### I_Love_YrNameCouldBeHere's blog

By I_Love_YrNameCouldBeHere, history, 7 weeks ago,

#### Hello Codeforces!

ssense and I are very glad to invite you to Codeforces Round #726 (Div. 2) which will take place on Jun/18/2021 17:35 (Moscow time). The round will be rated for participants with a rating lower than 2100. Division 1 participants are also welcomed to take part in the competition but it will be unrated for them.

The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.

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

We would like to thank the following people:

The scoring distribution is: 500 — 750 — 1000 — 1500 — (1250+1750) — 2000.

UPD: Editorial

• +690

 » 7 weeks ago, # |   +523 As a coordinator...
•  » » 7 weeks ago, # ^ |   +93 omg orz! monogon taking over cf!
•  » » » 7 weeks ago, # ^ |   +390 Eventually I will fire Mike so I can be the real top contributor. (Don't tell him I said that)
•  » » » » 7 weeks ago, # ^ |   +79 monogonforces orz
•  » » » » 7 weeks ago, # ^ |   +30 As a follower of Monogon, I am obligated to upvote this comment.
•  » » » » 7 weeks ago, # ^ |   +71 At this rate, that day isn't far when... WORLD WILL BE TAKEN OVER BY MONOGON
•  » » » » 7 weeks ago, # ^ | ← Rev. 3 →   +190 In the future, if you're gonna need an icon just contact me. I'll do it for free just a few upvotes.Should I resume share or okay this?Also, The guy who looks 34-35% similar to Mike - meme
•  » » » » 7 weeks ago, # ^ |   0 OFZ
•  » » » » 6 weeks ago, # ^ |   0 ORZ 1-gonforces
•  » » 7 weeks ago, # ^ |   +212 Congrats for the successful steal of Anton's Job.
•  » » 7 weeks ago, # ^ |   +46 whatever man just take my upvote :/
•  » » 7 weeks ago, # ^ |   -12 As an Upvoter...
•  » » 6 weeks ago, # ^ |   0 in this contest i had solved problem A in first 8 mins and i was checking for the results and it was showing "in queue" so i thought to submit it on m1.codeforces so that it get tested fast but now i have seen that both of my solution got passed but the system have rewarded me the marks for the second one which i have submitted 2mins after and also it has given me -50 . can you plzz say something or do something on this plzzzz first time in 6 months i was able to solve 4 prblms in div 2
•  » » » 6 weeks ago, # ^ |   +28 According to the official rules, only the last submission with pretests passed is considered for system testing. When your submission was in queue you should've just waited and moved on to read the other problems.
•  » » 6 weeks ago, # ^ |   0 Hi 1-gon Can we do something for questions getting leaked before contest and answers getting posted on YouTube...?? That's what happened in the contest...
 » 7 weeks ago, # |   +185 As a tester, I tested.
•  » » 7 weeks ago, # ^ |   +46 Just take my upvote
•  » » 7 weeks ago, # ^ |   -51 Yea we can see your name in tester's list,why say it again?
•  » » » 7 weeks ago, # ^ |   +144 Yes, but how do you know what he did as a tester?
•  » » 7 weeks ago, # ^ |   0 You must have tested problems. Right?
•  » » 6 weeks ago, # ^ |   +1 oz
•  » » 6 weeks ago, # ^ |   0 Well tested!! Strong tests were included in system tests.
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   +3 I had the same fate, my friend. just changing s to s+s corrected it
 » 7 weeks ago, # |   +77 newbie sens
•  » » 7 weeks ago, # ^ |   0
 » 7 weeks ago, # |   +44 As a participant, I hope to become purple in this round. Wish me luck !
•  » » 7 weeks ago, # ^ |   +10 All the best, as a participant I hope to get my confidence back
•  » » » 6 weeks ago, # ^ |   +3 As a guy , I just want to be happy
•  » » » 6 weeks ago, # ^ |   +1 I dont want to be hor*y anymore. I just want to be happy.
•  » » » 6 weeks ago, # ^ |   0 as a newbie I wish the same and I hope for all of us to get to that level and more hopefully :)
•  » » 7 weeks ago, # ^ |   +9 Good Luck! I hope you become purple after this round but I sincerely hope that it's not coming at my ratings' expense.
•  » » » 6 weeks ago, # ^ |   0 Seems the opposite happened lol, you got rating at my ratings' expense :O. Anyways, good show.
•  » » 6 weeks ago, # ^ |   0 Hello ^^
 » 7 weeks ago, # |   +20 As a tester, i'm proud of these guys.
•  » » 7 weeks ago, # ^ |   -44 Just take my upvote
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   -21 Probably copying previous comments does not help. People who plan on doing this can probably take note. Hehe
•  » » » » 7 weeks ago, # ^ |   -19 CF can be ruthless haha
 » 7 weeks ago, # | ← Rev. 2 →   0 Damn ! Author of the blog Yur Heat Map is Very InSpiRational !!!
 » 7 weeks ago, # |   +27 5 months later, I'm still annoyed that you didn't take kostia244's suggestions for your new handle :(
•  » » 7 weeks ago, # ^ |   0 :(
•  » » » 7 weeks ago, # ^ |   +6 The biggest scam of 2021
•  » » » 7 weeks ago, # ^ |   +2 I_Love_SlavicG
•  » » » 7 weeks ago, # ^ |   -8 When the author gets lesser upvotes than the setters.
•  » » » » 6 weeks ago, # ^ |   +3 Did all the testers downvote this? I guess it doesn't matter since Monogon will soon rule Codeforces.
 » 7 weeks ago, # |   +32 As a tester, I_Love_YrNameCouldBeHere ORZ PS
•  » » 7 weeks ago, # ^ |   +12 inspirational
•  » » 7 weeks ago, # ^ |   -13 It really motivated me I_Love_YrNameCouldBeHere
•  » » » 7 weeks ago, # ^ |   +4 what about this guy? tenkei RunTimeTerror Motivation overflow
•  » » » » 6 weeks ago, # ^ |   +1 tenkei retired
•  » » » » » 6 weeks ago, # ^ |   0 He already gave us so much motivation and showed us everything is possible if we try honestly and embrace hard work. Hope he enjoy his retirement. We will miss him. Orz.
