Добрый день!
В 17.10.2021 14:05 (Московское время) состоится Технокубок 2022 - Отборочный Раунд 1 олимпиады для школьников Технокубок 2022. Раунд будет длиться два часа 15 минут, участникам будут предложены X задач. По его результатам лучшие участники (но не более 45% от общего числа участников раунда) будут приглашены на финальный этап. Для регистрации на раунд и участия перейдите по ссылке. Не забудьте заранее зарегистрироваться на раунд! Для опоздавших будет открыта дополнительная регистрация (с 13:15 до 16:20).
Зарегистрироваться на Отборочный Раунд 1 →
Соревнование открыто для всех в виде отдельного раунда.
Для всех участников обеих редакций этого соревнования будет пересчитан рейтинг.
Параллельно с Отборочным Раундом будет проведен открытый рейтинговый раунд для обоих дивизионов, в нем могут принять участие все желающие.
Напомним, что согласно правилам раундов Codeforces во время соревнования ваши решения будут тестироваться только на претестах (предварительном и неполном наборе тестов), а системное тестирование состоится после окончания раунда. Обратите внимание, что претесты не покрывают все возможные случаи входных данных, поэтому тщательно тестируйте свои программы! После прохождения претестов у вас будет возможность заблокировать решение, тем самым получив привилегию искать ошибки и взламывать чужие решения, но отказавшись от возможности перепослать ваше решение при каких-либо обстоятельствах (например, даже если вы найдете ошибку или вас взломают). Со временем задачи падают в стоимости. После системного тестирования учитываются только полные решения. Подробнее про правила соревнований можно прочитать по ссылкам:
Регистрация на олимпиаду Технокубок еще открыта. Победителей и призеров олимпиады ждут значительные квоты при поступлении в престижные технические вузы России и ценные призы! Если вы — школьник 8-11 классов и пока не зарегистрировались на Технокубок, то самое время сделать это:
Зарегистрироваться на олимпиаду →
После регистрации на олимпиаду не забудьте зарегистрироваться на Отборочный Раунд!
В финал соревнования будут приглашены лучшие участники каждого из отборочных раундов (но не более 45% от общего числа участников раунда).
Авторы отборочного раунда: Tlatoani, malachi_toney_goat, rabaiBomkarBittalBang, qlf9 и gotexans! Спасибо antontrygubO_o, isaf27 и KAN за координацию.
Спасибо всем, кто тестировал наш раунд: 244mhq, AmShZ, dorijanlendvaj, Keshi, czhang2718, 2020akadaver, Magikarp1, Richw818, quantum8, RayLee234, SlavicG, Qualified, Monogon, smax, wxhtzdy, kassutta, namanbansal013, HackerMonk, Ari, PurpleCrayon, FieryPhoenix, Jellyman102, bWayne, H4ckOm!
Для тех, кто впервые на Codeforces: в таблице ниже вы можете найти примеры решений на всех поддерживаемых языках:
| Группа языков | Языки программирования / компиляторы | Примеры |
|---|---|---|
| C | GNU C11 | 10903473, 17029870 |
| C++ | GNU C++14, GNU C++17, GNU C++20, MS VC++, etc. | 23794425, 5456501 |
| C# | Mono C#, MS C# | 3195513, 3794163 |
| D | D | 5482410, 2060057 |
| Go | Go | 7114082, 21366098 |
| Haskell | Haskell | 455333, 1668418 |
| Java | Java 8, Java 11 | 25491359, 23678167 |
| JavaScript | V8 | 35963909, 35681818 |
| JavaScript | Node JS | 38344430 |
| Kotlin | Kotlin 1.4, Kotlin 1.5 | 25779271, 25204556 |
| OCaml | OCaml | 6157159, 1281252 |
| Pascal | Delphi, FPC, PascalABC.NET | 1275798, 1259434 |
| Perl | Perl | 2519448, 1277556 |
| PHP | PHP | 413942, 35875300 |
| Python | Python 2, Python 3, PyPy2, PyPy3 | 35883730 (Py2), 36179112 (Py3) |
| Ruby | Ruby | 1837970, 1289551 |
| Rust | Rust | 25180002, 35652442 |
| Scala | Scala | 35847980, 2456025 |
Удачи!
UPD1:
Разбалловка Технокубка: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$1750$$$ — $$$2250$$$ — $$$2750$$$ — $$$3000$$$ — $$$3500$$$.
