By pikmike, history, 2 weeks ago, translation,

Hello Codeforces!

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Ne0n25 Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 dzh_loves_mjy 7 204
2 neal 7 231
3 WZYYN 7 250
4 noimi 7 357
5 Um_nik 7 384

And finally people who were the first to solve each problem:

Problem Competitor Penalty
A Kirill22 0:01
B dzh_loves_mjy 0:04
C SSerxhs 0:05
D gleb.astashkin 0:17
E Pigbrain 0:19
F WZYYN 0:54
G OnlyG 0:16

UPD: Due to the issues with problems A and B the round is unrated.

UPD: Editorial is out

• -539

 » 2 weeks ago, # |   0 Just 15 points away from specialist!! Hope to cover in this round!!
•  » » 2 weeks ago, # ^ |   0 Good Luck bro!!
•  » » 2 weeks ago, # ^ |   +14 lol the rounds unrated. hope you get it next time tho!
•  » » » 2 weeks ago, # ^ |   +8 My Bad!! either way i was stuck at A for like 10 minutes because of changing statements so, thought of giving up already:(
•  » » » » 2 weeks ago, # ^ |   0 even I had 1377 was hoping to go back to specialist ,had even solved two questions real quick but never mind
•  » » » » » 2 weeks ago, # ^ |   0 Ohh, so you managed to solve A?
•  » » » » » » 2 weeks ago, # ^ |   0 yeah xD knew my logic was right and when they corrected it I confirmed with given output and submitted
•  » » » » » » » 13 days ago, # ^ |   0 can you explain the logic please?
•  » » » » » » » » 13 days ago, # ^ | ← Rev. 2 →   +8 look at how many sticks there must be, from that you can convert some to coals to have at least $k$ sticks and $k$ coals.$\lceil{\frac{k+yk-1}{(x-1)}}\rceil+k$ is the answer
•  » » » » » » » » » 13 days ago, # ^ | ← Rev. 2 →   0 t = int(input()) for i in range(t): x,y,k = [int(x) for x in input().split()] ans = (k*y) + k count=0 count = m.ceil((ans-1)/(x-1)) count += k print(count)this is the ans i submitted, but i got wa
•  » » » » » » » » » 13 days ago, # ^ |   +4 I had the same issue. I think the ceil function does not work properly for long long numbers.
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13 days ago, # ^ |
-38

# include<bits/stdc++.h>

using namespace std; int main(){ long long int t; cin>>t; while(t--){ long long int x,y,k,c,d; cin>>x>>y>>k;

c=(k+y*k-x)/(x-1);
d=(k+y*k-x)%(x-1);
if(d==0)
cout<<c+1+k<<endl;
else cout<<c+2+k<<endl;

}


} whats the prob wid this code?

•  » » » » » » » » » 13 days ago, # ^ |   0 When x>(k+y*k), we need to print k+1 as you will exchange one stick for x sticks and k sticks for k coals.
•  » » » » » » » » » 13 days ago, # ^ |   0 Cuz you have int as input for x,y,k
•  » » » » » » » » » 13 days ago, # ^ |   +1 thank you so much. It got accepted
•  » » » » » » » » » 13 days ago, # ^ |   0 yep it got lot of my time too
•  » » » » » » » » » 13 days ago, # ^ | ← Rev. 2 →   0 Nicely explained
•  » » » » » » » » » 13 days ago, # ^ |   0 NoobM, shaded I had the same issue. And you are right python's inbuilt ceil function and as a matter of fact square root function does not work properly in case of long long numbers.So, to solve this issue I found out a solution on stackoverflow. To find the ceil value of division of a by b you can simply do -(a//(-b)) in python and -(a/(-b)) in C++, i did this and got AC.
•  » » » » » » » » » 13 days ago, # ^ |   0 oh nice! Thanks for the solution!
•  » » » » » » » » » 13 days ago, # ^ |   0 i also do the same very easy and simple logic
•  » » » » » » » » » 13 days ago, # ^ |   0 you are right bro
•  » » » » » » » » 13 days ago, # ^ |   0 look you have 0 coals so you will need k steps for k coals.Now as for the sticks you need k sticks to match the coals for conversion into torch.You also need k*y sticks extra so they can be converted into coals later.So, total you need to add (k*y+k-1) sticks extra(since you already have one stick). Now in every trade for sticks you gain x-1 sticks,so, number of steps required is :- (val/(x-1))+min((int)1,(val%(x-1))) where val is (k*y+k-1). So,total no. of steps is (k+(val/(x-1))+min((int)1,(val%(x-1))))
•  » » » » » » » » » 13 days ago, # ^ |   0 thank you!
 » 2 weeks ago, # | ← Rev. 2 →   -46 deleted
 » 2 weeks ago, # | ← Rev. 2 →   +61 Just curious to know, How they have such a large collections of problems?they are soon going to make century of educational rounds
•  » » 2 weeks ago, # ^ |   -14 $Events\,\, seemingly\,\, repeat \,\,themselves \,\,between \,\,the\,\, years\,\, of\,\, the \,\,time\,\, loop$ — Dark
•  » » 2 weeks ago, # ^ | ← Rev. 2 →   +11 6344 problems as of now. And they are uploading for like 10 years not a single person's job.
 » 2 weeks ago, # |   +9 before: educational so unbearable will again be one mathNow: wow they are doing really interesting rounds of course I will write it
 » 2 weeks ago, # | ← Rev. 2 →   -9 Hope I will get some increase today...
•  » » 13 days ago, # ^ |   +19 hope i will get billion dollars today
 » 2 weeks ago, # |   +411
•  » » 2 weeks ago, # ^ |   -29 That's so real!