•  » » 7 weeks ago, # ^ |   0 Orz =))
 » 7 weeks ago, # |   +29 Score distribution suggests that the contest is All-Killable
 » 7 weeks ago, # |   -49 So blues are writing div2s now? whats next? greendians writing div1?
•  » » 7 weeks ago, # ^ |   +18 Chup chapri. Dont write these useless messages.
•  » » » 7 weeks ago, # ^ |   -39 keyboard_warrior Another bluedian defending your blue buddies? Hope even you get to write a contest next.
•  » » » » 7 weeks ago, # ^ | ← Rev. 2 →   +20 problemsetting is all about having original ideas, which does not strictly correlate with rating. even if the problems were not worth it, the testers and coordinators and other people involved with this contest (a lot of whom have a ton of experience with contest-making i'd assume) would have caught that and we wouldn't even be seeing this announcement. adding on top of the previous point, there are more than just the problemsetters who are there to make the whole contest experience better for us, like testers and such. this stuff takes time to make and improve, and i believe that these people spend a lot of their time doing awesome work to gift us these valuable contests. like come on, don't be a jerk valuing people just by some random internet point. that's not cool at all... (edit: fixed some grammar mistakes)
•  » » » » » 7 weeks ago, # ^ |   +3 What about purples/blues(not me) who have really good problem ideas and not able to propose a contest because of the rating barrier and end up in submitting them in less kown websites or just moving on with dissatisfaction.
•  » » » » » » 7 weeks ago, # ^ |   0 make unofficial round.
•  » » 7 weeks ago, # ^ |   +63 I have a dream, that one day authors may be judged by the quality of their problems, rather than the color of their handle.As a tester, I can say that the problems are nice, and I recommend participating :D
•  » » » 7 weeks ago, # ^ |   +22 THIS IS SPARTAAAAA
•  » » 7 weeks ago, # ^ |   +11 Well, if you look at their number of solved problems you can easily say they solved enough problems to make their own problems. Also some red coders tested/coordinated it so there is no reason to not make a round. And those "greendians" are better than an unrated person with -53 contribution.
•  » » 7 weeks ago, # ^ |   +17 mad?
•  » » 7 weeks ago, # ^ |   +5 And what is your rating, mister alt user?
 » 7 weeks ago, # |   +23 As a tester, I couldn't help much with review of problems except in the initial div3 phase. The round is well balanced and may positive delta come your way :blobheart:
•  » » 7 weeks ago, # ^ |   0 Upvoted! Hope your Blessing Comes true...Pupil, I am coming BACK!!
 » 7 weeks ago, # |   0 All the best to all the participants,may there be a positive delta... The scoring distribution is quite interesting.
 » 7 weeks ago, # |   0 Awesome 1-gon starts coordinating the round.
 » 7 weeks ago, # |   +7 Moldova power!
 » 7 weeks ago, # |   +19 As a tester, I highly advise you to read all the problems!
•  » » 7 weeks ago, # ^ |   -13 As a tester, I assure you that all problems are very interesting. Please do participate. All the best!
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   -9 I often see this comment whenever we have a problem which is divided into two( easy & hard) versions . Let me guess E1 is a bit easy than D :) and one more thing we have to do a bit more reading :)
•  » » » 6 weeks ago, # ^ |   +1 I got it right , E1 was easier than D :)
•  » » 6 weeks ago, # ^ |   -7 Don't they all suggest the same thing. I think the trick to acing the contest might be to not read all the problems and not get depressed to try the problems.
 » 7 weeks ago, # |   +42 These rules changed? Problem setter requirements
•  » » 7 weeks ago, # ^ |   0 I guess they can be count as a writer in the past because they made 2 unoffical div 4 contests and those rounds had a good amount of participants.
•  » » » 7 weeks ago, # ^ |   +44 No disrespect, but they are a bunch of some easy problems. So how do they count?
•  » » » » 7 weeks ago, # ^ |   -8 they said this meant to be a div3 round,so those problems were enough to give them a chance to write a round.
•  » » » » 7 weeks ago, # ^ |   +39 To reds, a Div. 2 is a bunch of easy problems. To oranges, a Div. 3 is a bunch of easy problems. So how do they count?Honestly, I think the current rules were just a way to limit the amount of contest proposals without disenfranchising anybody. So if you were lucky enough to get something in before the new requirements, then you basically get grandfathered in.
•  » » » » » 7 weeks ago, # ^ |   +20 Div 2 is not rated for oranges and above. Div 3 is not rated for blues and above. But this round is rated til 2100.
•  » » » » » » 7 weeks ago, # ^ |   +28 And since it was reviewed, coordinated, and tested by a bunch of people high rated, it should be reason enough to believe that the contest will have good enough problems for it to be rated for participants with rating upto 2100.
•  » » » » » » » 7 weeks ago, # ^ |   +15 Probably, still I'm also somewhat dubious about the ability of people to come up with the problem and solution ~500-1000 points above their own rating, and the hardest problem of the round should be at least that difficultNo offence, it's just peculiar
•  » » » » » » » » 7 weeks ago, # ^ |   +17 Rating doesn't measure your overall problem solving skills.
•  » » » » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   +64 A more accurate explanation is probably that it is quite possible to make a problem that is 1000 points above your rating by making a problem from the solution, instead of actually solving a problem you came up with. And, even for problems that you came up with, you have a lot more time to try to solve them than when in a contest, and you also probably have more chances to solve a hard problem(the participants can't notice all of the hard problems that you didn't solve when problemsetting).
•  » » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   +24 Btw, all rounds have coordinators, but not all of them are equally good, so it doesn't really mean anything, just saying
•  » » » » » » » » 7 weeks ago, # ^ |   +44 Well, I didn't want to drag this on, but let's do it. not all of them are equally good What is a good round / contest? one where you can solve 3 or 4 problems? you feel good after the contest, that maybe you solved every problem that you read during contest? Sorry to disappoint you that not every round can be good for you.Tbh, have you ever even thought about what it takes to prepare a round, or the effort put in by coordinators / testers. All you see is the "Codeforces Round #XYZ" blog on CF. Although, I have never authored a round myself, I want to make problems for a while, but I can't even think about them. Forget about preparation of statements, constraints, test cases, and many other things that I might not even know about right now. Even though some rounds are not well recieved by contestants, it doesn't trash all the effort that was put into it, nor does it mean that problems are shit. Hell,the problems were accepted by anton, and even as a joke, it is popularly known how high a bar he has set.