Разбалловка параллельного раунда: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$1750$$$ — $$$2250$$$ — $$$2750$$$ — $$$3000$$$ — $$$3500$$$ — $$$4000$$$.
see you next banner
Note the unusual timings :)
Added to the announcement :)
challenging problems.
9 challenging problems in just 2:15 hour? isn't that quite less?
hoping for a good contest :)
Hoping to be back to cyan. working hard to become expert.
Clash with Atcoder Beginner Contest :(
Damn 9 problems in 2:15 hrs
Can we participate without being from a Russian speaking country?
You can register for the contest based on this (#749)
May the force be with you. Good luck to you all :))
As a tester, good luck!
Почему-то некоторые из учеников говорят, что им не приходит письмо с подтверждением регистрации. Сталкивался ли кто-то ещё с такой проблемой?
Да. А теперь пришло, но не удается верифицировать аккаунт на codeforces. Пишет: Verification has not been completed. Please, try again to start the verification process.
MAy the OOds be ever in ur favor
Looking forward to solve problems and see a hike in my rating.
The only bad thing is the time of the contest. There is another contest in atcoder at the same time.
9 problems，so much!!!!
May be there will be some subtask problems.
Solve 9 problems within 2 hours and 15 minutes... The time limit is too tight!!!
Probably first few problems would be ez :) . Otherwise the time is very tight.
...
for LGMSs it's not so tight, for Div2 it's not big difference to solve first 3-4 problems in 2:15 or in 2:45 for example
I had only 10 minutes for the last three problems,so I just left and took a look at Atcoder Beginner Contest :(
great
Woah cheater has the guts to comment!
How are you cheater?
how he is cheater. I also need the third eye ;)
PS: Lol! he changed his comment from "Hope again we will solve at least 3,4 problems . best of luck everyone !" to "great" XD.
he/she adds random comments for avoiding plagiarism check.
Sad thing is, they have actually figured out how to circumvent plagiarism checks.
look at his solution for problems during the contest... there are more comments in code than the code itself
oh! got it. An expert cheater XD
His way of writing code is so terrible, he got rank 47 by cheating, another terrible news!
solvecopy pasteI have solved the problems like f, too.
.
I think I now know why "we" was used :|
May Mike be with us.
Me: How many coordinators reviewed your problems?
Authors: Yes
This contest clashes with AtCoder Beginner Contest 223 at 17:30 (UTC+5.5)
have you noticed these days less number of people participating compared to two months earlier.
maybe the college/school is starting... idk
daytime in China,finally dont't need to stay up late
hoping for positive delta and also:
1 манул)
Looks like companies wanted to see each other clashes
:( Clashes with The International final
Accroding to tradition,This Contest might be delayed:(
Contest begin, my province: shut down the electricity.
Tree problem in B? why...
note that m<n, which means that there is at least one vertex never occurs as $$$b_i$$$ among all restrictions
we can always find a such vertex and link all other vertices to it.
so it's like a greedy problem more than a graph/tree problem to me.
Div.1 + Div.2 = Div.1.5
floor((1+2)/2)Div2 rounds based on some Russian contest for school kids are always harder than usual Div2 rounds.
Imagine wasting 30+ mins on D because you forget to assert the values of $$$a_i$$$ in a query and print $$$a_i = n + 1$$$ in a certain case T_T.
But my mistakes aside, absolutely brilliant contest, one of the best I have seen in a while — Not a single boring problem from A-E! (can't talk about the rest since I haven't solved them, though F does look interesting as well)
E is a really really really cool problem — it looks insanely complex at first but decomposes to pretty easy ideas through some cool observations.
Nice C I got it but too late
What could be the testcase 3 for C?
Likely some mistake with not checking if cells which are $$$X$$$ could hypothetically escape if they weren't $$$X$$$. If that isn't the case you can't determine if it was initially $$$X$$$ or not. Forgetting to check that caused WA3 for me.
I looked for patterns like: _X \n X. where _ is any character, Here [2,2] can be both X or . , so this subgrid is indeterminable. Did I miss something?
You were supposed to look for the pattern
UPD: Oops, finding both patterns is the same thing, you are right
In problem C XX XX
query : can 1 2 is determinable ?
Do you mean XX \n XX? Then it's not determinable.
Y ?
Because, we can't determine the state of (2,2) cell exactly without looking at the grid. Suppose, instead of
We're given,
Then, still none are exitable, but one cell in the latter is empty. So, undeterminable!