•  » » 2 weeks ago, # ^ |   +44 strong pretest, short statement (×)short pretest, strong statement (√)reference : https://codeforces.com/blog/entry/80333
 » 2 weeks ago, # | ← Rev. 2 →   -85 A
•  » » 2 weeks ago, # ^ |   +41 There has been bad incidents in the past( Long queues, round being unrated, pretests weak/wrong) whenever some author didn't thank Mike. So nobody wants to take the risk you see XD
•  » » » 2 weeks ago, # ^ | ← Rev. 2 →   -59 A
•  » » » 2 weeks ago, # ^ |   -62 Maybe Mike was angry that they didn't thank him and cursed the round
•  » » » 2 weeks ago, # ^ |   0 There has been bad incidents in the past( Long queues, round being unrated, pretests weak/wrong) whenever some author didn't thank Mike. So nobody wants to take the risk you see XDafter thanking also, everything happened which you have stated(long queues didn't happened bcz everybody left after this much) lol.
•  » » » 13 days ago, # ^ |   0 Truth !
•  » » 2 weeks ago, # ^ |   +131 Mike unlinke Mark doesn't sell your data, maintains ad free site, we get quality problems completely free of cost, and without watching any ads like SPOJ. People thank Mike out of gratitude for him.
 » 2 weeks ago, # |   -11 Wish me luck, please!
•  » » 2 weeks ago, # ^ |   -18 Let's exchange
 » 2 weeks ago, # |   -22 Very excited :D
 » 2 weeks ago, # |   +41 Just curious to know, we don't have any problem tester in educational rounds? Or is the same set of people involved in testing the problems?
•  » » 13 days ago, # ^ |   +3 If only they were less greedy and given free contribution to testers, this round would have been rated
 » 2 weeks ago, # |   +7 I hope it has strong pretest and interesting problems!
 » 2 weeks ago, # |   -9 I would like to know which debuging tool are you using? Oh, I suddenly realized that we don't need in cf :D But I still want to hear suggestions
•  » » 2 weeks ago, # ^ |   +5 Visual Studio Code is my first choice.
•  » » » 13 days ago, # ^ |   0 I will give a try.
•  » » 2 weeks ago, # ^ |   0 If your using mingw/gnu then it comes with gdb. U can use that for debugging its pretty cool.
•  » » » 13 days ago, # ^ |   0 I also gdb now. But I think its too old to use terminal... Maybe I need to learn more about it.
•  » » 2 weeks ago, # ^ |   +51 Print Everything
•  » » » 13 days ago, # ^ |   0 cout :D
•  » » » » 13 days ago, # ^ |   0 Also cerr for fun
 » 2 weeks ago, # |   +3 Purple Rain
 » 2 weeks ago, # |   +14 Hey, this is my first time participating in the educational round and I have a simple question. I quite can't grasp the idea of the time penalty. (I searched up the rules but I still don't get it.)To the best of my understandings, is it one of the followings? 1) the final score is directly related to time consumed 2) total time of 2 hrs gets shorter by 10 minutes for each wrong submission
•  » » 2 weeks ago, # ^ | ← Rev. 3 →   +46 Let's say you submitted 2 source code. Submission 1: got the WA at 00:03 Submission 2: got the AC. at 00:07 Then you get just the same score who got the AC at 00:17.
•  » » » 2 weeks ago, # ^ | ← Rev. 2 →   +27 Thanks :) Have a wonderful day!!
 » 2 weeks ago, # |   +68 All the best Guys, noob here hope i will make some score today..
•  » » 2 weeks ago, # ^ |   +31 I think you are a famous noob as you are getting upvoted by people :)...otherwise lower rating participants always get downvoted here
•  » » » 2 weeks ago, # ^ |   +18 dont worry i upvoted you ;)
 » 2 weeks ago, # |   +44 This contest is a nightmare.
 » 2 weeks ago, # |   +4 so sad that this round will be unrated after all the effort to solve the first few problems
 » 2 weeks ago, # | ← Rev. 2 →   +17 Thank God! It got Unrated. Problem A made me feel so worthless today.
 » 2 weeks ago, # |   +4 Back to UnratedForces...
 » 2 weeks ago, # | ← Rev. 2 →   +14 problem A was so irritating before:( now it is good:)
 » 2 weeks ago, # |   -84 Please let this round rated back :((. Just a small issue from the problem A right ?
•  » » 2 weeks ago, # ^ |   +36 Not a small issue for those whose schedule and timing of the contest is completely ruined.
 » 2 weeks ago, # |   +90 It was obvious that round will be unrated when I saw those wandering trader deals.
•  » » 2 weeks ago, # ^ |   +17 Also, 1stick+1coal -> 1torch wtf?
•  » » » 2 weeks ago, # ^ |   +54 It's minecraft
•  » » » 2 weeks ago, # ^ |   +594 .
•  » » » » 13 days ago, # ^ |   0 hahaha XD, Noice usage of Stonkz!!!
 » 2 weeks ago, # |   0 Problem A strikes again.
 » 2 weeks ago, # | ← Rev. 2 →   0 It's unfortunate that the negativity started in the comment section, in particular this comment has eventually cursed the round and a good problemset. :'(
 » 2 weeks ago, # |   +47 Out of curiosity how did whatever mistake happened slip through on A? It seems to have been a major error considering that all the samples itself were wrong.
•  » » 2 weeks ago, # ^ | ← Rev. 2 →   +6 Anyone got it right when the data was not changed?I made the same mistake and passed all tests when the sample was not changed. Maybe the testers said "Hey, how could there be bugs in A?" and simply let it slip thru.
•  » » » 2 weeks ago, # ^ |   0 I submitted within a minute of the change, there were 800 submissions by that point. There were 400-500ish before that.
•  » » 2 weeks ago, # ^ |   0 really wonder how this obvious error can be passed in Polygon...
•  » » » 2 weeks ago, # ^ |   +3 Yeah, I would expect that there should have been at least a "should TLE" brute force. Moreover how did it pass testing? Did all the testers make the same mistake? Frankly this makes me think that educational rounds don't actually have a testing phase.
