### GreenGrape's blog

By GreenGrape, 3 years ago, translation,

Hello, Codeforces! Haven't seen ya for a while :D

Codeforces Round #461 will take place on Wednesday, 20:15 MSK. Note that the round is scheduled a bit later than usual.

The round will be rated for all division 2 participants. Division 1 is heartly welcome aswell :)

My gratitude to Kolya (KAN) for round coordination, Grisha (vintage_Vlad_Makeev), Oleg (xen), AmirReza (Arpa) and Senya (kraborak) for testing and Mike (MikeMirzayanov) for awesome Codeforces and Polygon. A special thanks goes to Dima (Dmitriy.Belichenko) and Camille (pseuda) for sharing a problem sketch with me.

The main character of this round will be Imp the playful monster. Yet the statements are guaranteed to be short :)

There will be six problems with the following scoring:
500 — 1000 — 1250 — 1500 — 2000 — 2750

GL & HF!

Note that the round has been moved 40 minutes later to avoid clashing with CSAcademy Round.

UPD. The contest is over. Well...

UPD. System testing is over! Editorial.

Congratz to the winners!

Div. 2:

Div. 1 (unofficial):

UPD. The complete editorial is out (D/F).

• +674

 » 3 years ago, # | ← Rev. 2 →   +44 Really loved your last round but hopefully this time, there will be a better difficulty distribution in the problems!
•  » » 3 years ago, # ^ |   +74 I tried to make it a bit more balanced this time :)
•  » » » 3 years ago, # ^ |   +45 I am still trying to solve your C up to this day.
•  » » » » 3 years ago, # ^ |   -38 How to download Test #15 of 650C Table Compression?
•  » » » » 8 months ago, # ^ |   0 Hey! This might help. Spoilers for problem C : Cave PaintingInstead of checking whether the remainder left by dividing n by 1 to k are different or not, lets find the max K such that when n is divided by 1 to n, it gives different remainder all the times. For e.g. whenever n is even, K = 1 Note that all these 1 to K will give remainders as 0,1,2,...K-1 as they are different and they don't have other choice.So just find K for the given n and check if it is greater than equal to k.Hope I am clear.
•  » » » 3 years ago, # ^ |   0 balanced such that b is easier than a
 » 3 years ago, # |   +2 codeforces contests are always great and amazing , i always try to attend the contests , but my bad , may be i will not be able to participate in this contest due to some travelling stuffs , and it is much painful for me , competitive programming is my passion
•  » » 3 years ago, # ^ |   -39 How to download Test #15 of 650C Table Compression?
 » 3 years ago, # |   +38 Ohh god CM problem setter again.
•  » » 3 years ago, # ^ |   +48
•  » » 3 years ago, # ^ |   -28 How to download Test #15 of 650C Table Compression?
•  » » » 3 years ago, # ^ |   +1 In input : 1000 1000 1002 1002 1002 1002..... (1000 times) In output : 1000 1000 1000 1000..... (1000 times) Coz you were asking so many times.
•  » » » » 3 years ago, # ^ |   +3 the other numbers must be n*m(1000*1000). And also if there is only 1002(1000*1000 times) the answer must be 1(1000*1000 times).
•  » » » » » 3 years ago, # ^ |   +7 oh but downloading the full test cases is not possible Moreover it is highly unlikely that you will be able to debug your code with such a large input /output file
 » 3 years ago, # |   -55 Is it rated?
•  » » 3 years ago, # ^ |   +40 Are You Retarded?
•  » » » 3 years ago, # ^ |   +64 it's not my comment. but i still want to apologize
•  » » » » 3 years ago, # ^ |   -14
•  » » » 3 years ago, # ^ |   -33 How to download Test #15 of 650C Table Compression?
 » 3 years ago, # |   +46 Obligatory comment: "Hope the problem statements are as short as the announcement"
 » 3 years ago, # | ← Rev. 2 →   +18 Have seen you being quite responsive in the last contest. Keep up the good work, and well, wish a quality and balanced contest ;)P/s: Yup, I still can't figure out the most optimal income when using Perun's ult xD
•  » » 3 years ago, # ^ |   +3 Thnx :D
 » 3 years ago, # |   +13 Oh... This contest is even later than others T_TToo bad I can't enter it because of... time zone problem...Wish you guys good luck and high ratings anyway ^_^
•  » » 3 years ago, # ^ |   +8 When it starts in your country?
•  » » » 3 years ago, # ^ |   +24 00:35……
•  » » » » 3 years ago, # ^ |   0 I just want to switch on the music Sound of Silence.
•  » » » » 3 years ago, # ^ |   0 its not a friendly time for UTC+8