Ok
Thanks for putting me out of my misery. I spent around 2 hrs struggling with this problem. Now that I know, no one deserves that -100 more than me, I am at peace. Thanks again.
A was much harder than B. Really disapointed about not reading problem B first
wtf, i insta-solved A and only managed to solve B with 2 minutes remaining
Same. Lool. I feel a little better now for struggling at B. Thank you.
B isn't that much tougher than A. I mean the statement literally highlights $$$m \lt n$$$ in bold which is the key idea to an easy enough pigeonhole idea.
oh wow! clearly i didn't notice the bolding
is this contest is for High-school ??
It's Russian highschool level which is roughly the same difficulty as a U.S. post-doctorate program or a Chinese gradeschool contest.
This isn't a joke, is it? :)
Yes, but also "It's funny because it's true"
Most probably not even a gradeschool contest Chinese contest because all the problems were well known tricks.
Problem E is such a problem, where you know that the first thing that comes to your mind has to be correct [path does not matter], otherwise it will be NP-hard or something.
What is wrong in this approach for Problem C
I check the minimum answer for each column with BFS and for each query I get the maximum for each column $$$C_l$$$ $$$C_(l+1)$$$ ... $$$C_r$$$, If it's less than L, print "YES", otherwise print "NO"
had the same idea, WA pretest 3 :(
I did something similar but got WA on test 3.
Maybe you did not handle if a blocked cell itself cannot exit the sub-graph if it was not blocked. In this case we will not be able to know if its N represents a blocked or an empty cell.
Can u explain why do I need to check that?
Are you checking if cells which are $$$X$$$ would be theoretically exitable if they weren't $$$X$$$ as well?
Suppose one of those cells is not exitable even if it wasn't $$$X$$$, then you wouldn't be able to differentiate between it being initially marked $$$X$$$ or not, so the grid isn't determinable.
I forgot about this and got WA3 because of that.
aah so you are saying that for n=2 m=2 XXXX query (1,2) should be NO instead of YES?
That's correct, because it could also be
XX
X.
Both situations are identical in the view of exitability, since all 4 cells are not exitable.
Well I understand now
I can't understand why I need to check If $$$X$$$ would be theoretically exitable if they weren't $$$X$$$.
I am checking if there is empty cell is not exitable, if there is atleast one then I can't differentiate.
Nice problems. Though it would be better (at least for me) to extend the duration of the contest... I just gave up when I finished reading problem G and found out that there was only 20 minutes left.
hint for B please
$$$m \lt n$$$ (strictly less)
I made edges a -- c for the given m restrictions. Then, to get the remaining edges I connected all separate components (checked using DSU) by an edge. Whats wrong with this? I kept getting WA2.
Connecting all edges a-c may result in a cycle even when m<n. Consider the following: 1 4 2 2 5 3 3 6 1
Consider this case,
4 3
1 2 3
3 1 2
2 3 1
In your final answer, the third constraint will not be satisfied.
Make b[i] leaves
It's given
1 < m < n.So there will be at least one node that can be between any two nodes, make it root. Now just connect all the other nodes to it.
My approach for B: I basically checked which node is not present as bi in the restrictions, then i used this node to create edge to every other node. but this gives WA on pretest 2 Can anyone explain where i am going wrong??
its correct its my solution too u probably implemented it wrong
Check if your bi is 0-indexed or 1-indexed. I had the same problem where I accidentally read in all bi's 1-indexed, and changing it to 0-indexed fixed it.
I also did same.. May be you have done some mistake in implementation
Nice problem set.
Anyone else getting RTE on pretest 25 on Problem E?
Update: Figured it out. Dumb mistake, I was trying to pick a starting point for the spanning tree by finding a vertex with a nonzero number of queries, but accidentally set the starting point to be the number of queries itself instead of the vertex id. In hindsight, could have just rooted the spanning tree at vertex 1.
https://mirror.codeforces.com/contest/1586/submission/132264873
can anyone help with finding bug in this solution for C?
found a bug, there was incorrect handling of invalid columns
here is the fixed submission if someone would struggle with similar bug https://mirror.codeforces.com/contest/1586/submission/132293465
Did someone see F before? ko_osaga?
F has very similar solution to COCI 2020-2021 Contest -4 Hop .Unfortunately I recalled that I had done this problem before just after the contest ended.