•  » » » » 2 weeks ago, # ^ |   0 literally likely. A previous error make that Problem B can be accepted by almost any way, but nobody noticed that and as a result I got unrated.
•  » » » » » 2 weeks ago, # ^ |   0 Wow, and it gets worse, B's checker today is also wrong.
•  » » » » » » 13 days ago, # ^ |   0 wish generator is right
 » 2 weeks ago, # |   +3 I thought for a long time why the test case came out like this.
•  » » 2 weeks ago, # ^ |   0 me too :(
 » 2 weeks ago, # |   -34 But i was doing so well :(((
 » 2 weeks ago, # |   +20 They should explain the sample test case of first question.
•  » » 2 weeks ago, # ^ |   +12 New way to make problem A hard to implement.
•  » » » 2 weeks ago, # ^ |   0 They explained it. LOL
•  » » » » 2 weeks ago, # ^ |   0 Yeah xD
 » 2 weeks ago, # |   -96 Damn these unrated rounds. Ruined my first contest.
•  » » 13 days ago, # ^ |   -100 Does this stepmother have cumshots on her boobs?
 » 2 weeks ago, # |   +6 I feel really sorry for setters. They have make good problemset using a lot of efforts. Contest has been ruined only by A problem.
 » 2 weeks ago, # | ← Rev. 2 →   -14 Unrated round because of a simple maths problem. Sad for setters and competitors, there should be a tester in edu rounds
 » 2 weeks ago, # | ← Rev. 2 →   +38 This is pretty sad. Even if educational rounds don't have testers (which they should), can't you at least have more than 1 person write model/correct solutions? I know how polygon works and it's not very hard to simply ask someone else to upload another solution and find any mistakes, especially in problem A.
•  » » 13 days ago, # ^ |   +3 I agree with MagentaCobra
 » 2 weeks ago, # |   -109 WHY UNRATED? IT WAS THE EDUCATIONAL ROUND IN WHICH I COULD HAVE IMPROVED MY RATINGS. WHY THIS ALWAYS HAPPENS WITH ME.GOD! AM I THAT BAD?
•  » » 13 days ago, # ^ |   +27 username checked.
 » 2 weeks ago, # | ← Rev. 2 →   -8 Does anyone check the correctness of the problem A??? Bad experience!
 » 2 weeks ago, # |   +6 Im sad because problems B and C were cool, usually I dislike them, but in this contest they were pretty nice :(
 » 2 weeks ago, # | ← Rev. 3 →   -10 Sorry, looks like its only my problemDELETED
•  » » 13 days ago, # ^ | ← Rev. 2 →   -7 this is the problem D screenshot from my screen, just notice to word "piles"also i asked some of my friends and they had same problemBTW, if you don't have this problem, sorry for unrelated comment
 » 2 weeks ago, # |   +26 Leaving aside A, other problems are very interesting, do give it a try guys :D
•  » » 2 weeks ago, # ^ |   0 But it is no point for me to stay overnight now... I've refrained from staying up for three days in preparation of this contest
•  » » » 2 weeks ago, # ^ |   +9 wow, that sounds so inspiring, but don't worry your hard work will pay off in the next contest for sure :D
 » 2 weeks ago, # |   -8 Damn that was the first time I've solved 2 problems in 20 minutes. But... it's unrated. So sad.
 » 2 weeks ago, # |   -8 hello everybody. I'm curious to know how to become a tester of the round. if there is a link or description about becoming a tester of the round please leave it under this comment. thank you.
 » 2 weeks ago, # |   +71 Please don't keep sample tests unexplained even if its problem A.
•  » » 2 weeks ago, # ^ |   +15 I think the author could have realized his mistake had he written the explanation of the sample tests once.
 » 2 weeks ago, # |   -24 Can we please have a div3 at least until saturday?
 » 2 weeks ago, # |   -25 Bad round and not clear statements -_-
 » 2 weeks ago, # |   +12 Can you please explain the test case 3 for problem A,
 » 2 weeks ago, # |   +59 ()
•  » » 2 weeks ago, # ^ |   +19 Tbh, problem A is pretty easy just like normal A's a simple one line formula for AC, only difference is, they messed up the sample test cases which made it a lot confusing, adding on to that they didn't provide any explanations.
 » 2 weeks ago, # |   -11 Every other educational round: me: if I can't solve B I definitely can't solve C. them: we shall make sure you don't solve B. me: end up solving just A. Today: me: solves A & B in under 20 minutes. them: unfortunately this round will be unrated. P.S: I know the frustration is higher for the writers when a round becomes unrated, so feeling sorry for them too.
 » 2 weeks ago, # |   +119
•  » » 13 days ago, # ^ |   0 lol
•  » » 13 days ago, # ^ |   0 This also happened to me..
 » 2 weeks ago, # |   0 I wrote 2 codes for problem B they gives same output to sample but one of them passes the sample one of them don't but they gives same output for output but when i submit it says wrong on this sample input why?
 » 2 weeks ago, # |   +25 Before the samples of problem A were fixed, there were already about 1K submissions. I wonder what solution they submitted?
•  » » 2 weeks ago, # ^ |   +5 Can I tell it now? I made the same mistake.
•  » » 2 weeks ago, # ^ |   +7 None of them did add the other operation and it was more likely that no one added.
•  » » 2 weeks ago, # ^ |   0 Maybe they did so without seeing the sample testcase and worked it on their own... I have seen many coders do that maybe out of competition or confidence
•  » » » 2 weeks ago, # ^ |   +16 If the jury's mistake was: the statement says you should count both type of operations or at least is ambiguous so it can be interpreted this way, but model solution counts only one of them, then IMO it's a little weird decision to change the tests rather than just make an announcement like: "you should count only operations of type 1, we have updated the statement to make it clear".
