Привет! В 01.08.2022 17:35 (Московское время) начнётся Codeforces Round 811 (Div. 3) — очередной Codeforces раунд для третьего дивизиона. В этом раунде будет 6-8 задач, которые подобраны по сложности так, чтобы составить интересное соревнование для участников с рейтингами до 1600. Однако все желающие, чей рейтинг 1600 и выше могут зарегистрироваться на раунд вне конкурса.
Раунд пройдет по правилам образовательных раундов. Таким образом, во время раунда задачи будут тестироваться на предварительных тестах, а после раунда будет 12-ти часовая фаза открытых взломов. Мы постарались сделать приличные тесты — так же как и вы, мы будем расстроены, если у многих будут падать решения после окончания контеста.
Вам будет предложено 6-8 задач и 2 часа 15 минут на их решение.
Штраф за неверную попытку в этом раунде (и последующих Div. 3 раундах) будет равняться 10 минутам.
Напоминаем, что в таблицу официальных результатов попадут только достоверные участники третьего дивизиона. Как написано по ссылке — это вынужденная мера для борьбы с неспортивным поведением. Для квалификации в качестве достоверного участника третьего дивизиона надо:
- принять участие не менее чем в пяти рейтинговых раундах (и решить в каждом из них хотя бы одну задачу),
- не иметь в рейтинге точку 1900 или выше.
Независимо от того являетесь вы достоверными участниками третьего дивизиона или нет, если ваш рейтинг менее 1600, то раунд для вас будет рейтинговым.
Спасибо MikeMirzayanov за платформы, помощь с идеями для задач и координацией нашей работы. Задачи были придуманы и написаны командой Университета ИТМО: MikeMirzayanov, myav, Gol_D, Aris, senjougaharin, мной Vladosiya.
Также большое спасибо yorky, Jostic11, turmax, oversolver, ivanz, antonis.white, molney, KerakTelor, andrey.starodubtsev, Ahmad45123, myway, sofiaasta, Muhammad98 за тестирование раунда и весьма полезные замечания.
Всем удачи!
UPD: Разбор
Good Luck everyone! I hope everyone has fun!
I agree. I wish you all the best success!
Right. Let's hope that this competition will be interesting and everyone will achieve success.
Let's
I dont wanna look 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.
So you mean that her ex is jqdai0815, her father is Petr and "the guy she told you not to worry about" is tourist
This contest was in fact fun, I can confirm. There is one thing on my mind though. When will ratings change?
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
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.
Follow the " Pay Attention " section of this site. The registration link will be available there before the contest.
Codeforces Round #811 (Div. 3) Registration
Just go to the "CONTESTS" page (codeforces.com/contests) and you will easily find a place to register.
Use this link: https://codeforces.me/contests/1714
Good luck everyone!
Good Luck to everybody.
I am competing tourist in parrallel universe.
That's great man, wish you high rating
Finally a div.3 round!!❤(●'◡'●)
yes lads we gray guys can finally answer more than just the first 2-3 questions
For the guys who solved D. How did you reach the prefix sum observation? Had you solved similar problems before?
Good Luck EveryOne !!!
All The Best EveryOne !!
Hopeful this contest will be more friendly to newbies. Best of luck.
I hope it will be a great contest for everyone
I hope everyone has fun Good luck for everyone I hope it will be nice contest
I hope everyone has fun Good luck for all participants I hope it will be nice contest
Good Luck
Ollie gave
All my dreams are of an AK. (It only happened on a div 4)
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! ;)
I kid you not, this actually happened with me a week ago.
I am so weak that I can only get higher rating by div3 or div4 :(
Why so many downvotes???I can't understand.
Because you even got rating down on div3, like these downvotes.
Hope me and everyone good luck. I wish to AK some contest... I'm so weak that I can only solve 1 problem in Div.2...
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?
No, only the rating while registering is considered. If you want to participate as rated, you can unregister and register
One person is missing is Vovuh :p BTW, Thanks Vladosiya and team for the contest.
How to do well in Div. 3? What's your opinion if someone fails to solve all of the problems..?
then try solve div 4
Assuming you meant to fail to solve any problems, knowing arrays, sortings, and elementary maths is usually enough for first problems.
swap F G!!!