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 It was supposed to start at 01:35 AM here, and it even became 02:15 :)
•  » » 3 years ago, # ^ |   0 I have been used to the time zone problem...
 » 3 years ago, # |   0 I hope my rating will be more than 1900. Then this contest is unrated for me. I don't like rated contest. My rating will down if I don't do well in this contest. CONTEST, CONTEST and CONTEST ... RATING, RATING and RATING ... TRAINING, TRAINING and TRAINING ... ......
•  » » 3 years ago, # ^ |   -22 if you cheat in this contest, this contest is unrated for you too, and if nobody discover it, you can have a high rating:)
•  » » » 3 years ago, # ^ |   +82
 » 3 years ago, # | ← Rev. 2 →   -22 Good luck to all let's see, can i be a Candidate master at the end of this round or not ... wish me luck... :)
•  » » 3 years ago, # ^ |   0 Good Luck!) Hope you will be a Candidate
•  » » 3 years ago, # ^ |   +5 why people doesnt like this post whats wrong about it
•  » » » 3 years ago, # ^ |   0 He didn't get cm :\
•  » » » » 3 years ago, # ^ |   0 sorry,what is cm?
•  » » » » » 3 years ago, # ^ |   +1 Candidate Master
•  » » » » 3 years ago, # ^ |   0 you're right, i was far away from my ideal performance last night ... too upset for it... :(
•  » » » » 3 years ago, # ^ |   +3 well ... now i'm :)
•  » » » » » 3 years ago, # ^ |   +5 congrats
 » 3 years ago, # |   -6 So sad that magic is over, can't troll with different colored grapes anymore :(
 » 3 years ago, # |   0 I wondered what is different from participant division 2 and division1
•  » » 3 years ago, # ^ |   +8 -The Blue Color and all the colors below it are for Div2 contestants. -the purple color and all the colors above are for Div1 contestants. -Div1 guys can participate in Div2 Rounds unofficially.
•  » » » 3 years ago, # ^ |   0 Thank you :D
•  » » » » 3 years ago, # ^ |   0 u r welcome.
 » 3 years ago, # |   0 really awesome, wait for some blossom
 » 3 years ago, # |   +33 A prime number contest!
•  » » 3 years ago, # ^ |   +1 Nice observation!
 » 3 years ago, # |   -19 not u again
•  » » 3 years ago, # ^ |   +52
•  » » » 3 years ago, # ^ |   +14 No way....
•  » » » » 3 years ago, # ^ | ← Rev. 5 →   0 → Literally JoJo video→ Inserts "reference meme"
•  » » » » » 3 years ago, # ^ |   +5 → Literally JoJo video→ Obligatory weeb comment
•  » » 3 years ago, # ^ |   +4 Look on the bright side, man, this time there are no pupil users as problemsetters :D
 » 3 years ago, # |   -23 How to download Test #15 of 650C Table Compression?
•  » » 3 years ago, # ^ |   0 Unfortunately , the large inputs can't be previewed completely. if you stuck with the problem you can see the tutorial of the contest (here). if you still stuck , you can check out the Accepted solutions (here).
 » 3 years ago, # |   +13 "Yet the statements are guaranteed to be short :)" this is the best announcement so far.
•  » » 3 years ago, # ^ |   -49 how to download test 15 on 650 C table compression
 » 3 years ago, # |   -8 Yet the statements are guaranteed to be short :)Thank you very much:)
 » 3 years ago, # |   +11 Unusual early scoring!
 » 3 years ago, # | ← Rev. 2 →   +23 "Codeforces и Polygon"Learning some russian.Learning everywhere. :)
•  » » 3 years ago, # ^ |   +6 Learning so much that you start to understand Croatian. (Source: COCI 2016/17 Round #6 solution, p. 4, line 8) Let Pa < Pb i k * Pa ≤ Pb < (k + 1) * Pa.
 » 3 years ago, # | ← Rev. 2 →   +2 Hope another great problem set from this amazing head. Big chance to rise up. CF is too good to stay away from. :)
 » 3 years ago, # |   -7 Can anyone help me with one problem? I have good answer on test 2 on my computer, but on codeforces I get WA no matter what I do. Obviously compiler settings or something isnt same. Just visit my problem and check last submissions, problem C from few rounds before. 34983468
•  » » 3 years ago, # ^ |   0 Test ur program on custom invocation and u will undrestand why got WA.
 » 3 years ago, # | ← Rev. 2 →   +14 It's quite strange no one noted that this round coincides with CSA round #68 tomorrow! Can't you just delay it 30 more mins? GreenGrapeUPD: the contest is delay 40 mins!
•  » » 3 years ago, # ^ |   +5 More like, CSA coincides with Codeforces Round 461.
•  » » » 3 years ago, # ^ |   +10 linguistically, is there a difference?!
