Vladosiya's blog

Codeforces Round #811 (Div. 3)
By Vladosiya, history, 21 month(s) ago, translation, In English

Hello! Codeforces Round 811 (Div. 3) will start at Aug/01/2022 17:35 (Moscow time). You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks.

You will be given 6-8 problems and 2 hours and 15 minutes to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:

take part in at least five rated rounds (and solve at least one problem in each of them), do not have a point of 1900 or higher in the rating. Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of our work. Problems have been created and written by ITMO University teams: MikeMirzayanov, myav, Gol_D, Aris, senjougaharin, me Vladosiya.

Also many thanks to yorky, Jostic11, turmax, oversolver, ivanz, antonis.white, molney, KerakTelor, andrey.starodubtsev, Ahmad45123, myway, sofiaasta, Muhammad98 for testing the contest and valuable feedback.

Good luck!

UPD: Editorial

  • Vote: I like it
  • +217
  • Vote: I do not like it

Comments Comments (193)
| Write comment?
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it +20 Vote: I do not like it

Good Luck everyone! I hope everyone has fun!

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    I agree. I wish you all the best success!

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      Right. Let's hope that this competition will be interesting and everyone will achieve success.

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +75 Vote: I do not like it

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I dont wanna look it.

      → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      The "You" bit should be in a different section so that it is assured that people will read it after the other 5 graphs. Otherwise it becomes a bit weird.

      → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it +14 Vote: I do not like it

      So you mean that her ex is jqdai0815, her father is Petr and "the guy she told you not to worry about" is tourist

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This contest was in fact fun, I can confirm. There is one thing on my mind though. When will ratings change?

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      it depends. some contests give rating points after only 12 hours, while other can take up to 36, or even 48 hours. Reason? I don't know. I only speak based on my contest participating experience

      → Reply
»
21 month(s) ago, # |
  Vote: I like it +5 Vote: I do not like it

Where is the link to register? I haven't participated in any of the rounds earlier. Can someone direct me, how to do so. Thanks.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +8 Vote: I do not like it

Good luck everyone!

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +8 Vote: I do not like it

Good Luck to everybody.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +14 Vote: I do not like it

I am competing tourist in parrallel universe.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

Finally a div.3 round!!❤(●'◡'●)

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yes lads we gray guys can finally answer more than just the first 2-3 questions

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +3 Vote: I do not like it

All The Best EveryOne !!

→ Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Hopeful this contest will be more friendly to newbies. Best of luck.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +3 Vote: I do not like it

I hope it will be a great contest for everyone

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope everyone has fun Good luck for all participants I hope it will be nice contest

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Good Luck

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Ollie gave

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

All my dreams are of an AK. (It only happened on a div 4)

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Agree with you! Me too having a dream to AK a div3!

    Now finally! I can participate in a contest.. after a long time (I had my exams). Hope this div3 would be last rated div3 for me! ;)

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I kid you not, this actually happened with me a week ago.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it -17 Vote: I do not like it

I am so weak that I can only get higher rating by div3 or div4 :(

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    Why so many downvotes???I can't understand.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Because you even got rating down on div3, like these downvotes.

      → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

I'm registered as out of competition due to rating rollback, when the contests start will I be considered as out of competition or what?

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    No, only the rating while registering is considered. If you want to participate as rated, you can unregister and register

    → Reply
»
21 month(s) ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

One person is missing is Vovuh :p BTW, Thanks Vladosiya and team for the contest.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it -10 Vote: I do not like it

How to do well in Div. 3? What's your opinion if someone fails to solve all of the problems..?

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    then try solve div 4

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Assuming you meant to fail to solve any problems, knowing arrays, sortings, and elementary maths is usually enough for first problems.

    → Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it -39 Vote: I do not like it

swap F G!!!

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +43 Vote: I do not like it

    I don't think it is a good idea to swap them during the round.

    Аfter all, it is ICPC rules, the order doesn't play huge role here.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
      ← Rev. 2   Vote: I like it +18 Vote: I do not like it

      I mean,F is more difficult to solve than G. not really swap them :)

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
          Vote: I like it +13 Vote: I do not like it

        finally I solved both of them :)

        → Reply
        • »
          »
          »
          »
          »
          21 month(s) ago, # ^ |
            Vote: I like it +21 Vote: I do not like it

          So you open an alter account for div3?

          → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    Sorry It's a typo that make me downvoted.

    I wrote E and F by mistake.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +2 Vote: I do not like it

Hardest Div3 ?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +64 Vote: I do not like it

I shouldn't work out problems from back to the front.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

oh,how to solve tast D and E?It's too difficult!

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Problem D:

    Since |T| <= 100, you can find all substrings.

    While finding the substrings, loop through each string s and see if it is equal to the current substring. If it is, store its index, start, and end in a vector. This takes O(N^3).

    Now you have a vector of segments. The problem then becomes: find the minimum number of segments needed to cover a line, in this case, T. You can do this greedily in O(N).

    For each segment left of the current position, find the one that goes farthest right. Keep doing this until you reach the end or it's impossible to solve this case.

    My submission

    Sorry if if the implementation is bad. I was rushing.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      i do the same but not think greedly how to take minimum of them Thanks for sharing your submission

      → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Nice approach. I didn't think about this. I use DP instead. :')

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Can u please explain your solution.

        → Reply
        • »
          »
          »
          »
          »
          21 month(s) ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Let $$$dp_i$$$ be the minimum required string to color string $$$t$$$ where letters at $$$t_i$$$, $$$t_{i+1}$$$, ..., $$$t_{|t|}$$$ are going to be colored (all the letters from $$$1, 2, ..., i-1$$$ must already been colored)

          Then I will try all possible $$$s_j (1 \le j \le n)$$$ that can fits position $$$i$$$ (in other words, I will try to color $$$t_i$$$, $$$t_{i+1}$$$, ... $$$t_{i+|s_j|-1}$$$ using the selected $$$s_j$$$). To check which $$$s_j$$$ match, I use a simple pattern matching logic

          Then for the transition for the next $$$i$$$ (call it $$$i'$$$), I will pick such starting index so that all the letters between $$$1$$$ to $$$i'$$$ already colored

          In other words the value of $$$i'$$$ lies between $$$i+1, i+2, ..., i+|s_j|$$$. I pick those who gives minimum value and construct the answer.

          Therefore : $$$dp_i = min(dp_{i'})$$$ for all possible $$$i'$$$

          You can look at my submission for more details

          → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Oh god, I was thinking that you have to choose the substring covering the highest amount of still black characters and struggled to find a bug in my code, thank you for the explanation.

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Problem E

    If we add the first digit a certain amount of times, we always get 2 as the last digit

    • 11 + 1 = 12
    • 12 + 2 = 14 + 4 = 18 + 8 = 26 + 6 = 32 (see this)
    • 13 + 3 = 16 + 6 = 2 2
    • 14 + 4 = 18 + 8 = 26 + 6 = 32
    • 16 + 6 = 22
    • 17 + 7 = 24 + 4 = 28 + 8 = 26 + 6 = 42
    • 18 + 8 = 26 + 6 = 32
    • 19 + 9 = 28 + 8 = 36 + 6 = 42

    The special cases are 0 and 5, because

    • 10 + 0 = 10 + 0 = 10...
    • 15 + 5 = 20 + 0 = 20...

    So if any number has a 0 or 5 in the last digit we must check if all numbers with 0 as last digit are equal, and if the last digit is 5, we must add 5 and check again that all are equal. If a number doesn't have 0 or 5 as last digit, we can't do it.

    If not we increase every number to have a 2 as last digit and check if all a[i]/10 (the number minus the last digit) have the same parity.

    The reason: if you look in the examples, secondly we have 12 to 32, and to go from a digit 2 to the next digit 2, we must add 20, if we ignore the last digit (as we do in a[i]/10) we must add 2.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      "If we add the first digit a certain amount of times, we always get 2 as the last digit"

      Actually, the digits 2, 4, 8, and 6 will all work for this.

      → Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

Please do a STRICT PLAG CHECK. According to me, D was quite tough , 2k submissions seems fishy

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Actually it isn't once you realize that it is a variation of merge intervals problem.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Interesting. I just brute-forced (kinda) b/c bounds were really small.

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's heavy implementation problem. Also A, B, C (and probably E) were really fast to implement once you got the right idea. So, in my opinion, that's why problem D got so many last second submissions (I've spent around 1 hour 30 minutes writing + debugging it and only got AC after the contest).

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +31 Vote: I do not like it

Why is C SAME as 1462C - Unique Number!?

Only difference is that in contest today x $$$<=$$$ 45 and not x $$$<=$$$ 50.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it -10 Vote: I do not like it

    It is the same problem. for x >45 the answer will be just -1

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +10 Vote: I do not like it

I think number of a div increased by 1 accidentally

→ Reply
»
21 month(s) ago, # |
← Rev. 4   Vote: I like it -15 Vote: I do not like it

Another Div 3 which I'm going to downvote. Authors think that they are preparing problems for Div 2.

D — good idea but shit ton of implementation.

E — couldn't solve but I think it's some casework which I absolutely hate, hence didn't bother.

Also, C = https://codeforces.com/contest/1462/problem/C. wtf?

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    E is just being able to observe a certain pattern that arises when you keep applying the operation on a number.

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    E is not about casework but observation

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    E was a great problem that taught you about cycles without having to go into any implementation

    → Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

isn't G just a simple dfs solution ??

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +7 Vote: I do not like it

I hate all those people who are expert or more and do not participate with their main account, But after all this was a round that I didn't like(just my opinion)

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

may i say this round is just a piece of shit ? div3 seriously?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve 1714E - Add Modulo 10?

This is my submission. I have feeling I am either very far away from solution or like one line away from solution.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    So basically there is a cycle of ones digits, and if u keep adding u will eventually enter 2, 4, 8, 6 etc, which is a cycle of size 20. You simulate for every number until it reaches the digit that you choose, 2, 4, 8, or 6, and the difference between all numbers must be divisible by 20, due to the cycle. The other case is 5 and 0, which, if it's a ones digit with 5, u add the 5 and see if all the values are equal, because the 0 can not change the number.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +6 Vote: I do not like it

Why is D so much implementation???

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

I have observed that number of submissions during last 10-20 minutes is too much high. Especially in this contest, I saw that submission rose too much.

Is this an indicator of cheating or are these really clutch submissions?

Also I don't understand how can problem E have more submissions than G. G was so straight forward, E atleast required some property of functional graph and heavy implementation.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Am I the only one who overkilled G with k-th ancestor and binary search stuff? I supposed there had to be an easier way but I didn't think about using lower_bound on a stack. Well, at least I solved it...

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +10 Vote: I do not like it

U THINK F INTERESTING ? I came up with it in 5 minutes and get WA on pt 2 all the time lol

→ Reply
»
21 month(s) ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

166589237

Can anyone tell why it is failing?

edit: nvm, got it

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    That's too much of implementation. You can solve it easily if you measure time only in minutes

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is my D giving MLE? 166596216 Will there be any more space efficient solution.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    "s.substr" will create a new string but not its address, that costs most of your memory limit

    → Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

did someone notice that problem A — 1714A - Everyone Loves to Sleep was difficult than normal A problems

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I'm stuck with this task

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    For sure, they could've swapped A and C in my opinion.

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It would be easy if you convert time entirely in minutes

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Tourist did that in first minute of contest

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Implementation hell. Took longer for me than most div 2 A's lol.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Solved ABCDE, got too tired to think about F or G lol.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

I wasted approx 45 minutes solving D and implementing it was a hell . but it gets TLE and from the submissions it looks E was easier than D.

Can anyone guide me how to improve my rating as i am stuck here since very long

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    E is implementation aswell, I don't know how did it got that many submissions, wait till hacking phase finishes then judge

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Are hacks rated?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Missed my AK~

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Why the same code got different results on c++14 and c++20? (https://codeforces.com/contest/1714/submission/166601208) (https://codeforces.com/contest/1714/submission/166600974)

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Solved B and C, got stuck at A !!!!