•  » » » » » » » » » 7 weeks ago, # ^ | ← Rev. 3 →   +27 Nah What is a good round / contest? one where you can solve 3 or 4 problems? Rly, do not assume that the other person is stupid. Whether my performance is good or bad has nothing to do with the round being good or bad. Say, LATOKEN round was quite good even though I got -50Model good round is a round with good problems, balanced problems, diversed topics, good problem statements and good editorials. Also without long queues and other technical issues but that is out of control of coordinators.It's all nice and everything. Yet again, some rounds are better than the others despite all the hard work of the authors you mentioned
•  » » » » » » » » » 7 weeks ago, # ^ |   +33 Have you realized that the first 5 requirements you stated are completely subjective? Sorry, this website doesn't cater to each and every one of your individual needs. The round is good, it was tested by a wide variety of colours, it was coordinated by reds and approved by Mike, I'm not sure what else you could ask for.
•  » » » » » » » » » 7 weeks ago, # ^ |   +18 Well put, I couldn't have said it better.
•  » » » » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 They're not so much subjective as you're trying to make it soundSay if the statement is long and confusing for a lot of people, it is a bad statement and there is little subjective about itIf one problem has difficulty 1000 and the next one is 2200, it is unbalanced roundEtcAlso I don't quite understand what you're arguing about, Do you disagree that some rounds are better than the others? Because that is pretty much the sole point of my messages here
•  » » » » » » » » » 7 weeks ago, # ^ |   +9 As the round has neither started nor you are a tester, then how can you say anything about a round when you literally don't know anything about the round I don't know if you read the testing part, but the round has testers from almost every color from pupil to LGM.So, it would be better if u shut up now and give your opinions after the round has ended. There is a comments section in editorial also
•  » » » » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   +41 Lets just end this thread by a well known standard result : Quality of contest depends on your rank in that contest QED
•  » » 7 weeks ago, # ^ |   +24 What actually happened was some combination of Mike liking the problems but not thinking it was Div 3, and having Anton and Monogon coordinate.
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   +39 They wrote an unofficial div4 contest on gym before, so it isn't a violation of the rules really. They had experience in problemsetting and messaged me (who started coordinating after that blog) about it, as it says in the last paragraph.
 » 7 weeks ago, # |   +24 As a tester, I can certify this is a great round with interesting and clear problems. Best of luck to all who are participating!
•  » » 7 weeks ago, # ^ |   0 Nice bro
 » 7 weeks ago, # |   +5 Both the problemsetters have a highest rating of expert. So how did they propose this round, as you have to be a master to write one?
•  » » 7 weeks ago, # ^ | ← Rev. 4 →   0 I guess they were writer in past.And 1-gon also mentioned that they have organised a unofficial div4 round in gym in past . Apart from that there are many high rated people who are coordinating the round , if they have no issues then you shouldn't have .
•  » » 7 weeks ago, # ^ |   0 You can suggest problems if you are an expert. I don't know if it's any different than proposing contests.
 » 7 weeks ago, # |   +1 As a participant I hope to get past C this round
 » 7 weeks ago, # |   0 I wish I could be expert. People how to get expert batch within a short time? Is it their outcome of lot of practice, How can I get expert batch? Please suggest me. (Sorry for unnecessary comment. But I need to know to develop my problem solving skill)
•  » » 7 weeks ago, # ^ |   +5 Literally just solve questions that are challenging enough, so basically problems that are rated +200 ish from ur current rating and try to understand and apply new concepts u encounter from editorials n blogs then u should be set.
 » 7 weeks ago, # |   0 I didn't knew that Blues can also host contest at codeforces.
•  » » 7 weeks ago, # ^ |   +3 Blue power exists. :)
 » 7 weeks ago, # |   +7 are the problem statement short mainly first 2 as i am unable to do after that in last rounds hope that i will be able to solve a,b,c this time.
•  » » 7 weeks ago, # ^ |   +33 As a tester I think that reading all problems in 2 hours is very well doable.
•  » » » 7 weeks ago, # ^ | ← Rev. 3 →   +5 So we can assume there can not be much more than 1000 words in each statement. Very reassuring
•  » » » » 7 weeks ago, # ^ |   +10 Yeah but tbh I've never found a contest where I wasn't able to read all problems. Reading them is not taking a large amount of time compared to solving them, regardless of the number of words
 » 7 weeks ago, # |   +79 If 1-gon can write a good 3500, which is 900 higher than his rating, then can't I_Love_YrNameCouldBeHere write a good 2600?2600 is the upper end of div2 difficulty. I think this will be an excellent round!
 » 7 weeks ago, # |   -19 I have no idea who Zoroland_ is, and how he/she copied my code for A,B,C problems. It maybe because i used idone in public settings during this contest. I will stop using idone at all after this.
 » 7 weeks ago, # |   0 Motivational speech: The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.
•  » » 6 weeks ago, # ^ |   +16 In other words, if you perform badly then you're even worse than you thought.
 » 7 weeks ago, # |   +4 Might be another speedforces coming up!!
 » 7 weeks ago, # |   +13 Ah I see, the tables have turned now!
 » 7 weeks ago, # | ← Rev. 2 →   0 Thanks, and I hope the statement will be short and clear.
 » 7 weeks ago, # |   +3 omg finally ssense ans slavic round, cant wait to participate!
 » 7 weeks ago, # |   +3 Hope to solve problem C this time.
 » 7 weeks ago, # |   +27 I wish I was never born
 » 7 weeks ago, # |   -69 Ban radewoosh,he is racist.
•  » » 7 weeks ago, # ^ |   0 OMG! What happen to Radewoosh?
 » 7 weeks ago, # |   +8 Anyone noticed that when you move your cursor to testers' handles, you can see text "Tester" or "orz Tester"?