•  » » » » 13 days ago, # ^ | ← Rev. 3 →   +10 It was worse than that as far as I understandThe jury's solution also had a rounding error. So they couldn't just correct the statement the way you suggestThis was the accepted solution at the time of the round  need_sticks = (y + 1) * k cnt = (need_sticks - 1) // (x - 1) print(cnt) 
•  » » 13 days ago, # ^ | ← Rev. 3 →   +22 I adjusted my solution to the samples without much thinking when I saw that it does not match the result. Added incorrect rounding and omited the second operation. I thought "I don't have time to think about it, will figure out why it has to be this way after the contest"And then I had to return everything back...
•  » » » 13 days ago, # ^ |   +6 Wonder whether author of second model solution did the same
 » 2 weeks ago, # |   0 Did a rejudge on B just happen? Some people's Bs got converted from pretests passed to WA on pretest 1.
•  » » 2 weeks ago, # ^ |   0 yeah
•  » » 13 days ago, # ^ |   0 What was the mistake though? Mine remained AC so i didn't get what they changed.
•  » » » 13 days ago, # ^ | ← Rev. 2 →   +13 A friend of mine (ViciousCoder) misread the question and wrote a solution in which the earliest negative point was minimized, not the latest, so I guess almost anything worked.
•  » » » » 13 days ago, # ^ |   0 Ohh shit. that was bad.
•  » » » » 13 days ago, # ^ |   +8 I wonder who that friend is :p
•  » » » » » 13 days ago, # ^ |   -8 Happy now?
•  » » » » » » 13 days ago, # ^ |   +8 Lmao I was just messing around.
•  » » » » » » » 12 days ago, # ^ |   0 I really should have put a ":P" at the end of the last comment, I wrote that jokingly as well.
 » 2 weeks ago, # |   -10 I am just curious, How is it happen. Is both setter and tester got the same idea for the solution.
•  » » 2 weeks ago, # ^ |   +3 Probably I actually got the wrong solution on my first try which matched their test case.
•  » » » 13 days ago, # ^ |   0 That's crazy level of coincidence
•  » » » » 13 days ago, # ^ | ← Rev. 2 →   +3 Well the mistake was not counting the 2nd kind of trades so basiclly we output ans-k instead of ans.
 » 2 weeks ago, # |   -25 Problem A like a nightmare:(
 » 2 weeks ago, # |   +33
 » 2 weeks ago, # | ← Rev. 2 →   +57 this is what happens when you push code to production without writing unit tests
 » 2 weeks ago, # |   +1 This contest is really cursed.
 » 2 weeks ago, # |   0 When I found the problems easy and solve them fast, the round became unrated. Bad luck for all :(
 » 2 weeks ago, # | ← Rev. 2 →   -13 I accidently submit wrong answer according to the previous test case of A and my rank was under 150 and i am not able to believe that a newbie is under 150 rank initially ,i realised i have become pro for sometime :) :)
 » 2 weeks ago, # |   -7 What a stupid contest !
 » 2 weeks ago, # |   0 there should have been testers for the round , as 2 problems had issues in them.
 » 2 weeks ago, # |   +21 Didn't anyone test this pre-test?
 » 2 weeks ago, # |   0 is the round unrated?
•  » » 2 weeks ago, # ^ |   0 yes:(
•  » » 2 weeks ago, # ^ |   0 yes
 » 2 weeks ago, # |   +3 The contest problems were good enough but only problem was with the checker.Enjoyed problem solving
 » 2 weeks ago, # |   +25 Waiting for announcement regarding Problem C
 » 2 weeks ago, # |   -14 now as the contest is unrated , will there be a hacking phase or we will go strait up to system testing?
•  » » 2 weeks ago, # ^ |   +22 Who cares? XD
 » 2 weeks ago, # |   +4 Every time I perform little better the contest gets unrated. ;__;
 » 2 weeks ago, # |   +8 BledDest for preparing A & B ? Still remember he writes his checker by casting INT to BOOL.
•  » » 13 days ago, # ^ |   +31 No, this time I prepared G
•  » » » 13 days ago, # ^ |   0 Great thanks to you for preparing EDU rounds and wish a better experience next time !
 » 2 weeks ago, # |   +38 ;(
•  » » 13 days ago, # ^ |   +1 i once solved and got hacked
•  » » 13 days ago, # ^ |   0 Same :(
 » 2 weeks ago, # | ← Rev. 2 →   0 I think there is a wrong output in Problem B 4th Case:Your ouput: -4 0 1 6 3p=[−4,−4,−3,3,6] k=3 But i have better Solution:1 0 -4 6 3p= 1 1 -3 3 6k=1.
•  » » 2 weeks ago, # ^ | ← Rev. 2 →   0 I think we just don't understand what the problem really want , I have the same case as you in my solution
•  » » » 2 weeks ago, # ^ |   +3 But I got AC on Problem B just ignored their output
•  » » » » 2 weeks ago, # ^ |   0 Good Job bro
•  » » 2 weeks ago, # ^ |   0 Let k be the maximum j (1≤j≤n).. k is not negative value count that is index.. but your solution is also correct max index is also 3 as in the test case !!
•  » » » 2 weeks ago, # ^ |   0 Ok..Thanks
•  » » 13 days ago, # ^ | ← Rev. 4 →   0 You don't have better solution (just a solution) :P...Create prefix array for [1, 1, -3, 3, 6] and you'll get it. P( j ) < 0, j = 3
 » 2 weeks ago, # |   0 I didn't read comments . Now I got to know the round is unrated.. __
•  » » 2 weeks ago, # ^ |   +1 you even did not get the notification ?? :/
 » 2 weeks ago, # |   0 Couldn't even solve one complete question. Need a lot of practice for me. Any good resources and motivation to learn more about competitive coding.
•  » » 13 days ago, # ^ |   +3 It is OK, go to the problem set sort the problems by difficulty and start practicing with the easier ones, you will be able to solve A problems in a short time
 » 2 weeks ago, # |   0 The contest seemed good enough. Easy A, B and C with correct difficulty gradient. However, D was not clear to me at all. What did it even want to say? What was the approach for it? In fact what did it want us too :|
 » 13 days ago, # | ← Rev. 2 →   +1 Problems were outstanding. But, A's sample test ruined it :(
•  » » 13 days ago, # ^ |   0 not only the sample test , the whole tests.