•  » » » 3 years ago, # ^ |   +4 I mean, CSA consistently hosts contests only in that time slot.I guess it went under the radar since they start 30mins earlier until recently.
•  » » 3 years ago, # ^ |   +1 CF round moved by 40 minutes.
 » 3 years ago, # |   +7 Glad to hear this: Yet the statements are guaranteed to be short :) Looking forward to it !!!
 » 3 years ago, # |   -11 How to download Test #15 of 650C Table Compression?
•  » » 3 years ago, # ^ |   +16 Okay, this should be the last time I see this anywhere.You cannot do that. At all.(Well, except if the problem setter provide the full package publicly).For convenience, all test cases are compressed in case they are too big, so large input/output files are near to complete omittance.Usually, if you get stuck with a big tests, it can only mean a few things: You ran out of memory (MLE verdict). You ran out of time / your algorithm was too complex (TLE verdict). You handled the undefined behaviors poorly (WA/RE verdict). These tests are mostly used to tests how a code works against large input streams, and not to be corner cases, so please check your own source code for places that can be optimized, not just spamming the blogs all the time.
 » 3 years ago, # |   +34 Normal guy's Imp.Codeforces member's Imp.
 » 3 years ago, # |   +117 Is it legal?
•  » » 3 years ago, # ^ |   +34 I registered him manually, just a prank ^^
•  » » » 3 years ago, # ^ |   +54 How is registering someone in a contest a prank?
•  » » » » 3 years ago, # ^ |   +8 It’s just weird so a little funny. Pseudo will be like “wtf, dude” .. hence.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +3 Seems someone also wanted to prank you and registered you manually :P See you on leaderboard :D
•  » » » » 3 years ago, # ^ |   +5 This will be the first time someone solve all problems in 1 minute :D
 » 3 years ago, # |   +11 I am gonna solve A, B and C today !
•  » » 3 years ago, # ^ |   0 Best of luck !
•  » » » 3 years ago, # ^ |   0 Thanks Izanagi !!! I Hope, you will also do well in this round.
•  » » 3 years ago, # ^ |   0 you will solve nothing we will come newbie
•  » » » 3 years ago, # ^ |   +4 o yeah ! I have solved A, B and C !
•  » » » » 3 years ago, # ^ |   0 only pretest passed
•  » » » » 3 years ago, # ^ |   0 only pretest passed**
•  » » » » » 3 years ago, # ^ |   0 Let's see after system testing !
•  » » » » » » 3 years ago, # ^ |   0 i hope we solve nothinh
•  » » » » » » 3 years ago, # ^ |   0 i hope we solve nothing
•  » » » » » » » 3 years ago, # ^ |   0 Problem — C Time limit exceeded on test 60 ! :'(
•  » » » » » » » » 3 years ago, # ^ |   0 be peptimistic
•  » » 3 years ago, # ^ |   0 Pretests passed
 » 3 years ago, # | ← Rev. 7 →   +55 Wow :O
•  » » 3 years ago, # ^ |   +15 And I thought heaven didn't exist.
 » 3 years ago, # | ← Rev. 2 →   +1 Two line summary of the announcement : 1. The statements are guaranteed to be short :) 2. There will be six problems with the following scoring: 500 — 1000 — 1250 — 1500 — 2000 — 2750
 » 3 years ago, # | ← Rev. 2 →   +1 The time is a little late for me,but I am still glad to take part in the round.It's really a good chance to train myself. :)
 » 3 years ago, # |   0 I just want to say it's too late for Asian contestants . Hope I will stay alive for being such late......
•  » » 3 years ago, # ^ |   0 One hour earlier, but yeah, I feel you, dude...12:15 AM in UTC+7 (Hanoi, Vietnam). Uh-oh...
 » 3 years ago, # |   0 Hope I can become Expert!
•  » » 3 years ago, # ^ |   +6 Happy!
•  » » » 3 years ago, # ^ |   0 Congrats :)
 » 3 years ago, # |   -8 Sanwli saloni, adaayein manmohni Tere jaisi beauty kisi ki vi ni honi Thande ki botal main tera opener Tujhe ghatt ghatt main pee lon Coca Cola tu ROUND GONNA BE AMAZING !
•  » » 3 years ago, # ^ |   +2 You Should Behave On website like codeforces , though fun is necessary , but people seach for good comments over here . I hope you understand it ! Good Luck for contest .
 » 3 years ago, # |   0 Let's see How this go?
 » 3 years ago, # |   +1 i think i will be green this time,thank you GreenGrape.
•  » » 3 years ago, # ^ |   +1 Pessimism level: Notali.haidar
 » 3 years ago, # |   -24 is it rated??
•  » » 3 years ago, # ^ |   0 i think you love negative contribution.
•  » » » 3 years ago, # ^ |   0 it increases my ego and confidence , yes.
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   0