•  » » 7 weeks ago, # ^ | ← Rev. 3 →   +13 I see 1-gon >> yoyo master. This is how it works- 1-gonand without actually tagging the person- monogonResult: monogonI hope 1-gon didn't recieve a notification for this.
•  » » » 7 weeks ago, # ^ |   +34 I didn't get a notification
•  » » » » 7 weeks ago, # ^ |   +24 Thanks! You never disappoint me, my lord. And you are fast.
 » 7 weeks ago, # |   +7 As a commenter , I wish +2 :) I_Love_YrNameCouldBeHere
•  » » 7 weeks ago, # ^ |   +12 I'm also in desperate need of +2
 » 7 weeks ago, # |   +8 I think 1-gon is gonna takeover codeforces completely !
 » 7 weeks ago, # |   -12 This is gonna be my first contest. Wish me luck guys :)
•  » » 6 weeks ago, # ^ |   0 All the best bro! Mistakenly I downvoted your comment so the best I could do is wish you luck!!
•  » » » 6 weeks ago, # ^ |   0 Haha im already into negative karma ;-; Is all good tho
 » 7 weeks ago, # |   +8 As a participant, "As a + something" jokes are not funny anymore...
•  » » 7 weeks ago, # ^ |   +78 That's not true. "Anymore" implies there was a time when they were funny.
•  » » » 6 weeks ago, # ^ |   +14 So you're testing if they will keep upvoting those I guess
 » 7 weeks ago, # |   +6 I want to be a tester. But I don't know any of the problem setters, and I don't just want to be friends for benefits just to become a tester. Is there any hope?
 » 7 weeks ago, # |   0 i want to speclist
 » 7 weeks ago, # |   +36 Score distribution says the total score of E is 3000, whereas the score of F is 2000. Does this mean E2 is harder than F? Interesting.
•  » » 7 weeks ago, # ^ |   +34 Not necessarily, it could also mean that the 2 subtasks of E aren't related, like 1092D1 - Великая Вовина Стена (Версия 1) and 1092D2 - Великая Вовина Стена (Версия 2)
•  » » » 6 weeks ago, # ^ |   0 Oh, that makes sence.
 » 7 weeks ago, # |   +13 The day is not far when MikeMirzayanov will sit and rest and 1-gon and other GMs and LGMs will take over Codeforces ...
 » 7 weeks ago, # |   +4 The person wrote this notice got lots of contribution :)
•  » » 7 weeks ago, # ^ |   +2 when you give monogon contribution, he repays you with contribution as well
 » 7 weeks ago, # |   +8 I hope that all those who work hard will get good rating
 » 7 weeks ago, # |   +19 I_Love_YrNameCouldBeHere Guys just have a look at his profile .... Respect ++ man :)
•  » » 6 weeks ago, # ^ |   +8 also the cool profile pic !
 » 7 weeks ago, # | ← Rev. 2 →   -6 .
 » 7 weeks ago, # |   +12 Im so proud of you guys for making this round !!! Way to hyped for tomorrow :yayy:
 » 6 weeks ago, # |   -34 It's coinciding with WTC Finals ToT
•  » » 6 weeks ago, # ^ |   -40 No one cares
 » 6 weeks ago, # |   0 Did cf change their rule? I thought yellow was necessary to set problems!
 » 6 weeks ago, # |   +24 I think either D or E is going to be an interesting game problem.
•  » » 6 weeks ago, # ^ |   +5 Well done myself!!Anyone here who wants to get a prediction of their future cf rating?? ASk me!! (or better, I should write a blog abt it :D)
•  » » » 6 weeks ago, # ^ |   0 What about mine
•  » » » » 6 weeks ago, # ^ |   0 Around 1450 after this contest and then you're gonna reach expert in a month or so.
•  » » » » » 6 weeks ago, # ^ |   0 What about mine.
•  » » » » » 6 weeks ago, # ^ |   0 How do you say this confidently?..what,s logic behind that?
•  » » » 6 weeks ago, # ^ |   0 Can you please predict my future?
•  » » 6 weeks ago, # ^ |   +1 :D
•  » » 6 weeks ago, # ^ |   0 thanks for div3)
 » 6 weeks ago, # |   0 I am curious to know 'who invented orz ?' tag that person :o
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +17 who invented orz? Answer: orz's parents. And if you want to know about them just ask orz. Maybe he'll ask them to join codeforces (just like SecondThread asked his mom) and then you can tag them. ✌️
•  » » » 6 weeks ago, # ^ |   +6 answer is better than expected :p
 » 6 weeks ago, # |   -60 Hello! I would like to propose the next Div. 1 please. I know I am unrated and have not given any contests but please do not judge me by my rating (or the lack of it lol) and give me a chance to write problems for an official contest. Thank you.
•  » » 6 weeks ago, # ^ |   -24 Tagging 1-gon for better reach.
•  » » » 6 weeks ago, # ^ |   +40 You should either get the required rating to propose it, or demonstrate that you have sufficient problemsetting experience.
 » 6 weeks ago, # |   0 expecting my first codeforces contest!
 » 6 weeks ago, # | ← Rev. 2 →   +3 i hope i can solve atlest 1 problem todaY all right i solved a and almost b.thank you for this contest author.boosted my confidence
 » 6 weeks ago, # |   -12 wtf bro!! why you just didn't upvoted me
 » 6 weeks ago, # |   -8 Hi All, My rating is 1309, Am I not allowed to appear in this contest.
•  » » 6 weeks ago, # ^ |   0 no
 » 6 weeks ago, # |   +13 Long Queue?
 » 6 weeks ago, # |   +6 Queueforces :(
 » 6 weeks ago, # | ← Rev. 2 →   0 I have submitted problem A 4 minutes ago.It's still in queue..
 » 6 weeks ago, # |   0 Omg already fear of WA plus this queue making me go wild and nervous
 » 6 weeks ago, # |   +3 waiting for my submissions to deque :)
 » 6 weeks ago, # |   0 In problem A is each appended number also limited by $10^4$ or any non-negative numbers are okay?