A classic result is that if a directed graph with chromatic number $$$\chi$$$ is edge-colored with $$$r$$$ colors and $$$k^r \lt \chi$$$, then there exists a monochromatic path of length $$$k$$$. Problem F is just the construction to show that this result is sharp.
Reference: https://www.tau.ac.il/~nogaa/PDFS/Publications/Monochromatic%20directed%20walks%20in%20arc%20colored%20directed%20graphs.pdf
The second sample kinda gives away the solution. They should have made it more obscure.
I mean if you already came up with something like k-nary numbers, then probably it might help. But I'm not sure you can deduce something without prior knowledge.
EDIT: looking at the editorial, it seems more clear what did you mean by that.
I conjectured that the solution will be something like this.
Btw why are you tagging me??
6 min between submits, also you are a legend :)
thanks for the compliment!!
I liked F very much, thanks for the round :)
I think problem F is more like a MO problem than a OI problem.
(Deleted)
The parallel AtCoder abc223 had better quality of programming problems.
well, the quality of the problemset does not depend on whether you can solve them or not.
Chinese problemsetting forces invading Russia lol
Problems are great, but I personally think the duration is a bit short, especially for data structure problems like H. Sadly I wasn't able to debug my code in time :(
more sadly, I succeeded in debugging it 10 minutes after the contest ended :(
Did anyone else get WA on test 6 for problem E?
Your DP part is probably incorrect.
Output should be 3.
I see thanks!
Is C a segment tree problem?
More like dp-problem
No, much simpler than that.
Hint: Consider under what situations will a single cell not be determinable.
It can be used (I used it) but not necessary
The subgrid is bad if there is
Nope, its dp combined with a sliding window like idea.
The core idea is that if a cell can't exit the grid due to being blocked by cells above / to the left of it, it would be non-exitable regardless of whether it was $$$X$$$ or not. So all cells in a subgrid with $$$1 \leq i \leq n$$$, $$$x_1 \leq j \leq x_2$$$ must not be blocked for the answer to be yes.
Basically let $$$minrow_{i, j}$$$ and $$$mincolumn_{i, j}$$$ be the smallest row and column reachable from a cell $$$(i, j)$$$.
Then the minimum $$$x_1$$$ for a given $$$x_2$$$ is $$$max(mincolumn_{i, j})$$$ where $$$1 \leq i \leq n$$$, $$$1 \leq j \leq x_2$$$ and $$$maxrow_{i, j} \gt 1$$$.
This can be calculated quickly for all $$$x_2$$$ with an easy observation — Since the function is max, $$$x_1$$$ never decreases as $$$x_2$$$ increases. So we can just iterate first on increasing $$$j$$$ then on $$$i$$$ to calculate the answer in $$$O(n \times m)$$$ time.
My solution is if there's an empty cell on column i, and it is blocked, then it is bad.
So I use dp to check the furthest left a cell could and store the max for all cell of that column, then use segment tree to check max(l,r) <= l
I aolved with seg tree tho idk it was the only soultion that came into my mind
Does anyone have a rigorous proof why arbitrary spanning tree works for E?
It suffices to prove that, for a given tree and $$$u,v,w \in V$$$, edges which appears in exactly one of the simple path $$$uv$$$ and $$$vw$$$ forms the simple path from $$$u$$$ to $$$w$$$. This should be verified easily.
First let's think of it as a dfs tree after getting a dfs tree.
what you can do if you have a range of edges of ones on this tree is that you remove this range and replace it with one edge with 1 which is a back edge this edge will also have a 1 and since this edge is mapped to a range of edges it's obvious that you can't make it into a 0.
I think it's the same with spanning trees the difference is the edges are not back edges but they are still mapped to one range of edges.
The existence of an example is equivalent to all nodes participating in an even number of paths as endpoints. => is trivial, <= can be shown easily to hold for any tree (consider any edge, if it is used an odd number of times, the sum of usages in one of the subtrees would be odd as well).
path from u to v = (path from u to root) xor (path from v to root)
What is the solution for D? Best idea I got was a randomized algorithm with expected number of queries to be exactly 2*n(which didn't pass as expected)
Suppose we can find $$$p_n$$$ in at most $$$n$$$ queries, then we can find all other values in at most $$$n$$$ queries by increasing $$$a_1, \ldots, a_{n - 1}$$$ or $$$a_n$$$ by the required amount for it to match each value $$$1 \leq k \leq n$$$, one at a time.