Any hints for A ? Stuck at test 3

My Submission

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    Convert times in minutes. Determine what happens if the current time is before/during/after each alarms and compute in minutes the difference of the current time and the alarm. Output the minimum difference in hours-minutes format.

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    It will be easier if you convert everything to minutes and then output $$$ans÷60$$$, $$$ans$$$ $$$mod$$$ $$$60$$$ .

    → Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Amazing trick for looping through array or string backwards (used in todays B):

for(i=n-1; ~i; i--)

If you use ~i it will return false for $$$i=-1$$$ and true for all other $$$i$$$.

→ Reply
»
21 month(s) ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Problem D = debugging infinite loops for an hour, at least thats what I felt and still couldnt find out whats causing it

Thats what happens when you nest while loops

and problem G, which I would have solved in 20-30 mins on LC using the standard recursive dfs approach, I didnt do it because I knew it wouldnt pass and idk how to implement that iteratively. Thats why my 2330 LC rating using pure python wouldnt matter here lol :/ same for two of my friends who used python only have 2400+ LC and 5 stars codechef but couldnt hit expert on cf

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Wait, am i on next level or smth? Why so many people talk about A being harder then usually? Is it really that hard to observe, that you need to sum this day time left and new day time if m < M? Or what is the difficult?

No offence to ones who didn't solve it, but i am really curious.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Usually A is one simple observation, and simple implementation. For this one you had to consider multiple cases and also figure out a way to quickly convert the hours into minutes and back again.

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Its just that, its a little bit confusing/harder to implement than B & C in todays contest.

    → Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

In today's contest, these two users have the exact same code for problems E and F (the slightly harder ones), Chahel and goelarnav12. Please take action, as this is against the spirit of individual competition and is blatant plagiarism.

In fact, this is goelarnav12 's second offence. This user has plagiarised in the past. You can look at the "skipped" submissions. Kindly take strict action against these people.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +3 Vote: I do not like it

So, I registered, participated. Couldn't solve a single question. Bahaaha, almost hysterical.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

Can anyone explain the dp method for problem D? Thanks in advance

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Could someone try hacking my D? (https://codeforces.com/contest/1714/submission/166556061) Feeling paranoid now since I'll finally get to expert with this round probably, but seeing a lot of TLEs.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    "I want to get expert. Can somebody please prevent me from getting expert?" lmao

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      All submissions will eventually be tested against hacks anyways, so I'd just like some confirmation so I can celebrate/grieve in peace.

      → Reply
»
21 month(s) ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

someone please help me in G my complexity is o(2*nlogn) but I get TLE submission : https://codeforces.com/contest/1714/submission/166597686 basically for every node , I increment the b as much as I can

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If your tree is a path then height is no lomger log(n) and your time complexity hits O(n^2).

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I tought it is still o(n) , like A traverses the tree and B comes afterwards , so there are a total of 2 traverses , what is wrong with that ?

      → Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    When you have a long path with short sum of b's for vertex v, which has a lot of children each requiring adding up all b's, yours becomes O(n^2).

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      freehandle I didnt get it , can you giva a case , I will run jt on my machine too see how many operations does it make , btw I want to point out that : if vertes V's ancestors b is k , then to start searching for V's b , we will start the search from k , so it wont start from the root for every node to find their b's

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Exactly what I wrote is the case. Start a path with a = 1, b = 10^8, make a long path of length n/2 with a = b = 1, then make n/2 children at the end with a = 10^9, b = 1. For each children, you have to add up n/2 b's.

        → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

What is the point of such tasks as C? The stupidest precalc (166535760) solves them.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is my submission for Problem D failing on test 5? 166614128

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +1 Vote: I do not like it 
    1
abababac
2
aba
bac

    Your solution outputs 4, but you can solve it in 3 iterations (1 1; 1 3; 2 6).

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

C using recursion 166615609

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It can also be done with bitmasks.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +2 Vote: I do not like it

In E Problem, I thought that if the last 2 digits are the same, those values can become the same after some finite operations, but it is giving WA. Can anybody help why my solution is wrong

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +3 Vote: I do not like it

Why am getting TLE in question B?