 » 6 weeks ago, # |   0 In Problem B, can Anton entre the same room twice?(According to the Visualization, he can't, but the problem didn't say it.)
•  » » 6 weeks ago, # ^ |   0
•  » » » 6 weeks ago, # ^ |   +3 There is ask a question section where all problems are shown...Ask there
 » 6 weeks ago, # |   0 can you er-open the registration pls ?
 » 6 weeks ago, # |   0 As a loyal antontrygubO_o supporter, I upvoted 1-gon . Have you?
•  » » 6 weeks ago, # ^ |   0 my upvote only to 1-gon sir
 » 6 weeks ago, # |   0 wa2forces...
 » 6 weeks ago, # |   -52 I think this round should be unrated.The reasons: queue forces. F is similar problem ,it was solved 200+! (include Low level players (like me).).
•  » » 6 weeks ago, # ^ |   +19 lol what
•  » » 6 weeks ago, # ^ |   -19 The same problem should be banned from such kind of contest!Strongly support this proposal.
 » 6 weeks ago, # |   +94
•  » » 6 weeks ago, # ^ |   0 actually it should be the other way around
•  » » » 6 weeks ago, # ^ |   +14 How? Its pretty intuitive that the optimal string you use in E is going to be a repetition of prefixes, so just try all of them, felt as easy as A-C at least in my opinion.
•  » » » » 6 weeks ago, # ^ |   +6 D is to realize the "good" and "bad" states — if you just throw out some prime factorizations on paper you'll realize that as long as you have an even power of 2, or at least 1 "2" and another number, it's a good state and Alice wins. (if i get fst, rip)Else bob.Anyways for E, im bad at cpp so I used java and had a bunch of stupid bugsFinally, I think C is the hardest problems :P
•  » » » » » 6 weeks ago, # ^ |   0 Yes, and I brute first 1000 values of D and solved it the same way. But it actually took time to do that. At least for me E1 felt far more obvious.C would actually be tricky if they didn't give away both the cases in the samples lol. In terms of difficulty I'd say:B < A < C < E1 < D < E2 < F
•  » » » » » » 6 weeks ago, # ^ |   0 geniosity orz, thinks B is easier than A
•  » » » » » » » 6 weeks ago, # ^ |   +2 My solution to B was just printing "1 1 n m", sure it took me a tad bit longer than A, but even I found it easier :P
•  » » » » » » 6 weeks ago, # ^ |   0 How did you solve E2 , can you just give a basic idea
•  » » » » » » » 6 weeks ago, # ^ |   0 I gave the rough idea in this comment
•  » » » » » » » » 6 weeks ago, # ^ |   0 Can u please explain how u even brute forced D,I tried but i couldn't.
•  » » » » » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 logicLet's call a position good if the played playing at that position win.dp[i]=1 // if position i is good base case dp[0]=dp[1]=dp[2]=dp[3]=0; because if some player lands into these positions he, don't have any divisor to subtract.now, loop for all i's in the range 4 to 1000suppose currently i=12. check for all divisors of i which we can subtract and the new number we get after subtracting is:new_no = i-divisor_that_we_subtracted.These new numbers that we get after subtracting are the position we can send the other player to. if any of the new positions is bad, we will send the other player to that position and he will lose. So, dp[i]=1 if any of dp[i-divisor(i)]=0;now to check for any n in range 1-1000if dp[n]=1 // alice winselse bob wins because alice is starting the game codeint dp[1000]; vi getDivisors(int n) { vi ans; for (int i = 1; i*i < n; i++) { if (n % i == 0) ans.push_back(i); } for (int i = sqrt(n); i >= 1; i--) { if (n % i == 0) ans.push_back(n/i); } return(ans); } void initialize() { dp[0]=0; dp[1]=0; dp[2]=0; dp[3]=0; for (int i = 4; i < 1000; ++i) { vi div=getDivisors(i); int good=1; for(auto ash:div) { if(ash==1 || ash==i) continue; good&=(dp[i-ash]); } dp[i]=!good; } } void solve() { for (int i = 1; i < 1000; ++i) { if(dp[i]) { cout<<"Alice"<
•  » » » » » » 6 weeks ago, # ^ |   0 how did you brute force 1000 values? logic?
•  » » » » » » » 6 weeks ago, # ^ |   0 refer this
•  » » » » » » » » 6 weeks ago, # ^ |   0 Thank you
•  » » 6 weeks ago, # ^ |   +3 I still don't know how you're supposed to reach the solution of D, I just brute forced for small values and observed the general and edge case (and then proceeded to submit only the edge case, not the general one and get confused when it got WA)
•  » » » 6 weeks ago, # ^ |   +17 I think the general idea is: plot out some initial values, make a conjecture, then try to prove your conjecture.Conjecture: For any $n \gt 0$, the first player has a winning move iff $n$ is even and not an odd power of 2.Proof by strong induction. The base case $n = 1$ is obvious. For the inductive step, let $n \gt 1$ and suppose the claim holds for all $m \lt n$. We can show the claim is true for $n$ by case analysis:Case 1: $n$ is an odd prime. Then there are no moves and the claim holds.Case 2: $n$ is an odd composite. In this case all divisors of $n$ are odd. Suppose the player plays some divisor $a$, and let $b = n / a$. Then the new value is $m = n - a = a \cdot (b-1)$. Since $a$ and $b$ are odd, $m$ is even but not a power of 2, so by our inductive hypothesis $m$ is a winning value. So $n$ has no winning move and the claim holds.Case 3: $n$ is an even number, but not a power of 2. Then $n$ has an odd proper divisor $a$. Let $b = n / a$ Then $b$ must be even so $m = n - a = a \cdot (b-1)$ is odd, and by our inductive hypothesis a losing value. So $a$ is a winning move for $n$ and the claim holds.Case 4: $n = 2$. $n$ is an odd power of 2\$ and there are no moves, so the claim holds.Case 5: $n = 2^k$ for some $k \gt 1$. Then any valid move is of the form $a = 2^i$ for some $1 \le i \lt k$. Let $j = k - i$ and let $b = n / a = 2^j$. If $j = 1$ then $m = n - a = 2^{k-1}$, which by our inductive hypothesis is a losing value when $k-1$ is odd. If $j \gt 1$ then $m = n - a = a \cdot (b-1)$ is even, but not a power of 2, so by our inductive hypothesis this is a winning value and $a$ is a losing move. So when $n$ is an even power of 2 then $a = 2^{k-1}$ is a winning move, otherwise there is no winning move, and the claim holds. This completes the induction.