 » 13 days ago, # |   0 Very sad to know that the round is unrated after I tried so hard to solve D&E..
 » 13 days ago, # |   +17 Rating is just a number. The problems were interesting and instructive. Thank you all problem setters! Waiting for the editorial.
 » 13 days ago, # |   +25 Thanks to GOD that the contest was unrated....... The problems are hard and hard to understand.......
•  » » 13 days ago, # ^ |   0 For the first time, I solved 4 problems without any WA. sedlyf
•  » » » 13 days ago, # ^ |   0 What a pity for u that the contest was unrated...
•  » » » » 13 days ago, # ^ |   0 yes bro!
 » 13 days ago, # |   +320 I'm really sorry about the issue with the problem A. This was completely my fault. I re-checked the formula yesterday and discussed it with others and nobody found any mistakes.1) I didn't round up the result of the division.2) I forgot about $k$ trades you need to get coals.So, my solution was completely incorrect. I will not blame anyone else because this was my problem and I failed its preparation. I didn't want this round to be unrated and this issue right at the start of the contest just completely ruined me. Sorry guys.
•  » » 13 days ago, # ^ |   +38 its ok, human are not perfect
•  » » » 10 days ago, # ^ |   0 Or are they?
•  » » 13 days ago, # ^ |   0 you are talking about taking the ceil right?
•  » » 13 days ago, # ^ |   +15 aw man
•  » » 13 days ago, # ^ |   +51 Mistakes happen. You are a great problem setter. I've always liked your problems <3
•  » » 13 days ago, # ^ |   +14 It is OK man, it is kind of encouraging for us beginners to see that even strong coders sometimes make mistakes in problem A ;-)
•  » » 13 days ago, # ^ |   -9 wouldn't it be nice idea to extend the time of round by 10 more minutes instead of making it as unrated, because the problems are extremely good , feeling sad for this round becoming unrated
•  » » » 13 days ago, # ^ |   +11 Sadly, if the first problem is wrong, it affects almost anyone. And also this was not the only issue because the checker of the problem B was incorrect. Cursed round. Really cursed.
•  » » 13 days ago, # ^ |   +2 Well mistakes do happen, but the problem itself was good.Just curious how did it slip away from testers?
•  » » 13 days ago, # ^ |   0 So,the outputs for the sample test cases are incorrect?
•  » » 13 days ago, # ^ |   0 This is a pretty bad mistake, but you handled it pretty well during the competition and now by admitting to it. I'm surprised that nobody noticed the issue, did anyone actually solve the problem (in code) other than you?
•  » » » 13 days ago, # ^ |   0 One person solved it before the contest. Sadly, he just tried to find the formula which passes the example. This is fun and sad at the same time.
•  » » » » 13 days ago, # ^ |   +8 Well don't you think it's a good idea to have multiple testers on every problem regardless of its difficulty? If testing status were always like this, this situation were bound to happen.
•  » » » » » 13 days ago, # ^ |   +1 There are reasons Educational Rounds are pretty rarely tested and these reasons almost don't belong to me. Moreover, I'm pretty rare guest in Educational Rounds now because I'm mostly busy with Div.3.
•  » » » » » » 13 days ago, # ^ |   0 That's kinda sad to hear. One might expect sponsored rounds to be more prepared than the regular ones but it sounds like it's not the case.
•  » » 13 days ago, # ^ | ← Rev. 2 →   0 But i am surprised to see about 700 people solve the problem before correction. Looking at this i thought i might got something wrong.
•  » » 13 days ago, # ^ |   +69 No rated contest should have problems that are not stress-tested with a naive slow solution.But yeah, shit happens.
•  » » 13 days ago, # ^ |   0 did you give it to newbies to check?
 » 13 days ago, # |   0 Problem D — For input [1,3], we move 1 to 2 and 3 to 2. This makes both of them together.Kindly correct me, if I got it wrong !!Can anyone explain the test case [1,2,4,8,9,100].How are we moving the piles to make them together?
•  » » 13 days ago, # ^ | ← Rev. 2 →   +3 for [1,3] you just keep each one in its place. The problem says to put all of them in at most 2 piles, not necessarily 1.
•  » » 13 days ago, # ^ |   +3 For input [1,3] the answer is 0, you can leave them at their original positions because you need at most two pilesFor case [1,2,4,8,9,100], you move all the piles except the one at 100 to position 4, which takes 8 moves
 » 13 days ago, # | ← Rev. 2 →   +3 It seems C was like 1400 and D 2000Still, sad that it is unrated
•  » » 13 days ago, # ^ |   +2 D was simple with set and multiset. It is O(q*log(n))
•  » » » 13 days ago, # ^ | ← Rev. 2 →   -8 I wonder if my solution is overcomplicated or I just suck in c++. I couldn't come up with a solution with datastructures available in python and had to implement it in c++
•  » » » 13 days ago, # ^ |   -24 use SpoilerPlease
•  » » » 13 days ago, # ^ | ← Rev. 2 →   +6 In the end my estimate was nearly perfect and off by 100 Though it maybe affected by the fact that the round ended up unrated
 » 13 days ago, # |   +3 The problems were really interesting, so unfortunate it became unrated, just curious, how the problem A thing happen? all samples were wrong.
 » 13 days ago, # |   -8 Will this round be unrated?
•  » » 13 days ago, # ^ |   0 Yes, unrated!
 » 13 days ago, # |   +7 I admit it is fair to unrate the contest because of the issue of problem A but I still feel unhappy because I get a pretty good result.Anyway, thanks to the problem setters. Bad things happen all the time. Do not worry about it too much.
 » 13 days ago, # | ← Rev. 3 →   -23 Man, Can't catch a break.I do feel bad for Problem Setters though. The problems were definitely good.