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.
I mean,F is more difficult to solve than G. not really swap them :)
finally I solved both of them :)
So you open an alter account for div3?
Sorry It's a typo that make me downvoted.
I wrote E and F by mistake.
Hardest Div3 ?
Yep A>B,C
It was the easiest I have ever written
I shouldn't work out problems from back to the front.
oh,how to solve tast D and E?It's too difficult!
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.
i do the same but not think greedly how to take minimum of them Thanks for sharing your submission
Nice approach. I didn't think about this. I use DP instead. :')
Can u please explain your solution.
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
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.
Problem E
If we add the first digit a certain amount of times, we always get 2 as the last digit
The special cases are 0 and 5, because
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.
"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.
True.
Please do a STRICT PLAG CHECK. According to me, D was quite tough , 2k submissions seems fishy
Actually it isn't once you realize that it is a variation of merge intervals problem.
Yeah Actually discussed with someone got to know that it is based on some LEETCODE prblm . But just said this in last 10-20 mins the submissions got tremendously inc.
if you know the leetcode problem name, could you please share?
Merge Intervals from leetcode
Thanks!
Interesting. I just brute-forced (kinda) b/c bounds were really small.
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).
Why is C SAME as 1462C - Unique Number!?
Only difference is that in contest today x $$$<=$$$ 45 and not x $$$<=$$$ 50.
It is the same problem. for x >45 the answer will be just -1
I think number of a div increased by 1 accidentally
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.me/contest/1462/problem/C. wtf?
E is just being able to observe a certain pattern that arises when you keep applying the operation on a number.
E is not about casework but observation
E was a great problem that taught you about cycles without having to go into any implementation
isn't G just a simple dfs solution ??
Yep 166625826
Also
Binary lifting + binary search
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)
may i say this round is just a piece of shit ? div3 seriously?
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.
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.
I did smth similar
Thank you.
Why is D so much implementation???
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.
I thought g was straightforward, which is why I started to work on it before E, but I couldn't implement a correct solution, so I went to E since it is mathy.
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...
For G you can keep a reference to the prefix sum vector in the dfs and preform a binary search on that.
https://codeforces.me/contest/1714/submission/166575758
That is what I meant I didn't realize during the contest. If you do it like that, you don't actually need to write your binary search, since you can use the C++ function lower_bound. See for example tourist's submission
no, you aren't :D i did the same thing
U THINK F INTERESTING ? I came up with it in 5 minutes and get WA on pt 2 all the time lol
I don't think this problem meaningful.
I agree with u. I think F has so many details.
I accepted it just now by seeing test cases... It is not suitable for a codeforces round because there are many ways easy to think,but most of them aren't easy to achieve.
Actually I just divide it into $$$4$$$ sub problems, it is pretty easy
which case did I miss :( This is such a torture not knowing the missing case https://codeforces.me/contest/1714/submission/166587100
Take a look at Ticket 15966 from CF Stress for a counter example.
it's one of the worst problems.
166589237
Can anyone tell why it is failing?
edit: nvm, got it
That's too much of implementation. You can solve it easily if you measure time only in minutes
Why is my D giving MLE? 166596216 Will there be any more space efficient solution.
"s.substr" will create a new string but not its address, that costs most of your memory limit
Thanks Learnt a new thing today !!
did someone notice that problem A — 1714A - Everyone Loves to Sleep was difficult than normal A problems
I'm stuck with this task
For sure, they could've swapped A and C in my opinion.
I've solved B and C before problem A
It would be easy if you convert time entirely in minutes
Tourist did that in first minute of contest
Implementation hell. Took longer for me than most div 2 A's lol.
Solved ABCDE, got too tired to think about F or G lol.
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
E is implementation aswell, I don't know how did it got that many submissions, wait till hacking phase finishes then judge
there was a very small error in my D which lead to infinite loop feeling very frustated as i could gain some rating but sadly not this time too happening last 4 rounds as i keep on loosing my rating and look like this time i will again become newbie
Are hacks rated?