•  » » 6 weeks ago, # ^ |   +6 That's obvious coz D:1500 and E1:1250
•  » » 6 weeks ago, # ^ |   0 i was solving e1 and e2 for an hour)
 » 6 weeks ago, # |   +3 lot's of FST's coming :(
 » 6 weeks ago, # |   +3 Weak pretest in E1 — my AC code fails on n=8,k=8 [dcdadcdb]
•  » » 6 weeks ago, # ^ |   0 FST coming :(
•  » » 6 weeks ago, # ^ |   0 What is the answer of this test case?
•  » » » 6 weeks ago, # ^ |   0 dcdadcda
•  » » » » 6 weeks ago, # ^ |   0 Can you explain, how?
•  » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 Remove the last 4 characters (dcdb) using operation 1 four times. Concat the rest of the string (dcda) twice using operation 2.
•  » » » » » » 6 weeks ago, # ^ |   0 Yeah, my bad! I am getting dcdadcda
 » 6 weeks ago, # |   0 What's the logic for D? I just observed the pattern from a brute-force solution and got an AC without any logic.
•  » » 6 weeks ago, # ^ |   0 Due to induction. For numbers > 9 and not power of 2, you can see that it if it's even then Alice will win else bob. If even number has odd factor, then we can subtract odd factor from it and resultant number will be odd and use answer of previous numbers.
•  » » » 6 weeks ago, # ^ | ← Rev. 3 →   0 rananjay23 Can you explain a little bit more the logic of D?
•  » » » » 6 weeks ago, # ^ |   +1 I started writing down numbers from 1 to 21 and found following : Bob wins for 1,2,3,5,7,8,9,11,13,15,17,19,21 Alice wins for 4,6,10,12,14,16,18,20For even n, if it's not power of 2, then there is some odd divisor say o>1. Hence n-o is odd and we know result of it . Example : n=22, then 22-11=11 and we know who wins if n is odd and less than 21.Similar reason you can find for odd numbers > 21. For even numbers with power of 2, It's always alternate i.e for 8 bob will win, for 16 Alice, for 32 Bob. Here also explain (just subtract divisors and see) in similar way.
•  » » 6 weeks ago, # ^ |   0 I wrote out a proof here: https://codeforces.com/blog/entry/91835?#comment-805791
 » 6 weeks ago, # |   +9 The pretests for E1 are trash!
 » 6 weeks ago, # |   0 I kicked myself for spending 1 hour on C :(
 » 6 weeks ago, # | ← Rev. 2 →   +32 F is a easy version of https://www.luogu.com.cn/problem/P6185
•  » » 6 weeks ago, # ^ |   +8 Line graph version of F also appeared very recently in https://atcoder.jp/contests/arc119/tasks/arc119_cI feel like the operation of adding x to two adjacent cells also appeared on many other CF problems before. They are almost always solved by two coloring.
 » 6 weeks ago, # | ← Rev. 4 →   0 Nice Round. Thanks for this :)Not good for me. Solved E1 after 4 WA's just because of min/max string silly mistake. Rank dropped from 500 in first half to 1400 till end. And left with no time to try E2Moral:- Don't submit again and again by changing small part just erase whole solution do that again. Hopefully that silly mistake will be resolved.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   0 B failed in system testing..Reason — 1 "=" sign My bad
 » 6 weeks ago, # |   +15 The problems are cute, but don't you think this is closer to Div2.5 than Div2? My first full solve of a Div2 and its around what I would have got for a fairly fast n — 1 problems in a Div2 normally.
•  » » 6 weeks ago, # ^ |   +3 Yess, the authors have mentioned in the above blog:The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.which is probably why it felt easier than the rest :)
 » 6 weeks ago, # |   0 Found the OEIS sequence for D in the last 3 minutes. Completed my code 10 seconds after the contest got over due to a stupid mistake.Now I feel sad.
•  » » 6 weeks ago, # ^ |   0 Wait same, but I got when $\omega(n)^{\phi(n)} \equiv 1 \pmod{n}$, which is apparantally wrong.
•  » » » 6 weeks ago, # ^ |   0 I got this one OEIS
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   +3 That one doesn't seem to be the answer though. I'm not sure but I get that the answer is Odd -> Bob winEven we have two cases 2^k with odd k -> Bob win Otherwise -> Alice win This passes the pretest but I need to wait for the system test for confirmation.
•  » » » » » 6 weeks ago, # ^ |   0 Odd -> Bob win?What about n=9? (Given as an example in the problem). Also, 25. Or are those just a special cases?
•  » » » » » » 6 weeks ago, # ^ |   0 n = 9 is not in the example.Also, check this! https://codeforces.com/blog/entry/91835?#comment-805566
•  » » » » » » » 6 weeks ago, # ^ |   0 whoops... guess my brute forcer was borked.
•  » » » » » » 6 weeks ago, # ^ |   +4 n=9 was in an explanation and it was Bob's turn to move, and he lost. So n=9, Alice's turn, she'll lose.
 » 6 weeks ago, # |   +6 I didn't get A for most of the contest because I'm an idiot, but this was a fun contest!For E, what you do is take some prefix and concatenate it until it's at least length $k$. In E1, you can brute force every prefix. In E2, is the approach some kind of z-function thing?
 » 6 weeks ago, # | ← Rev. 2 →   0 Weak pretests for E1. :( . whereas good pretests for E2 :( .Test case on which your E1 might fail12 30dbdbdadbdbdbCorrect Outputdbdbdadbdbdadbdbdadbdbdadbdbda
 » 6 weeks ago, # |   +3 Was E2 = E1 + suffix array for the reversed string?
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +9 I thought to do that, but then I passed pretests by greedy...And now it failed. lmao.
•  » » 6 weeks ago, # ^ |   +5 I used rolling hash values in combination with binary search to find out which part of the subsegment of the string is different from the prefix string.