Trying to do it O(n) only :[

Solution

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    You are failing this test:

    $$$t = 10^4$$$

    $$$n = 1$$$ for each test

    And for every testcase you create $$$10^5$$$ array, that means you do $$$t \cdot 10^5 = 10^9$$$ operations

    → Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    There are a few optimizations for your code. Don't use vector<int> freq(200001, 0). You can use map<int, bool> freq or set<int> freq.

    You might be thinking: but I cannot use those data structures because freq[a[i]] can be $$$2$$$. You actually can if you first check if freq[a[i]] == 1 (in this case same as freq[a[i]]) and if it isn't then increase freq[a[i]] by $$$1$$$.

    This way time complexity is $$$O(NlogN)$$$ because grabbing elements in std::map takes $$$O(logN)$$$ and std::set.count() also takes $$$O(logN)$$$.

    Your accepted solution using std::map: 166667859.

    Your accepted solution using std::set: 166668032.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

I highly recommend watching neal's explanation for problem E if you don't know how to solve it.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi MikeMirzayanov! The two packages seem to be very similar.

https://codeforces.com/contest/1714/submission/166564267

https://codeforces.com/contest/1714/submission/166596662

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

Did anyone noticed that lots of the hacked solutions for D have almost the same code ?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone tell me what's wrong with the problem E? This is my code here.

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    When a[i]%10==2 for all i, f doesn't get set

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      If a[i]%10!=0 or 5, then all a[i] can be transformed into a[i]%10==2 by using the operations. In my code, f means if a[i]%10==2 exists, then if there is another a[i]%10==0 or 5, the answer should be no.

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        Yes, f is meant to indicate whether a[i]%10==2 exists after the transformation, but f is not being set when a[i]%10==2 for all i in the original array.

        → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain the solution for. problem E??

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    There is a small observation.Say the current number is x. If x%10 is 2, and you perform operations you get a series. 2->4->8->6->2 so after performing 5 operations from any of these numbers you get back to the number itself with a sum of 20.Now suppose x%10 is 1,3,7 or 9. again if you perform one operation you reach 2,6,4,8 respectively.So the numbers become a part of the series.Now the only numbers which cannot be a part of these series are 0,5.So there are three cases.

    1. The array has numbers of the form 0,5 and at least one number from rest of the numbers.In this case the answer is No.
    2. The array has numbers only of form 0,5. In this case for the numbers having x%10=5 do the operation once.Now all elements end with 0 in the array, so no operation can be don further.Now all elements in the array must be equal for an answer to exist.If all elements are equal answer is Yes.Else answer is No.
    3. Now if all elements are of the form 1,2,3,4,6,7,8, or 9.For numbers having 1,3,9 do the operation once.So now all numbers are of form 2,4,6,or 8 in the end.Now as mentioned we can add 20 to any of these numbers without any issue. So we add 20 to all numbers such that the difference between all numbers of 20 is within 20.Now we try to make all elements of this array equal to the maximum element of the array using operations on each of the element. suppose the maximum element of the array is 38, We would try to make all elements of the array equal to 38 46 52 or 54.If is it possible to make all elements equal to any of these elements answer is Yes else answer is No.Note :we are not checking for 58 as if we can make all elements equal to 38 we can also make all elements equal to 54.

    For implementation details check this out 166574350

    → Reply
»
21 month(s) ago, # |
  Vote: I like it +10 Vote: I do not like it

HAVE BEETHOVEN97 DONE MOST HACK IN CODEFORCES HISTORY?

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Hacking is not a very easy thing to do, it takes patience to look through everyone's code and attempt to hack it, and hacking so many solutions in every contest is pretty impressive. The man's a legend and i hope he doesn't stop.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

"Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you."
Does this mean I could get some points from this contest if my rating is less than 1600?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +1 Vote: I do not like it

Problem D can be treated as a graph problem and can be solved by greedy and DFS or similar. Just store the adjacent strings that satisfy the condition, making it a directed graph, do BFS, and backtracking. Though the implementation is awkward and the cases are trivial at least to me LOL. Checking out this solution if you want. 166652986

→ Reply
»
21 month(s) ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

When will ratings come?

p.s. hopefully i get pupil :3

edit: yay i am pupil now :)

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Why Rating not updated yet?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it +3 Vote: I do not like it

Hello everybody! Can anybody tell me why my E problem changed it’s green color to red? Yesterday during the round it passed all tests and got “full solution” status… I suppose that it’s connected with “hacks” but I am new here and still barely understand the matter of that word.