•  » » » » » 3 years ago, # ^ |   -11 if you really get annoyed because of some online comments you need help:)
•  » » » » » » 3 years ago, # ^ |   0 may be this site is not just a random website for many people
•  » » » » » » » 3 years ago, # ^ |   0
•  » » » » » » » » 3 years ago, # ^ |   0 who knows :D
•  » » » » 3 years ago, # ^ |   +3 I want a 'HaHa' button(as in fb) in codeforces, people have become too funny :v
•  » » » » » 3 years ago, # ^ |   -6 yes i would spam that button on every comment of yours
 » 3 years ago, # |   0 can't wait.....
 » 3 years ago, # |   +11 Again a round with near 7000 participation :)
 » 3 years ago, # |   +3 I hope the long queue is fixed before the contest starts
•  » » 3 years ago, # ^ |   +12 I hope the int vector is fixed setprecision(10) after the contest end ;!
 » 3 years ago, # |   +1 How's this possible?!
•  » » 3 years ago, # ^ |   +5 What's wrong? I thought you can lock the problem as long as you have the "Pretest passed" verdict, irrespective of whether you get hacked or not. Correct me if I am wrong.
•  » » » 3 years ago, # ^ |   +3 Nope, you can't lock after you got hacked. But I think it happened because times are very close.
•  » » » » 3 years ago, # ^ |   0 Oh okay, that's interesting. This probably happened because of the long hack queues today.
 » 3 years ago, # | ← Rev. 2 →   0 Contest of hacks... :D
 » 3 years ago, # |   +8 very long queue in hacks
 » 3 years ago, # |   +6 HackerRank will be proud of this contest
 » 3 years ago, # |   +27 I think this contest are more "Successful hacking attempt" messages than "Accepted"
 » 3 years ago, # | ← Rev. 2 →   +15 A round full of hacks. Even one of my friends hacked me, nice contest
 » 3 years ago, # | ← Rev. 2 →   0 Hackforces in a nutshell...Anyway, to be serious, maybe it was just me, but I think the scoring is still a little bit underrated. This should be 500-1000-1500-1750-2250-3000. C is sufficient to be called a Div2C, D is somehow quite twisted (got 4 WAs verdict on the 4th problems, all on pretest 4).Not really sure about my opinion about E and F, coz' I have just read it for a short while.After all, it was fun actually. And these Imps are just cute ^^UPD: I took my words about D back. Somehow I wish I could think sharper...
•  » » 3 years ago, # ^ |   +4 Even you hacked me. This is first time in my life when I got 3 hacks on A
•  » » » 3 years ago, # ^ |   +5 Everyone has their own bad days...I also faced the same situation as you, once at Codecraft-18 and Codeforces Round #458 (Div. 1 + Div. 2, combined).
•  » » » 3 years ago, # ^ |   0 bravo
 » 3 years ago, # | ← Rev. 2 →   +3 Can someone please provide explanation for third testcase of problem E.
•  » » 3 years ago, # ^ |   +18 I think you are confused just like I was — summoning a bird doesn't increase your mana, it increases you maximum mana.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +3 Ohh Thanks.Same mistake
•  » » 3 years ago, # ^ |   +8 Take 1 bird from the first nest (you don't have enough mana to take two birds), take 10 birds from second nest (your mana capacity is 17 and you have 15 mana so you can take them all).
 » 3 years ago, # |   0 Problem D is solvable with radix sort?
•  » » 3 years ago, # ^ |   +7 I just used normal sorting with a comparing function that returns numS[a] * numH[b] > numS[b] * numH[a]
•  » » » 3 years ago, # ^ |   0 Hell man! :'( I tried to do it with normal comparison too but that condition didn't pop up to my mind! How can you prove this works ?
•  » » » » 3 years ago, # ^ |   +8 Given a string s, denote f(s) to be the number of subsequences 'sh'. Now given strings s1, s2, ... sn, it is easy to prove that if we denote xi, yi as the number of 's' and 'h' in string si, that . To maximize our desired value, all we need to look is the value . For n = 2, it's trivial that the comparing function wewark described works.We will induct on n — say the comparing function works for n and we prove it works for n + 1. Clearly, , as after deciding which sn + 1 to choose, we are left with optimizing the value with n strings. (induction hypothesis is used here)Now, all we need to do is prove that we need to put sn + 1 as the string with the highest value. This is simply greedy, easy to prove.