•  » » 6 weeks ago, # ^ | ← Rev. 3 →   +15 Not really, its a simple greedy. Clearly as we proceed from the starting of the string (after the first character) its optimal to take any $s_i \lt s_0$ since it will lead to a lexicographically smaller result than appending the string which will add $s_0$ again at that point. But what about if $s_i = s_0$? Then we have to check further characters. We can see that it will produce a better answer if in the first position where $s_{i + k} \ne s_{k}$, we have $s_{i + k} \lt s_{k}$. We can maintain this by just keeping a counter for this k when we reach an equal element and resetting it when we get a smaller character (or breaking if its a larger character). Additionally we have to be careful to store this temporary string during the equal part somewhere else and only add it once we reach this k-th position.Code — 119871715
•  » » » 6 weeks ago, # ^ |   +3 "We can see that it will produce a better answer if in the first position where si+k≠sk, we have si+k
 » 6 weeks ago, # |   0
 » 6 weeks ago, # |   0 I often take Virtual Practice on CF and am used to facing a lot of hack tests. So since I saw the small number of the pretests, I have been very worried to be hacked!
 » 6 weeks ago, # |   +9 How to prove for D?
•  » » 6 weeks ago, # ^ |   +15 I think in most case the number will be (odd->)even->odd....->odd.But when $n=2^k$ it is special.
•  » » 6 weeks ago, # ^ |   +6 yeah, in particular when $n$ is odd, Bob has a symmetry strat (always pick the same divisor to subtract that Alice does, and things work out bc of parity)
•  » » 6 weeks ago, # ^ |   0 I wrote out a proof here: https://codeforces.com/blog/entry/91835?#comment-805791
 » 6 weeks ago, # | ← Rev. 3 →   0 I solved E (E1+E2) using 2 Pointer approach. 119902053 Edit :- Got WA on Test 93, Weak Pretests :(
•  » » 6 weeks ago, # ^ | ← Rev. 3 →   +3 why didn't you just  n=min(n,k);  instead of  if(k<=n)do(k);  else do(n);
 » 6 weeks ago, # |   -36 A lot of plagiarism happened in Round; Many posted solutions on youtube. Better to make round Unrated.
•  » » 6 weeks ago, # ^ |   +5 give link
•  » » » 6 weeks ago, # ^ |   +12 After seeing this comment, I searched and look what I found https://www.youtube.com/watch?v=e9i8c6JKmXs here you go :)
•  » » » » 6 weeks ago, # ^ |   +77 Went through the linked video, the guy is an absolute legend This guy trying to hide his username be like
•  » » » » » 6 weeks ago, # ^ |   0 xD
•  » » » 6 weeks ago, # ^ |   0 This is the solution video of problem A which was uploaded when the contest was still running. Link The user goes by the name Blunder_AY and his submission for the problem A can be found under 119812636
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +36 Then every round will be unrated :(
 » 6 weeks ago, # |   +5 Missed E2 by 1 sec :(
 » 6 weeks ago, # |   0 How to solve D ?
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +3 i actually wrote a recursive solution and then i could see the pattern of even odd.Next i just validated for even odd. Found that for power of two that are odd , answer is false.Hope that helps Spoilerboolean rec(int x) { if ( x == 1 ) return false; if (ok[x] != -1) return (ok[x] == 1) ? true : false; boolean isPrime = true; for (int j=2 ; j*j<=x ; j++) { if (x%j == 0) { isPrime = false; break; } } if (isPrime) { ok[x] = 0; return false; } boolean canWin = false; for (int j=2 ; j*j<=x ; j++) { if (x%j == 0) { canWin = canWin | !rec(x - j); if ( x/j != j ) { canWin = canWin | !rec(x - (x/j)); } } } ok[x] = canWin ? 1: 0; return canWin; } 
•  » » » 6 weeks ago, # ^ |   0 Research again ) Nice approach.
 » 6 weeks ago, # |   +62 meet a new cheater This is how kedos123 bypasses Plagiarism testing.I am watching him from so many contest , He has done this today and in previous contest, and I am sure he must have done it multiple times before as well. People like kedos123 are spoiling the sport. I don't understand where would cheating take them in life. They will never get anywhere in life but always remain what they are i.e cheater. He should be banned from the platform as soon as possible . MikeMirzayanov sir pls ban him and skip his solutions .todays submission[submission:119865102] 119855748 saw his submission time vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++; vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++; vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;
•  » » 6 weeks ago, # ^ |   +6 Biggest template I have ever seen
•  » » » 6 weeks ago, # ^ |   +4 Looks like he took the template of every LGM (including Benq, he has a huge template) and mashed it into one.
•  » » 6 weeks ago, # ^ | ← Rev. 3 →   0 hello kedos123 , everyone ask about you code snippet , can you also copied it from somewhere ..... lol mf
 » 6 weeks ago, # |   +13 What's the idea behind E2? Can it be solved using Z function?
•  » » 6 weeks ago, # ^ |   0 Rolling hash with binary search got me pass the pretests
•  » » 6 weeks ago, # ^ |   0 I used same idea for E1 and E2. I just checked for the character which exceeds any character in the prefix.
 » 6 weeks ago, # | ← Rev. 3 →   0 Was the TL for E2 really tight? My O(N) soln kept giving TLE on case 12.https://codeforces.com/contest/1537/submission/119909827
•  » » 6 weeks ago, # ^ |   0 Try this input: 50 100 zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfeda
 » 6 weeks ago, # |   0 Spent 1 hour on B due to my stupid mistake . It was a nice round
 » 6 weeks ago, # |   +3 Can E2 be solved without suffix array??
•  » » 6 weeks ago, # ^ |   0 I solved it without suffix array. Waiting for the system test.
•  » » » 6 weeks ago, # ^ |   0 Can you please explain your solution?