•  » » 13 days ago, # ^ | ← Rev. 2 →   +3 Your Graph does not justify this, but Ya, lol
•  » » 13 days ago, # ^ |   +19 I bet 10 bucks and say if the contest was rated and had no problems, this 46 would be more than 4600.
•  » » » 13 days ago, # ^ |   +11 I'm not denying this. Not at all. Just wanted to post this. It is not meant to be taken seriously.
 » 13 days ago, # |   +1 pikmike I know this doesn't come under rules but is it possible for you guys to discard the hacking phase as the round is unrated it won't affect much and we don't have to wait for the editorials.
•  » » 13 days ago, # ^ |   +11 They could post the editorials if they want to, it does not depend on the hacking phase.
•  » » » 13 days ago, # ^ |   0 Ohhh, I didn't know about that.
 » 13 days ago, # |   -57 Honestly, did not liked the problems much. A, B, C where long and fairly unclear statements. D was long, too, but at least clear.E, F, G might be to hard, I assume less than 20 rated participants would have solved any of them.
•  » » 13 days ago, # ^ |   -25 Just noticed that I missunderstood D, it is not that we can move at all positions x, we can move all objects at one position x. So this is is shitty misunderstandable statement, too.
 » 13 days ago, # |   0 Did D have some elegant implementation? There was too much case work in my solution.
•  » » 13 days ago, # ^ |   0 92816134 This is most probably what u wish I guess
 » 13 days ago, # |   0 where did we receive notification regarding making round non rated?
•  » » 13 days ago, # ^ |   0 while receiving notification for correction in problem A
 » 13 days ago, # |   +5 Can anyone explain how to solve G because the question was pretty straightforward ??
 » 13 days ago, # |   +8 How to solve F and G?
•  » » 13 days ago, # ^ | ← Rev. 5 →   +29 F: Fix $d = gcd(x_1, x_2)$ where $x_2$ is yet to be determined, also the condition $l \le x_1y_1 \le r$ gives some bounds $L \le y_1 \le R$.If $x_1 = dk$, $x_2 = ds$ then we need $s \mid y_1$ and $k < s \le \frac{n}{d}$. We have two cases: If $x_1 \le \frac{n}{2}$, then consider $x_2 = 2x_1$. If $y_1$ has at least two possible values then at least one is even and we win. (Choose $x_1, y, 2x_1, y/2$). Otherwise, there's only one possible $y_1$, and it's easy to check if it works. If $x_1 > \frac{n}{2}$, then $R \le 2m$. Keep a BIT over $a[i] = \text{ number of multiples of } i \text{ in } [L, R]$, you can keep this updated by doing a total of $O(m \log m)$ updates, and you can find possible choices for $s$ by querying the interval $[k + 1, n / d]$. There's a total of $O(n \log n)$ choices for $d$ (counting over all possible values of $x_1$), so there's also this many queries. This gives a $\log^2$ solution.
•  » » 13 days ago, # ^ | ← Rev. 9 →   +36 My solution to problem G, AC after 8 minutes of the contest :(Let's iterate over all R from left to right, and count how many L such that the subarray [L..R] is good for each fixed R.When we are now on a fixed R, we need to know for each number X, what is the range of valid Ls such that the frequency of X in range [L .. R] is 3.We can easily do that by storing the indices of number X in a vector V, let's say that the vector size is K, so the valid range of Ls for this X is [V[K-4]+1, v[k-3]], we will use a lazy segment tree to increase / decrease this range each time we see X.Let's form an array called can[i], can[i] equals how many distinct numbers in range [i .. R] having frequency equals 3.Now our task becomes, for each R, count how many Ls, such that can[L] equals to the number of distinct integers in range [L .. R]. Let's say that distinct[i] = number of distinct integers in range [i..R]. We can easily update distinct[i] while we are iterating over R, when we find a number X we just need to increase distinct[j] by one, such that lastPostition[X] + 1 <= j <= i. We can do that using the same lazy segment tree.So we need to count how many (can[i] — distinct[i] == 0). Since (can[i] <= distinct[i]), then (can[i] — distinct[i] <= 0), so we just need to know what is the max number in the lazy segment tree, and if it was 0, we will add the frequency of the max to the answer. So we just need to support querying the max number and its frequency in the segment tree.My code: 92841078
 » 13 days ago, # |   0 It does not matter if its unrated but the problems were awesome
•  » » 13 days ago, # ^ |   -8 Would like to give me some instruction to solve the problem?
 » 13 days ago, # |   0 **After seeing the pretest 1 answer 9 for problem A , i was like time to say goodbye to this round XD. **
 » 13 days ago, # |   0 Can anyone explain the solution of C?
•  » » 13 days ago, # ^ |   0 There is two cases.1) The first character is 1, then friend have to use the skip(obviously). 2) There is substring "111". For any other case we can avoid using the skip
•  » » » 13 days ago, # ^ |   0 I first tried to implement greedy, but was not able to make it work :/
•  » » » » 13 days ago, # ^ |   0 Look at my submission: 92819400
•  » » » 13 days ago, # ^ |   0 Can you point out what is wrong in my approach https://codeforces.com/contest/1418/submission/92837513
•  » » » » 13 days ago, # ^ |   0 Just do a simple dp of the form dp[i][j]=Minimum skip points required to complete the game starting from index i, and such that player j starts the game. Try to get the recurrence relation from there.
•  » » 13 days ago, # ^ |   0 I used a dp.dp[i][pl][isSecond] is min number of skip points to kill ith boss by player pl with isSecondsth move.92828592
•  » » 13 days ago, # ^ |   0 problem C was basic Dp for every i store the minimum skip point his friend would use from 1 to i in array. and finally print min(dp[n],dp[n-1],dp[n-2]).minimum skip points for every i can be stored as follows. dp[i]=min(dp[i-1],dp[i-2],dp[i-3]). because maximum kill that can be done by anyone is 2 and for n<=3 just print a[0].