•  » » » » 3 years ago, # ^ |   0 Let's assume we have two strings a = sshhh and b = ssshhh.we can do a + b = sshhhssshhh.And b + a = ssshhhsshhh.How many sh's we got after this concatenation? We can note that this value is numS[a] * nums[b], because we didn't lose any sh's that were in a or in b, so what we got is .The same logic works for .
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 Do you have a formal proof of such comparator?UPD Just came up with your idea.
•  » » » 3 years ago, # ^ |   0 Got it, thank you all.
 » 3 years ago, # |   +3 The 1st problem made for the hack..
 » 3 years ago, # |   +1 Idea behind C ?
•  » » 3 years ago, # ^ |   0 For small Ks (probably <= 1000) you will find two exact values, so just implement
•  » » 3 years ago, # ^ |   0 just do it with brute force :)
•  » » » 3 years ago, # ^ |   0 Ok. Limits are very large so how do you guarantee fast running time ?
•  » » » » 3 years ago, # ^ |   0 it will find a pair soon
•  » » » » » 3 years ago, # ^ |   +2 Thats what my questions is. How to prove that ?
•  » » » » » » 3 years ago, # ^ | ← Rev. 2 →   +7 It is known that (a very important consequence/equivalence of prime number theorem) . For n, k to be good, (n + 1) must be a multiple of . So n must be like ek. This should be good enough, as running time is .
•  » » » » » » » 3 years ago, # ^ |   +13 Or, say primes >10 will multiply, so it be over 10^18 fast...
•  » » » » » » » 3 years ago, # ^ | ← Rev. 3 →   0 Bruteforce works. :) This was my approach.gcd(1,2,...,k) = 1 Therefore, lcm(1,2,...,k) = 1*2*3*..*k Also, since the remainders have to be distinct, one observation is that for any i between 1 to k, n%i should be equal to i-1.This means: n = i*C + i-1 ==> (n+1)%i = 0. Therefore n+1 should divide all the numbers from 1 to k. This also means that n+1 should be equal to the least common multiple of 1,2,..,k (in this case its just the product.) or any multiple of this lcm. Looping from 1 to k and stopping the loop whenever product crosses n+1. Product crosses n+1 easily by the time we reach i=50 in the loop. We return "No" in this case. If k! is equal to n+1 then we return "Yes"
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   -7 try thisbool f;int top=0;for(int n=1;n<=100000;n++){for(int k=1;k
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 People have this idea:Since you need to have all unique values till k. They should be obtained serially from 1,2,3,..k like this: 0,1,2,3,4,5...k-1so, n should be of the form: 2x+1, 3x+2, 4x+3, 5x+4, 6x+5, 7x+6,... all at the same time. But if something is of the form 5x+4 .. it cannot be of the form 2x+1 .. (coz one is odd and other is even, for even x, similarly chose a pair for odd x) Edit:The conclusion seems flawed. Maybe we can continue the logic like this:Number has to be of form: 2x+1, so 3,5,7,9.. 3x+2, so 5,8,11,14.. ... ....It kind of creating a sieve of eratosthenes, which generates a prime number.I don't what that gets us :)
 » 3 years ago, # |   +9 Locking after being hacked
 » 3 years ago, # |   +6 how to solve D?
•  » » 3 years ago, # ^ |   0 sorting : )
•  » » » 3 years ago, # ^ |   0 merge or bubble
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 consider to string a and blet the number of 's' in a be a.S and the number of 'h' be a.Hif you arrange a and b be abthe you will get a.S*b.H pairsso we know that if a.S*b.H>b.S*a.H then a should be in front of b
 » 3 years ago, # |   0 What was the hack for A?
•  » » 3 years ago, # ^ |   0 if you donnot have a copy,you cannot get more copy
•  » » 3 years ago, # ^ |   0 Most common one: x 1 (while x is a positive even number).Expected answer: "NO" (as there are no initial copies to generate new ones)
•  » » 3 years ago, # ^ | ← Rev. 2 →   +3 0 00 11 03 02 1
•  » » 3 years ago, # ^ |   0 2 1output NO
•  » » 3 years ago, # ^ |   0 something!=0, y=0 and something!=0, y=1
•  » » 3 years ago, # ^ |   0 4 1
 » 3 years ago, # |   +5 Problem A just burn my room
 » 3 years ago, # |   0 What is wrong this this solution for E?Let dp[i][j] be the max mana we have at tree i with j summoned in the trees before us. The max capacity is always w + j·b for a fixed j so we don't have to store that. I used long long int here and for dp table so no overflow. The answer is the largest i such that dp[n][i] is valid. I kept getting WA on pretest 5.
•  » » 3 years ago, # ^ |   0 it's correct
•  » » 3 years ago, # ^ |   0 Did you use sliding window maximum?