No in div 3
Why the same code got different results on c++14 and c++20? (https://codeforces.me/contest/1714/submission/166601208) (https://codeforces.me/contest/1714/submission/166600974)
Solved B and C, got stuck at A !!!!
Any hints for A ? Stuck at test 3
My Submission
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.
It will be easier if you convert everything to minutes and then output $$$ans÷60$$$, $$$ans$$$ $$$mod$$$ $$$60$$$ .
Amazing trick for looping through array or string backwards (used in todays B):
If you use
~iit will return false for $$$i=-1$$$ and true for all other $$$i$$$.Is the purpose of A to scare people away from CP!??
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
same with me too but after seeing a testcase i figure out where it was happening
https://codeforces.me/contest/1714/submission/166613393
the recursive dfs approach with binary search to find the index will pass the time limit
You can get AC of any recursive solution in python just by increasing threadsize. Solution Of G in python using dfs approach. Just Wrap your whole code in a function and call it using threads.
welp it did. but not all the time. I have gotten wrong answer using threading before. I should not be too scared of fst (which made me try D first) and just bravely do it. lessons learnt. sigh. I tried to do iterative for a while (which I could not implement) and didn't even touch recursive
https://codeforces.me/contest/1714/submission/166657533
The memory limit is so tight omg, almost MLE
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.
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.
Its just that, its a little bit confusing/harder to implement than B & C in todays contest.
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.
submission : https://codeforces.me/contest/1714/submission/166593483 O(2*nlogn) gives TLE in the G!
my approach was , for every node get as much as you can for b we do a dfs , a goes from the front , and makes the path for the b and then for every node , while sumb < suma we increase the b , b can be updated at most n times , and so as a so the total time complexity is O(2*nlogn) (logn because I use maps for storing edge weights)
So, I registered, participated. Couldn't solve a single question. Bahaaha, almost hysterical.
Что такое "фаза открытых взломов"?
By the way, check out my problem C solution:
https://codeforces.me/contest/1714/submission/166538483
Можно взламывать решения других участников, т.е искать в них ошибки.
Но разве тут не на полном наборе тестов проверяется сразу? При отсылке пишет не "Претесты пройдены", а "Полное решение".
Завтра утром будет проверка на полном наборе тестов, а взломы тоже есть
Проверка идёт на полном наборе авторских тестов, но каждый удачный взлом по задаче добавляется в список тестов, и по завершении фазы посылки перетестируют с учётом взломов.
Can anyone explain the dp method for problem D? Thanks in advance
Neal wu did dp, hope he posts a video on this round
Could someone try hacking my D? (https://codeforces.me/contest/1714/submission/166556061) Feeling paranoid now since I'll finally get to expert with this round probably, but seeing a lot of TLEs.
"I want to get expert. Can somebody please prevent me from getting expert?" lmao
All submissions will eventually be tested against hacks anyways, so I'd just like some confirmation so I can celebrate/grieve in peace.
ok, well good luck
Thanks man
congrats, you've reached your goal! (me too :D)
Congrats man :). Feels weird seeing my handle in blue after so long haha
someone please help me in G my complexity is o(2*nlogn) but I get TLE submission : https://codeforces.me/contest/1714/submission/166597686 basically for every node , I increment the b as much as I can
If your tree is a path then height is no lomger log(n) and your time complexity hits O(n^2).
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 ?
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).
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
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.
oh thanks , I got it now , thanks so much !
What is the point of such tasks as C? The stupidest precalc (166535760) solves them.
I used binary search
https://codeforces.me/contest/1714/submission/166511626
I think the point is greedy from the back, which can be sometimes essential.
Why is my submission for Problem D failing on test 5? 166614128
Your solution outputs 4, but you can solve it in 3 iterations (1 1; 1 3; 2 6).
C using recursion 166615609
It can also be done with bitmasks.
How to solve F?
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
There is counter example: $$$6$$$ and $$$16$$$.
Oh yeah, got the error where i was failing, thanks a lot!!
Why am getting TLE in question B?
Trying to do it O(n) only :[
Solution
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
Aah I seee! thanks fren.