•  » » » 13 days ago, # ^ |   0 Thanks dude. I had used the same approach but fucked up with the case of n < 4
•  » » 13 days ago, # ^ |   0 I used greedy: For every turn, if its the friend's turn, kill the current boss and see if the next boss is easy, then kill it as well. If its your turn, then kill the current boss and see if the next boss is difficult, then kill it as well.My solution
•  » » » 13 days ago, # ^ | ← Rev. 2 →   +1 sh_maestro I did the same but got WA on test 2. my code... my code
•  » » » » 13 days ago, # ^ |   0 funny, same error as mine ;)"wrong answer 55th numbers differ — expected: '0', found: '1'"
•  » » » » » 13 days ago, # ^ |   0 If you figure out that test case , please tell . I am also getting wrong on same test case.
•  » » » » » » 13 days ago, # ^ |   +4 Most probably 55th Test of 2nd test case is-50 0 0 1 1
•  » » » » » » » 13 days ago, # ^ |   +1 Thanks!
•  » » » » » » » 13 days ago, # ^ |   0 How did you find the testcase?? Amazing!
•  » » » » » » » 13 days ago, # ^ |   0 hey, I also found a similar test case during contest when my similar greedy solution failed on tc-2, but I want to ask that how can we prove the greedy solution wrong by using some mathematical argument or Is finding a counter-example is the only way? as sometime it may be hard to find a counter-example and also takes a lot of time.
•  » » » » » » » » 13 days ago, # ^ | ← Rev. 4 →   +1 Greedy solution does work, though it uses a little bit of coming back a stepgreedy solution with insightsCheck the link solution
•  » » » » » » » » » 12 days ago, # ^ |   0 umm thanks I got it know it's kinda some new insight.
•  » » » » » 13 days ago, # ^ |   0 In the code by sh_maestro, he is skipping the case when it is your turn and a[i]==0 then he is skipping the boss.  if(turn) { if(a[i]) { ret++; } i++; if(i>=n) break; if(!a[i]) i++; } else { if(a[i]) //Why adding this line makes this code AC? i++; if(i
•  » » » » » » 13 days ago, # ^ |   0
•  » » » » 13 days ago, # ^ |   0 I guess the only difference is that I don't kill a boss every turn (I leave the easy bosses for the friend) — since either of us has to kill 1 or 2 bosses every turn. But if there's a hard boss, I always kill him (and the subsequent one — should that be a hard boss as well).
•  » » » 13 days ago, # ^ |   0 I tried to implement exactly that, but got WA. Can you see where I did wrong? 92817183 thanks
•  » » » » 13 days ago, # ^ |   0 u can look at my submission 92849330
•  » » » » 13 days ago, # ^ | ← Rev. 2 →   0 The problem statement is a bit vague and there is room for error — after reading your solution, I re-read it and can see why someone would go with that approach. The key difference is that it is not essential for both you and your friend to kill a boss in your turns. So my solution was something like this: The friend starts. He kills the first boss (regardless of hard/easy). Then if the next one is easy, he kills that as well. It's my turn now. If it's a hard boss, I kill it. If there is another one and it's a hard boss, I kill that as well. However, if I had an easy boss, I would just skip my turn, so my friend could kill it in his turn.
•  » » » » » 13 days ago, # ^ |   0 Sorry, I still do not see the difference.The friend uses his second hit if current boss is 0, I use my second hit if the current boss is 1. Where differs your solution from this?I did understand that this logic fails for 0 0 0 1 1
•  » » » » » » 13 days ago, # ^ |   +3 "During one session, either you or your friend can kill one or two bosses" — sounds like either my friend or I (or both) have to kill 1 or 2 bosses each round. Though the lines that follow imply that both of us have to alternate and have to kill 1 or 2 bosses in our turn.If the statement implies that its mandatory to kill a boss in each turn (as opposed to kill 1-2 bosses in either my friend or my turn), then my solution is wrong (and yours and the other solutions shared here who got WA are correct). The problem writer would need to clarify what the intent was.
•  » » » » » » » 13 days ago, # ^ |   +6 I think I just got why your solution works. The key realization is when you are skipping, you can actually imagine that as transferring the previous job to me(basically take it away from friend). Then it, doesn't matter whether it was a 1 or 0. It will not affect the answer. The special case is when there is a 0 at the second position. This time you cannot take the previous job, so we imagine that our friend finished the first two bosses. If in the third position, we have a one, "I" will take care and if it is a zero, then we again imagine a redistribution of work in which first job was done by our friend, second one by me and third one by friend again. This way, the solution always yield the optimal answer.
•  » » » » » » » » 13 days ago, # ^ |   0 That's an interesting insight! One thing to think about is what happens when this renegotiation ripples to the beginning (i.e. after the negotiation, the first boss needs to be killed by me, however the solution requires the first boss to be always killed by the friend).
•  » » » » » » » » » 13 days ago, # ^ |   0 We will not have the ripple back effect because we are always alternating turns and each player can at max handle 2 jobs. When you get a 0 in second place, the friend will take care of it, if we get one more 0, we redistribute as follows:-First(it may be 0 or 1) one is done by friend, second by me and third one by friend again.If you have any counterexamples in mind, please share them?
•  » » » » » » » » » 12 days ago, # ^ |   +8 Not a counterexample, but how would we explain this for 1 0 0 1 1? I.e. what does the friend take first, then us, etc.
•  » » » » » » » » » 12 days ago, # ^ | ← Rev. 3 →   +8 Okay, so the working will be as follows:- First our friend finishes the first two bosses "1 0". When we see a "0", we exchange the last boss finished, i.e. "1" one done by friend, the next "0" is done by me and the third "0" is done by friend. This way, in the next turn we get to finish the next two hard bosses "1 1" and we are done. I wrote a code for this idea with comments:-link to submission. Did I explain your question?