•  » » 3 years ago, # ^ | ← Rev. 2 →   +3 problem is that max summoned birds numbercan be ci * n = 103 * 104 = 107, so your array will have size dp[103][107].EDIT: I didn't read task carefully :{
 » 3 years ago, # |   +1 How to solve B? Seems easy cause lot of people solved it, but I only solved C and D.
•  » » 3 years ago, # ^ |   +7 ans = 0 for a = 1 .. n for b = a .. n c = a ^ b if b <= c <= n and a + b > c ans++ print ans 
•  » » » 3 years ago, # ^ |   +3 Wow.. I'm a fool :) Thanks for your reply! (And all the others too)
•  » » 3 years ago, # ^ |   +3 can you please tell me how to solve C?
•  » » » 3 years ago, # ^ |   0 Try 1~50
•  » » » 3 years ago, # ^ |   0 n mod 1 , ... , n mod k has to be different. but you can easily notice that n mod 1 is always 0. So, n mod 2 must be equal to 1, because n mod 2 is 0 or 1, but it cannot be 0. ... n mod i must be equal to i-1, which means that n+1 is divisible by 1, 2, ..., k.
•  » » 3 years ago, # ^ |   +3 For a <= n, b <= n, let c = a ^ b, if all other conditions hold, then increment answer
•  » » 3 years ago, # ^ |   +3 a^b^c = 0 implies a^b = c. So bruteforce for a, b, and check if the 3 numbers can form sides of a triangle.
•  » » 3 years ago, # ^ |   0 You can iterate through all possible cases of a and b. c can be calculated by bitwise xor:As a xor b xor c = 0, we can agree that c = a xor b.If c is not lower than a or b, and not greater than n, and a,b,c forms a non-degenerate triangle (simply check that if all 3 inequalities of triangles are proved right), then you can add 1 to the answer.
•  » » 3 years ago, # ^ |   +3 Simply iterate over all a, b values . Check a + b > c, a + c > b and b + c > a
•  » » 3 years ago, # ^ |   +1 I actually passed it with brute forcing which is O(N^3). It gave me 780ms even though it doesn't seem to be the intended solution.
•  » » » 3 years ago, # ^ |   +3 Your condition ( c < a + b ) in the third loop in your code makes it about O(n^2) i belive
•  » » » » 3 years ago, # ^ |   0 it is still O(n^3) with very small constant.
 » 3 years ago, # |   +1 Test for hack C: 1 2
•  » » 3 years ago, # ^ |   0 Isn't it "yes" ?
•  » » » 3 years ago, # ^ |   +7 Yes. Many people (me too) have writen: if( n < k ) cout << "No";
•  » » » » 3 years ago, # ^ |   0 Bad luck, man
•  » » » » 3 years ago, # ^ |   0 fortunately I had written n<=k so it failed pretest and I was able to correct it. and now i'm purple.
•  » » 3 years ago, # ^ |   0 pretty sure this is the case that got me
 » 3 years ago, # |   0 How to solve F? I tried taking all N numbers initially and consecutively removing the one which will decrease f(I) the most but such that f(I) ≥ k will remain true. Just plain intuition, didn't work.
•  » » 3 years ago, # ^ |   0 I bruteforced n<=17, implemented it for larger n, it worked well for all local inputs. Still kept getting TLE on the 7th pretest. I just wonder whether it was my complexity, I/O, special countertest designed to make this kind of solution fail...
•  » » 3 years ago, # ^ |   0 Explanation — Problem F
•  » » » 3 years ago, # ^ |   0 Man, them noise and echo are simply unbearable which makes your speech unrecognizable.
 » 3 years ago, # |   +19
•  » » 3 years ago, # ^ |   0 Is it so interesting to post one picture several times?
 » 3 years ago, # |   +1 In question D, what is the pretest 4?
 » 3 years ago, # | ← Rev. 2 →   +46 I think there should be a limit on the number of hacks for each problem... Yeah hacks are a part of CF, but when a hacker with ABC has more points than a non-hacker with ABCD, I think there is a problem.
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 I think so too.On today's A,same hacker hacks same participant many times…For example,Hack 0 1 -> Hack 1 0 -> Hack 0 0…At least,some fix is need to this system…
•  » » » 3 years ago, # ^ |   0 what is the output for test case 0 1???Isn't it "YES"??
•  » » » » 3 years ago, # ^ |   0 Yes
•  » » » » » 3 years ago, # ^ |   0 For 0 0 and 1 0???
•  » » » » » » 3 years ago, # ^ |   0 NO NO
•  » » » » » » » 3 years ago, # ^ |   0 Everything is ok. but how/why my code is hacked??
•  » » » » » » » » 3 years ago, # ^ |   0 for 10001 1? should be No
•  » » » » » » » » » 3 years ago, # ^ |   0 found "NO"
•  » » » » » » » » » 3 years ago, # ^ |   0 Is case sensitivity a problem? NO vs No ?
•  » » » » » » » » 3 years ago, # ^ |   0 3 1 -> NO…?