As for finding $$$p_n$$$, suppose I set $$$a_1, \ldots, a_{n - 1} = 1$$$ and $$$a_n = x$$$, then we will only get $$$0$$$ as a response if $$$p_n + x \gt n + 1$$$ (or $$$p_n + x \gt (n - 1) + 1$$$ if $$$p_n$$$ is $$$n$$$). So we can just iterate on all values of $$$x$$$ from $$$2$$$ upward and see the first value for which the condition fails to hold to exactly identify $$$p_n$$$.
Hint: Create an undirected graph with N vertices where edge u-v indicates abs(p[u]-p[v])=1.
If you figure this out it's simple. All that is left is running a dfs.
wait, what? XD I need to think about it
aaah, smart
For any given $$$p_i=v$$$, suppose there is some $$$j$$$ such that $$$p_j=v+1$$$.
- If $$$i \lt j$$$, this $$$(i,j)$$$ pair can be found by a query with $$$a_j=-1$$$ and all other $$$a_k=0$$$, which will return $$$i$$$.
- If $$$i \gt j$$$, this $$$(i,j)$$$ pair can be found by a query with $$$a_i=1$$$ and all other $$$a_k=0$$$, which will return $$$j$$$.
So doing $$$n$$$ queries with each $$$a_i=1$$$ and another $$$n$$$ queries with each $$$a_i=-1$$$ gives us the exact ordered pair of indices at which each pair of consecutive values $$$v$$$ and $$$v+1$$$ occurs, as well as the exact location of $$$1$$$ and $$$n$$$.
Technicality: each query needs to have all $$$a_i \gt 0$$$. Just add an offset to all query values.
You can also solve it with a tighter query limit using binary search:
The idea is if you find the last element, you can find every other element in $$$n$$$ queries. We can find the last element by performing a binary search to find $$$p_n - min(p_{n - 1}, p_{n - 2}, ...)$$$. If this value is negative, it means the last element is 1. So we only end up consuming $$$n + logn$$$ queries :)
Can someone point out my mistake on C? https://mirror.codeforces.com/contest/1586/submission/132256767
I think, you should delete this line:
EDIT: It still struggles with test 4, Idk what else is wrong.
Oh , but if it's a X, then we can include that column rt?
Like this,
In such a construction of two diagonally adjacent 'X' cells, the cell between them will always be non-exitable, regardless of what is written in it. Thus, we can change its state and get a different initial table with the same set of exitable cells, so we have to consider the mentioned case too and include that column.
Ah my bad, didn't understand the statement clearly.
Alex_Wei it's Unsuccessful hack for a reason :D
Its 99.99% purposeful unsuccessful hacks to get last place. I've done it in the past when I felt I was getting too worried about my rating to the extent where the nervousness was ruining contests for me.
Its trivial to copy paste the same input and change 1 value (can't use the same input twice on the same person) and make 20-30+ hacks / minute to quickly send yourself to last place.
liked D very much ... though was'nt able to implement my logic in time , but felt really good after solving it! :D thnx authors for such good problems!
132242048 C isn't a multitest problem but why I don't clear the array it makes my code FST ? After I clear the array I got accepted.132265702. Why ? When I change C++20 to C++17 in my FST code I have wrong in test 19 instead of 16. Why ?? :>
You are defining an array locally (inside of a function) rather than globally. The locally-defined array can contain anything at initialization.
ya. It will be a big experience for me. Thanks
the array you defined is not global, if you didn't clear it then it just contain garbage :D
I am able to solve only 2 problems in most of the div 2 contests. can anyone give me suggestions to become a specialist
Try to solve 3 problems.
Thanks. I will try.
Okay to be serious, here are some suggestions that may be helpful for you:
Try to have a good format of code, this is helpful because it allows you to debug easier and code faster.(This means indentations and spaces)
Try to think more before you code. Try to have an idea of what algorithm you will implement, and what are the details that may cause problems.
Solve more problems in general. To be able to solve 3 problems, you need to be fluent in the language you code and be able to think logically to reach a solution. The third problem often needs some thinking, it isn't as direct, so it may need more practice.
If you are sure that your algorithm is correct, and you fail to debug, try writing it over.
Hope this may be helpful for you.
Does it mean that the first two problems require no thinking at all? :)
Jokes aside, seeing how many people here (even high rating ones) completely blacked out on B, nope.