•  » » » » » » » » 12 days ago, # ^ |   +8 Amazing realization, thank you for that, it really did help understand the greedy approach:)
•  » » » » » » » 13 days ago, # ^ | ← Rev. 4 →   0 no comment
•  » » » » » 13 days ago, # ^ |   0 As far as I understood the problem, it's mandatory to kill at least one boss in any session (friend's turn or my turn). How can you skip "my" turn?
•  » » » 13 days ago, # ^ |   0 In the code you have provided in the link, if its my turn and the a[i]==0 then, you have skipped the testcase. I can't understand that part.
•  » » » » 13 days ago, # ^ | ← Rev. 2 →   0 I think that's because you would like to (greedily) give the 0 values to your friend , to reduce the count. If you yourself use that value 0 ,then next values may be like 1,1 and friend will need 1 more skip in that case.
•  » » » » » 13 days ago, # ^ |   0 What about this line: During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins....?
•  » » » » » » 13 days ago, # ^ |   0 I am not sure about his code , but i surely didn't skip anyone's chance. my submission
•  » » » » » » » 13 days ago, # ^ |   0 Exactly! When we don't skip it shows WA at Testcase 55.
•  » » » 13 days ago, # ^ |   0 I think your solution is correct but explanation is slightly wrong bro, Consider the given case : 1 0 0 1 1optimal moves -> Friend, Me, Friend, Me, Mebut according to your explanation your friend will see that second boss is easy and therefore he will kill it and that will result in not being an optimal choice, isn't it?
•  » » » » 13 days ago, # ^ |   0 The way it works right now would be something like ("F" = Friend, "M" = Me) : 1 0 0 1 1 F F F M M In the above example, on the third boss, I see an easy boss and would skip it (so the friend can kill it)Some more examples: 0 0 0 1 1 F F F M M 0 0 0 1 1 1 0 1 0 F F F M M F F M F In the last example, I can only kill 2 bosses at most, so my friend is forced to use a skip point at boss number 6.
•  » » 13 days ago, # ^ |   0 You have to find islands of 1 starting from index 1 and 2 ,then divide it by 3 my soln
 » 13 days ago, # |   +8 I couldn't solve a single problem. I hope editorial will help me.
 » 13 days ago, # |   0 For problem A,I got the first 3 sample test cases right,but got the last two wrong,Can anyone tell me why?
•  » » 13 days ago, # ^ |   0 Did you use long long int? int might overflow.
•  » » » 13 days ago, # ^ |   0 I used python
•  » » » » 13 days ago, # ^ |   0 Ohhh Sorry, I don't use python, maybe someone with python can help :)
•  » » » 13 days ago, # ^ |   0 Also,is my logic correct?I first calculated the required number of sticks,then calculated the number of trades required to get to the number of sticks. Finally I added the number of trades with k.
•  » » 13 days ago, # ^ |   0 it's probably because of floating point see that 92843760
•  » » » 13 days ago, # ^ |   0 Aww man. I should have checked it.Thank you.
 » 13 days ago, # |   +20
 » 13 days ago, # |   -26 ok, what I've found out from the round. I should not have divided doubles I should have read the tasks very carefully (super carefully, extremely carefully).
 » 13 days ago, # |   0 Looks like there were no testers for this round.
 » 13 days ago, # |   0 Oh my god, 2 questions with minecraft references, who made those questions dude???
 » 13 days ago, # | ← Rev. 2 →   +3 Can someone please help me out to find flaw in my logic for question D?-> At the end for each query we are just interested in dividing the array(sorted array) into two subarrays.-> Answer would be diff (max element and min element of first subarray) + diff(max element and min element of second subarray).-> Also dividing subarray would be optimal when we divide around mid value(max element + mid element)/2Attached my solution(not sure why its failing on test case 3)92839897It would be great if someone can please provide a counter example so that I can figure out why my algo is incorrect. Thanks In Advance.
•  » » 13 days ago, # ^ |   0 1 4 1 1 100 1000000
•  » » » 13 days ago, # ^ |   0 Sorry man, I am not able to understand your test case. What I understood from question, all the pi will be different and in above test case there are two ones.
•  » » 13 days ago, # ^ |   0 Also dividing subarray would be optimal when we divide around mid value(max element + mid element)/2 Try 1,3,4,5,6; optimal division point is at 2
•  » » » 13 days ago, # ^ |   0 What is the expected correct answer? I can see my solution gives answer as 3. Is that incorrect?
•  » » » » 13 days ago, # ^ |   +3 The answer is end value — starting value — (maximum distance between any two consecutive elements after sorting ofcourse)
•  » » » » » 13 days ago, # ^ |   0 I see, thanks.
•  » » » » » » 13 days ago, # ^ | ← Rev. 3 →   +3 I made the same mistake :) That is optimal for the 2 sides separately but not combined. Eg: 1, 20, 40, 60, 100. Optimal for 60 to go left.
•  » » » » » » » 13 days ago, # ^ |   0 It was kind of easy to think, but my bad I was just trying to correct my above solution by thinking counter example. I would have thought for a different approach.
 » 13 days ago, # |   +5 Can anyone hint me on E?
•  » » 13 days ago, # ^ | ← Rev. 2 →   -7 $O(n*m)$ is trivial if you know about linearity of expectation. To optimize it observe that for a fixed shield all $d_i$ smaller than $b_j$ have same contribution(number of ways such that we can $d_i$ damage to $j'th$ shield) and all $d_i$ greater than or equal to $b_j$ have same contribution. Do some counting and prefix sum and binary search and that's it.
•  » » » 13 days ago, # ^ |   0 I can form the expression. Evaluating it takes O(n) for each sample can't see of a way to optimise it further. Would be great if you could check this out — If (count of no of d(j) >= b(i)) <= a[i](Let this be a for the moment) then the answer is zero. Else from linearity of expectation the total damage can be expressed as D = D[1] + D[2].... + D[n]. Clearly E[D[k]] = 0 for k <= a. For any other k > a it remains to evaluate E[D[k]], which can be obtained by using the definition of Expectation and the associated probabilities can be