the absolute difference between each pair of subtracks. If n==1, there is only one subtrack and hence no pairs => inconvenience = 0.
Can someone please tell me if this round is unrated??
is this contest unrated?
The ratings have been preliminary updated. Later, I will remove the cheaters and recalculate the ratings.
Orz
looking at the standings, me who did not participate in this round, telling myself " dude what happened here!"
This D1 solution was accepted https://codeforces.me/contest/1543/submission/121690053
My same D1 solutions were all rejected https://codeforces.me/contest/1543/submission/121633283 https://codeforces.me/contest/1543/submission/121631556 https://codeforces.me/contest/1543/submission/121634285
Is it time to move to C++? the last one even passed the pretest...
submit your same code in python 3 instead of pypy 3 and see the magic
https://codeforces.me/contest/1543/submission/121700436
how do you test or debug interactive problems? I've written soln for D2 but no idea what's wrong or how to start looking for the bug.
I used something like this for debugging my D1 and D2 solutions (a mostly untested conversion from D language):
If I pass
interactive_judgefunction pointer to thesolvefunction, then it uses stdout/stdin as required by the codeforces platform. But if I passsimulated_judgefunction pointer to it, then everything gets tested locally and can be debugged just like any other non-interactive problem.So this contest proved to be one of the worst contest.LOL