Sometimes those "easy" problems actually requires you to figure out a specific trick, you get the problem if you do and you don't get it if you don't. Which is exactly what B is.
okay.thank you so much
The contest is nice ,problem D,E,F are interesting . It is sad that I can solve G if 5 more minutes are given :(
It is sad that I can solve D if 5 more minutes are given :D
also 9 problems deserve way more than 2:15
issue solved
To not keep you waiting, the ratings are updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!
==
Рейтинги предварительно обновлены. Через некоторое время я удалю читеров и пересчитаю рейтинг снова!
This contest, I got rank 1328th (my previous rating is 1531) and my friend got rank 1446th (his pre rating is 1650) but my rating increased 75 and his rating increased 95. Can someone explain me this??? Many thanks
The rating calculation for new accounts is different. Maybe this is the reason.
I'm curious about the special judge for problem B.How to check an output is valid or not quickly(I can only thought about a data structure which is O(n log^2 n))?
if b is on path between a and c, than
dist(a,b)+dist(b,c)=dist(a,c)And?
this help's to check if the tree is valid
Finding distance is not any faster than simply checking nodes on the way to the LCA.
How is it relevant to the comment you replied to?
If there are $$$m$$$ restrictions, we can check the validity of output in $$$O((n+m)logn)$$$. this is faster than $$$O(nlog^2n)$$$
Can you elaborate?
first, we precalculate LCA(binary lifting) for each node & dist from root to each node. this takes $$$O(nlogn)$$$ time&memory. next, for each of $$$m$$$ restrictions like $$$a \space b \space c$$$ we check if $$$dist(a,b) + dist(b,c) = dist(a,c)$$$. we do this in $$$O(logn)$$$ time by finding the LCA for each pair of nodes. if so, the output is invalid.
I suspect that I read the C question incorrectly. Yesterday I wrote this question in a very complicated way, which always showed errors. Today I understood the ac code, but I entered what I was confused about yesterday, and something went wrong. https://codeforces.ml/contest/1586/submission/132311045 This is the code of my wa yesterday. https://codeforces.ml/contest/1586/submission/132313973 This is ac code. But I can't understand the ac code. 3 3 XXX XXX XXX one 1 3 NO
4 5 ..XXX ...X. ...X. ...X. one 4 4 YES I hope someone can help me.
Sorry, i was going to write that on commentary of 748 div.3, but i chose the wrong contest.
Hello please help me out. Just now I received a mail that my solution of problem 1586B of Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1) conicides with user 1928311. First of all there is no chance of my solution getting leaked to any one because i didnt shared it. Secondly, kindly see the submission time of his solution , I submitted that at 17:21 and he submitted at 17:45. How could I cheat then. Please help me regarding this.
How could I cheat thenyou shared your solution.i didnt shared itany proof?it can't be coincidence, the solutions are very similar.
Arey how could one prove that he/she didn't shared his solution. Its not possible na. Atleast I proved that I didn't cheat. Please help me. Else it will simply demotivate me from CP
have you used an online editor like ideone?
also for proving that you didn't cheat you need to prove that you didn't share any code, and that is almost impossible
No I used vscode. Wow got -113 without any fault :).I am impressed by your team and judgements for simply disqualifying a participant for not cheating :) Very good.
i am just a cf user not an admin or something else.
So whom should I ask for this problem??
no one.
i know that you cheated.
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this definitely can't be just coincidence.
even a function that wasn't needed was copy pasted.
You must be saying about the template.That is the xor function. I have a discord a server where I have shared my template with others. Many people use that
also for proving that you didn't cheat you need to prove that you didn't share any code, and that is almost impossible. Thats what i am saying na, you just say what kind of proof u need from me that i didn't shared the code. I am giving you.
There will be a person in charge in each round. You can go to 749 (div1+div2) comments to find out. One person said that he would launch a new ranking again within a few hours or days after the game. I think this person should be responsible for the ranking of the game.
Hope it can help you
just another indian cheater getting caught and still doesn't feel any guilt of it
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It's amazing that he's still not ashamed of himself even with his EXACTLY same code.
Yes that's because I didn't cheat. I was just explaining my solution to some of my friends in my hostel and one of them simply took the snap of it and wrote it as it is
that's called CHEATING
First of all there is no chance of my solution getting leaked to any one because i didnt shared it
and you lied :) Anything else you want to say
That's simply because I solved the qs myself n I didn't wanna loose my ratings. Nyways I am guilty for what I did and paid it with my ratings down
U better see the submission time and then decide who cheated