There are a few optimizations for your code. Don't use
vector<int> freq(200001, 0). You can usemap<int, bool> freqorset<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 iffreq[a[i]] == 1(in this case same asfreq[a[i]]) and if it isn't then increasefreq[a[i]]by $$$1$$$.This way time complexity is $$$O(NlogN)$$$ because grabbing elements in
std::maptakes $$$O(logN)$$$ andstd::set.count()also takes $$$O(logN)$$$.Your accepted solution using
std::map: 166667859.Your accepted solution using
std::set: 166668032.Or I can just declare a vector of size n+1 as i did in this solution solution
Yes. With map and set you use less memory, though.
I highly recommend watching neal's explanation for problem E if you don't know how to solve it.
Can you providd the link?
I provided. It's behind explanation for problem E.
https://youtu.be/pP_CBnmsdNM?t=2690
Hi MikeMirzayanov! The two packages seem to be very similar.
https://codeforces.me/contest/1714/submission/166564267
https://codeforces.me/contest/1714/submission/166596662
I think Codeforces system will detect this cheating automatically
It seems like I faced the same issue as some other people with F. Got Runtime Error on 2nd test case. I suspect it had to do with the case $$$n=3$$$ (from my own testing).
Happy to get E early though!
Did anyone noticed that lots of the hacked solutions for D have almost the same code ?
Can anyone tell me what's wrong with the problem E? This is my code here.
When a[i]%10==2 for all i, f doesn't get set
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.
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.
was D intervals problem ?
I solved by using intervals
Can anyone explain the solution for. problem E??
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.
For implementation details check this out 166574350
HAVE BEETHOVEN97 DONE MOST HACK IN CODEFORCES HISTORY?
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.
"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?
Why is my first submission of problem A wrong answer but passed the hack?
Sorry, I read it wrong.
Problem D can actually be treated as a graph problem using greedy and DFS or similar to solve. Just store the adjacent strings that satisfy the condition, do BFS, and backtracking. Though the implementation is awkward and the cases are trivial at least to me LOL,
166652986
Ну и зачем всё так усложнять?)
я вообще сканлайн написал...
у меня тоже нечто на это похожее но решение на столько костыльное, что боюсь оно не пройдёт сист. тесты
удачи! мое должно пройти, честный куб вроде как, и особо без костылей
только жаль что так долго идёт тестирование(
Это тестирование такое долгое? Раньше же вроде все было быстрее. С чем это может быть связано? Я просто только пару месяцев на этом сайте.
С большим количеством посылок наверное связано, ведь в div. 3 задачи решать легче, чем в div. 2...
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
When will ratings come?
p.s. hopefully i get pupil :3
edit: yay i am pupil now :)
Why Rating not updated yet?
And there is no editorial too
а че раунд как нерейтинговый отображается?
потому что только через пару часов все АС сабмиты протестируют на взломах, и где-то ближе к вечеру обновится рейтинг — так всегда с Div3, Div4 и Educational раундами :)
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.
Why am getting TLE in G link: https://codeforces.me/contest/1714/submission/166682000
the dfs is supposed to have TC of O(2*n)
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.
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.
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.
Ok i will dry run it on few tc Thanks for explanation
Hey your explanation was great could you also review this ? https://codeforces.me/contest/1714/submission/166712292 thanks in advance
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.
Does someone know what is test case 354 from test 2 at problem F?
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.me/contest/1714/submission/166555725 https://codeforces.me/contest/1714/submission/166575418
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.me/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.me/contest/1714/submission/166560427 his: https://codeforces.me/contest/1714/submission/166571248
anything happened? is the issue resolved , i too have the same problem and got all my code plagiarised
He didn't plagiarize my code, we just refer to code from a same blog that was posted before. The problem is not solved.
My code is not working for a particular test case. Can anyone check! Thanks here's my solution: https://codeforces.me/contest/1714/submission/166584533
Your second last for loop misses the final element (the condition should be i < n). It gets AC after fixing this.
Can anyone look and tell why is my solution for G is giving RTE for testcase 3? 166727901
Edit 1 : Thanks, I figured it out...
Can anybody kindly help me with this https://codeforces.me/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...).