•  » » » » » » » » » 3 years ago, # ^ |   0 found "NO"
•  » » » » » » » » » 3 years ago, # ^ |   0 4 1 should be NO
 » 3 years ago, # |   0 Can anyone give me some idea for 922F - Делимость
•  » » 3 years ago, # ^ |   +6 for i = 1 .. n for j = i .. n, j += i d[j]++ if sum(d) < k print no otherwise, kick out some numbers > n/2 - since they do not have multiples - so that number of pairs will decrease to k 
•  » » » 3 years ago, # ^ |   0 will this work for very small k?
 » 3 years ago, # |   0 can someone help me find why i was hacked in problem C? 35018691
•  » » 3 years ago, # ^ |   +8 n=1 k=2 answer:yes
 » 3 years ago, # |   0 Short statements not means a lot of hacks.
 » 3 years ago, # |   +11 This was the biggest slaughter since I started to participate at contests
 » 3 years ago, # | ← Rev. 2 →   +3 When will the system testing start ?Update: System testing started :)
 » 3 years ago, # |   +37 When someone asks how life is going on?
 » 3 years ago, # |   +13 While everybody is discussing today's contest, I just found my love <3Thanks Mike :)
 » 3 years ago, # | ← Rev. 2 →   +75 Isn't it obvious that current hacking system sucks? It's a brilliant idea, but really bad realization. Hacking should give you points (maybe even more than 100) if you find hidden not obvious bug in someone's code. But if you just take n = 0 corner case in div2A and it gives you 1500 points that's so unfair: you get that much points not because you are smart or your debugging skills are better than average, but because you have gray guys in your room who always do this mistake.That's why I think that hacking points must be not a constant, but a function, which depends on how much hacks there are on the problem.
•  » » 3 years ago, # ^ |   +2 if problem difficulty is 500 points so on one hack your receive 100 is 20% but if problem difficulty is 2500 points so on one hack your receive 100 is only 4%
•  » » 3 years ago, # ^ |   +3 til i am a grey
•  » » 3 years ago, # ^ |   0 Agreed. I hacked 18 solutions (Problem A) using only two test cases And guess what? My own solution missed one corner case and I got WA in system test. But 1800 points gave me an unfair advantage.
 » 3 years ago, # | ← Rev. 2 →   +12 Many among you would be disappointed because of all the hacks, but I feel that the problems were super exciting and I enjoyed all the brainstorming while getting hacked.+1 to GreenGrape
 » 3 years ago, # |   0 I don't know why, but my submission got stuck at test 39 for last 10 min.s. I am on page 6 of filtered list
 » 3 years ago, # |   +57 When you find one of the testcases for A
 » 3 years ago, # | ← Rev. 3 →   0 I think that there is a problem with D testcases. This is my AC submission.The problem with my solution is that the comparison is made entirely on integers, and therefore it could overflow.I'll try to generate the testcase, I think that: 2 sssss...ssssssh hhhhh...hhhhhhs is enough.EDIT: oh yes, I think it was enough. My solution gave 50000, but when reversing gave 2499950000. I don't think I'm the only one who failed for this.
•  » » 3 years ago, # ^ |   0 That's actually quite strange since tests 11 and 12 are made exactly to counter this.
•  » » » 3 years ago, # ^ |   +3 I would like to remember that overflow is an undefined behaviour, so anything could happen, even getting an incorrect AC (A guess for why this happens is that the compiler may do some simplifications like x*y/x = y — avoiding the calculation that gets the overflow — maybe something in that line is what happened)
•  » » » » 3 years ago, # ^ |   +3 I tested it using custom invocation, and it gives the incorrect output there (50k instead of 2499..etc).
•  » » » 3 years ago, # ^ |   0 Oh, I think I know what's happening here. My submissions treats different the strings consisting of only 'h' or only 's', so that's why my submission passed.
•  » » » » 3 years ago, # ^ |   0 lucky you didn't loose 100 points on tc 11.:)
•  » » 3 years ago, # ^ |   0 35024915 this solution gets tle on your test, if i am not mistaken. if so, D should be reevaluated
•  » » » 3 years ago, # ^ |   0 Looks fine to me. Total characters are 200000.
•  » » » » 3 years ago, # ^ |   0 yeah, i realized that i tested it wrong right now, sorry
 » 3 years ago, # |   +3 long long contest! #define int long long accept my C and D :)