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Why am getting TLE in G link: https://codeforces.com/contest/1714/submission/166682000

the dfs is supposed to have TC of O(2*n)

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    This solution is not actually O(n) since the two pointer system goes back.

    Imagine if the current position was 50k and the pointer was at position 1,

    Now you enter node at depth 50001 and the value is large enough to allow the pointer to move from 1 to 50k, that needs 50k operations, now you get to the end and go back in dfs to 50000, the pointer is back at 1 and when you another node at depth 50001 with a large value the pointer needs to move back to 50000 again.

    Worst case performance is (n^2)/4 or O(n^2) giving TLE.

    Edit: If you take a look at the test case that fails you can notice that the pointer will be stoped at value 1 while the tree continues going down in a tower with small values. At some point the values will become large again and the pointer will need to move back down a lot of times.

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      But still the pointer will not visit any node twice right? Suppose we are at noode root then in all subtrees of root the pointer will only visit to the depth only once.

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        This is not correct, when going from root to 50k the pointer will visit all nodes, when the dfs goes back the pointer will return to 1 and can visit the same nodes again.

        → Reply
        • »
          »
          »
          »
          »
          21 month(s) ago, # ^ |
            Vote: I like it +1 Vote: I do not like it

          Ok i will dry run it on few tc Thanks for explanation

          → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Hey your explanation was great could you also review this ? https://codeforces.com/contest/1714/submission/166712292 thanks in advance

      → Reply
      • »
        »
        »
        »
        21 month(s) ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        You seem to have a similar issue, the while loop inside your function that moves the pointer behaves in the save way that caused the issue I described above.

        → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Does someone know what is test case 354 from test 2 at problem F?

→ Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

My submission for G was considered as plagiarism even though I don't even know the other person. Compiled and tested solution locally. Please look into this MikeMirzayanov

https://codeforces.com/contest/1714/submission/166555725 https://codeforces.com/contest/1714/submission/166575418

→ Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Attention!

Your solution 166560427 for the problem 1714D significantly coincides with solutions aldol_reaction/166560427, Cx330_L/166571248. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

MikeMirzayanov Hello, the main function part of my code comes from https://blog.csdn.net/mashizuren/article/details/113345347

Me and cx330_ L is not consistent. I use KMP and he uses bruteforce. In interval merging, we use the merging method of this article. You can see that my code style is always the same for all the topics in this field. Please restore my place and return my score. I didn't cheat.

my: https://codeforces.com/contest/1714/submission/166560427 his: https://codeforces.com/contest/1714/submission/166571248

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    anything happened? is the issue resolved , i too have the same problem and got all my code plagiarised

    → Reply
    • »
      »
      »
      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      He didn't plagiarize my code, we just refer to code from a same blog that was posted before. The problem is not solved.

      → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

My code is not working for a particular test case. Can anyone check! Thanks here's my solution: https://codeforces.com/contest/1714/submission/166584533

→ Reply
  • »
    »
    21 month(s) ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Your second last for loop misses the final element (the condition should be i < n). It gets AC after fixing this.

    → Reply
»
21 month(s) ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone look and tell why is my solution for G is giving RTE for testcase 3? 166727901

→ Reply
»
21 month(s) ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Edit 1 : Thanks, I figured it out...

Can anybody kindly help me with this https://codeforces.com/contest/1714/submission/166761762 , this is my code for G which getting memory limit exceeded on test 3, I think either the way I am making the graph is taking more memory or the vector for B_i's I update everytime in dfs is taking alot of memory but I can't figure out how to do the problem without using the vector for B_i's and if anybody can see the tutorial kindly send me the link cause I can't see it (maybe its not out yet...).

→ Reply