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   +4 I took a random string. Let's say-"pabccepabcdf" I ran a loop over the string length. At the instant, where the character is greater than the 1st character ("p" in this case), I printed the string till the last character. At the moment, the character is equal to the 1st character, I checked further characters, if any of them is greater than the corresponding character in the prefix of string, I printed the string till the point I found a matching character for the first character. For example:- In the example, I have taken, the moment I found the 2nd p, I started checking the further characters with the corresponding character in the prefix. "a" with "a", "b" with "b" and so on. I found that in the d in pabcd is greater than the corresponding c in "pabcc". So, I stored the prefix "pabcc" and printed it required number of times. UPD: It passed. Submission
•  » » » » » 6 weeks ago, # ^ |   0 So overall you figured out that "pabcce" is a better prefix than "pabccepabcd" and then you broke the loop.But what about the prefixes in between this matching window? "pabccep", "pabccepa", "pabccepab", "pabccepabc" Why are you not comparing with these as well? to figure out the best prefix in this window
•  » » » » » » 6 weeks ago, # ^ |   0 "pabcce" is lexicographically smaller than "pabccep".
•  » » » » » » » 6 weeks ago, # ^ |   0 In this case, yes. But you are not comparing these prefixes in your solution in general(you are just ignoring them). I guess there is a way to prove that all these prefixes will not lead to a better result as compared to the current prefix.
•  » » » » » » » » 6 weeks ago, # ^ |   +1 Yes, it can be proved that those will not lead to a better result. I can't give you formal proof but I will try as much as I can.It is sure that there isn't a character greater than "p", else we would have used break and printed the loop with prefix just before that point. So, those prefixes must have a second occurrence of "p". The character just before "p" will be surely smaller than "p". So, that prefix will be lexicographically smaller
 » 6 weeks ago, # |   +5 Problem D have some ideas in common with a problem which was part of the recently first round of the Tournament of Towns Mathematical Olympiad
•  » » 6 weeks ago, # ^ |   +8 Moreover! Problem D was Problem 6 Argentian TST Cono Sur Mathematical Olympiad 2019
 » 6 weeks ago, # |   +27 D is very similar to MIT PRIMES 2019 G5Even the names are the same...
 » 6 weeks ago, # |   0 AE1BE2DC would have been a nice ordering (for me, at least).I think that's weird, isn't it?
•  » » 6 weeks ago, # ^ |   0 A,C,B,E1,E2,D :')
•  » » » 6 weeks ago, # ^ |   +10 Ah! That hurts.
•  » » 6 weeks ago, # ^ |   0 In B, we just had to print 1 1 n m :)
•  » » » 6 weeks ago, # ^ |   +5 Oh, I checked for (1,1), (1,m), (n,1) and (n,m) :'(
 » 6 weeks ago, # |   0 It was a cool round.
 » 6 weeks ago, # |   +1 Sometimes Java is worse.problem C . I used LinkedList. Costs me TLE . like seriously dumb Changed to ArrayList and AC . lolProblem D . Create a set of long for checking powers. Got WA — because i was comparing int with the long values in set Took me hell lot of time to bebug what went wrongGood problems. Sadly i coudnt perform well :(
 » 6 weeks ago, # |   0 Only solved A T_TGet the logic of C but TLE, my approach is sort the array in ascending order and rotate the array while the next rotation make |H1-Hn| minimum.
 » 6 weeks ago, # |   +18 E is a standard application for Z-array.
•  » » 6 weeks ago, # ^ |   +3 i tried using lps as in KMP, by i guess was missing some cases.
•  » » » 6 weeks ago, # ^ |   0 Try to append z character n times in the given string and then apply the same thing. This worked for me but I used Z algo.
 » 6 weeks ago, # |   +3 My Sol. for B is (1 1 n m) But i see large codes of ppl?
•  » » 6 weeks ago, # ^ |   +1 It's correct,
•  » » » 6 weeks ago, # ^ |   0 how? can you please explain?
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   +1 The maximum number of steps would be 2*[(n-1) + (m-1)] (first testcase was a hint). No matter what the starting position is, If the yo-yo's are at (1,1) and (n,m), You will always get maximum number of steps.Proof- There are only 2 ways 1. (i,j)->(n,m)->(1,1)->(i,j) 2. (i,j)->(n,m)->(i,j)->(1,1)->(i,j)Calculate the number of steps for these, you will always get maximum number of steps.
•  » » » » » 6 weeks ago, # ^ |   0 brilliant thought process!
•  » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 How did you think of or get 2*[(n-1) + (m-1)]. Sorry to ask but i am a newbie.Also thanks so much for the solution. I get it.
•  » » » » » » 6 weeks ago, # ^ |   0 During the contest, I thought of the following problem- In a n*m grid we have to go from point A to B and then come back to A. Maximum number of steps would be 2*[(n-1) + (m-1)] for A=(1,1) and B=(n,m). This is very clear I guess. Now if I throw in a point C in between, you can see (1,1)-C-(n,m)-(1,1) will always return in 2*[(n-1) + (m-1)] steps. I then verified the claim against sample test cases.
•  » » » » » » » 6 weeks ago, # ^ |   0 Thanks a lot!
•  » » 6 weeks ago, # ^ |   0 same as my solution! it also can be n,1 and 1 , m it will also produce same answer.
 » 6 weeks ago, # | ← Rev. 2 →   +8 After bricking C and getting it in the last 4 seconds, read D. These alice and bob problems always get on my nerves. Any suggestions on how to improve on game theory?
•  » » 6 weeks ago, # ^ |   +6 Lol same, im not too good in game theory but here i just brute-forced and coded the pattern
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 can you please share your core on brute force -><- I tried writing it but i'm not getting the expected output. Edit: got it :)
 » 6 weeks ago, # |   0 In problem C, what should the answer be for the test case belowGiven = 1 1 3 3 6 6Shouldn't the answer be 3 6 1 6 1 3 , coz the absolute diff b/w first and last is min =0 and also the difficulty level is 3
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   0 1 3 3 6 6 1
•  » » » 6 weeks ago, # ^ |   0 My bad, i didn't count the equality. It becomes tricky if the author changes the condition from h[i] <= h[i+1] to h[i] < h[i+1].Then for the test case i mentioned, your sequence will give difficulty = 2 whereas mine difficulty = 3. Am i sounding naive ??
• </ » » » » 6 weeks ago, # ^ |   0 Yeah, I made the same mistake. I was trying to solve the h[i]<=h[i+1] problem.