 » 3 years ago, # |   0 Problem A states: "Initially, Imp has only one original toy. but system test 20 tests for 0 (ZERO)!!!!!!
•  » » 3 years ago, # ^ |   0 Print "Yes", if the desired configuration is possible, and "No" otherwise. so when input is 0 you should say no because he have 1
•  » » » 3 years ago, # ^ |   0 I have read again the problem before seeing your reply and come back here to delete my post, but now i leave it. Of course you are right.
 » 3 years ago, # |   +4 Can't believe my STUPID code for C got pretests passed even and finally failed on test 76th!!! I used k instead of i in my loop.Failed.vsAccepted.
 » 3 years ago, # |   +13 Couldn't proof the solution for C during the contest. But now, I think I got a proof.Correct me if I'm wrong plz.If the answer for n is "yes", then for all i from the set [1,k] will have distinct remainders.We can easily see that for all i [1,k] n mod i == i — 1,so, we can say (n + 1) mod i == i == 0 ;That means all i [1,k] must divide (n+1) simultaneously.Which means (n+1) must be divisible by the lcm(1,2,3,........,k).Now the lcm(1,2,3,.........,k) would be > 1e18 when k is > 41(or something) .That's why the brute-force solution worked.
 » 3 years ago, # |   0 ohuenniy round
•  » » 3 years ago, # ^ |   0 ploxo materit'sa
 » 3 years ago, # |   +5 who is sorry-haghani ?!Haghani ...
•  » » 3 years ago, # ^ |   0 I think he is Deemo !
•  » » » 3 years ago, # ^ |   0 it's possible ... :)
 » 3 years ago, # |   0 Can anyone explain? :( 35036774 35011492
•  » » 3 years ago, # ^ |   +13 never use >= or <= on a cmp function. this is your accepted code 35038659
•  » » » 3 years ago, # ^ |   0 <= got Accepted 35037198could you please explain why should we avoid <= or >= in compare function?
•  » » » » 3 years ago, # ^ |   +6 you can read why here
•  » » » » » 3 years ago, # ^ |   0 Thanks :)
•  » » » 3 years ago, # ^ |   +5 Oh thanks:(
•  » » 3 years ago, # ^ |   +10 don't know why but writing this line in cmp function gives AC 35038641 return a.cnt[1]*b.cnt[0]
 » 3 years ago, # |   +5 Solving the B in O(n^3) no problem... i would say a O(n^2) solution was intended but anyway: http://codeforces.com/contest/922/submission/35039399
•  » » 3 years ago, # ^ |   0 lol. Does it pass with int as well?
•  » » » 3 years ago, # ^ |   +8
 » 3 years ago, # |   +8 Hello, world!
 » 3 years ago, # |   0 Why is my code giving runtime error on test 5 in D : Link
•  » » 3 years ago, # ^ |   0 as ajecc mentioned above never use >= or <= on a compare functionhttp://codeforces.com/blog/entry/57538?#comment-412900
 » 3 years ago, # |   +1 This was one of the best and most interesting contests on codeforces that i have given. So much of hacks and good difficulty distribution i should say ,awesome GreenGrape
 » 3 years ago, # |   +10 GreenGrape has never been Green on Codeforces.
•  » » 3 years ago, # ^ |   +8 Actually he was during new year magic
 » 3 years ago, # |   -29 contest was shit , weak pretests
 » 3 years ago, # | ← Rev. 2 →   0 I don't understand A number problem. please, anyone, help me to understand this problem. problem link : http://codeforces.com/problemset/problem/922/A thanks
•  » » 3 years ago, # ^ |   0 you start with o. Then you do magic copy to get additional o and c. Now you have : o,o,cNext you do magic copy on c and get additional c and c , so a total of 3 c,c,c. Likewise.
•  » » » 3 years ago, # ^ |   0 thanks brother
 » 3 years ago, # |   0 http://codeforces.com/contest/922/submission/35048494 why i am facing runtime error thanx in advance
•  » » 3 years ago, # ^ |   +1 try swapping these statement in your comp function if (!h1) return true; if (!h2) return false;
•  » » » 3 years ago, # ^ |   0 But what is logic behind this swap statement
•  » » » » 3 years ago, # ^ |   +1 If 2 equal objects are passed it should always return false. before it could return true.
•  » » » » » 3 years ago, # ^ |   0 Thanx alot
 » 3 years ago, # |   0
 » 3 years ago, # | ← Rev. 2 →   0 Problem B, O(n^3) without pragma 35154154....using pargma only make it a little faster..35129118
•  » » 3 years ago, # ^ |   0 exactly. this n is too small.
 » 3 years ago, # |   0 In Problem B, why (1,2,3)is not a valid set?
•  » » 3 years ago, # ^ |   0 Because it's a degenerate triangle (1+2 = 3) the
•  » » » 3 years ago, # ^ |   0 Oh! A triangle! thanks, I should learn to read the question carefully!