Добрый день!
В воскресенье, 20-го декабря в 18:05 по московскому времени состоится Отборочный Раунд 3 олимпиады для школьников Технокубок 2021. Раунд будет длиться два часа, участникам будут предложены 7 задач. По его результатам лучшие участники (но не более 45% от общего числа участников раунда) будут приглашены на финальный этап в Москву. Для регистрации на раунд и участия перейдите по ссылке. Не забудьте заранее зарегистрироваться на раунд! Для опоздавших будет открыта дополнительная регистрация (с 18:15 до 20:05).
Зарегистрироваться на Отборочный Раунд 3 →
Соревнование открыто для всех в виде отдельных раундов для первого и второго дивизионов.
Для всех участников всех трех редакций этого соревнования будет пересчитан рейтинг.
Параллельно с Отборочным Раундом будут проведены открытые рейтинговые раунды для обоих дивизионов, в них 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 — neckbotov и anta.baka. Cпасибо budalnik и KAN за координацию. Кроме того, хочу выразить благодарность тестерам, без помощи которых этот раунд не состоялся бы: isaf27, 300iq, Kaban-5, low_, Shinchan01, wiwitrifai, Ashishgup, spar5h, Nemo, Gauravvv, Vax, XLor, Um_nik, ADJA, jiufeng, RobeZH, Daryusz, antontrygubO_o, Normie28, kalki411, gratus907, manik.jain!
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 C, GNU C11 | 10903473, 17029870 |
C++ | GNU C++, GNU C++11, GNU C++14, GNU C++17, MS C++, 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 | 25491359, 23678167 |
JavaScript | V8 | 35963909, 35681818 |
Kotlin | Kotlin | 25779271, 25204556 |
OCaml | OCaml | 6157159, 1281252 |
Pascal | Delphi, FPC, Pascal.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!
В Отборочном Раунде Технокубка будет 7 задач, предварительные стоимости:
500 — 1000 — 1750 — 2000 — 2250 — 2750 — 3250.
В Div. 2 будет 6 задач, предварительные стоимости:
500 — 1000 — 1750 — 2000 — 2250 — 2750.
В Div. 1 будет 6 задач, предварительные стоимости:
750 — 1000 — 1250 — 1750 — 2250 — 3000.
Опубликован разбор.
Раунд завершен, поздравляем победителей!
Технокубок 2021 - Отборочный Раунд 3
Codeforces Round 692 (Div. 1, основан на Отборочном раунде 3 Технокубка 2021)
Codeforces Round 692 (Div. 2, основан на Отборочном раунде 3 Технокубка 2021)
))
Don't forget to notice the unusual start time!!!
QAQ
That's not KIND!
I think some registration cleanups need to be done after the rating update from the previous contest
https://codeforces.net/contestRegistrants/1465?order=BY_RATING_DESC
https://codeforces.net/contestRegistrants/1464?order=BY_RATING_ASC
also, please update the colors, MikeMirzayanov.
Thanks, I'll do it tomorrow before the round.
nice observation, never thought about this.
I am unable to type in the city name in the Technocup registration page(the input box isn't working ).Can anyone help?
As a person who is familiar with the anta.baka and neckbotov, I think that Russian-speaking high-school students are in good hands. I can assure you that everything will be all right with the contest, because anta.baka's passport is in my pocket.
Please return my passport. I couldn't buy beer for six months.
RIP passport...
How can I improve?
Practice. Lots of it.
learn and practice every day
I hope that I reach Pupil after this contest.
I think I will also reach pupil after this contest xd
Spread the greenery!
As a tester, I say that Problems are Interesting :)
Every tester says the same thing.
and every participant contradicts it, lol.
In letter from Mail.ru was said that contest will start 20.00 MSK, but here there is other time.
sad != said
Sorry, I don't know English very well :)
.
There are literally thousands of blogs related to your question (Google?) and why are you asking this here in the contest announcement blog? :/
P.S: registered 7 months and still new KEKW
надеюсь, отбор тк не короткий тур открытки, и я решу больше одной задачи(
Scoring distribution??
please anta.baka, neckbotov
Elimination Round will feature 7 problems, preliminary costs are 500 — 1000 — 1750 — 2000 — 2250 — 2750 — 3250.
Div. 2 will feature 6 problems, preliminary costs are 500 — 1000 — 1750 — 2000 — 2250 — 2750.
Div. 1 will feature 6 problems, preliminary costs are 750 — 1000 — 1250 — 1750 — 2250 — 3000.
I am gonna reach pupil.
good luck
thanks a lot
Is it just me, or does the scoring distribution of Division 2 seems a bit scary?
just you.
Well, now notice the difficulty gap between B and C. There's a reason why I mentioned that.
is it rated?
Div. 1 and Div.2 editions are open and rated for everyone. Register and enjoy the contests!
*Me after doing A and B: Let's check comment section
Same :((
This is why you are not improving. You should try C
Pretest 4 ruined the party for me... Looking forward to the solution to this one! :D
I have a doubt, for a problem if i did wrong submissions and didn't solve that question in the contest, will these wrong submissions have any effect on my ranking?
No.
What's test 5 in div2C (after contest)
Tell me test 4 (after contest)
try 6 5 1 2 2 3 3 1 5 6 6 5
Try this-6 5 4 1 3 2 2 3 5 4 4 5 Ans — 7
If you mean this
The input is not correct, Row 1 and 5 can not have rook at Col 4.
This is what i fixed to pass pretest 5 personally
Ans is 2. Isn't it?
Yeah. My problem was I was only considering movements in one direction, so this broke my code.
My code gives correct ans for this- 2 but still fails pretest 5.
I am getting 2 for this, what is the correct one?
thanks , seems that i have found whats wrong with my code
try 6 5 1 2 2 3 3 1 5 6 6 5
Getting 7
is ok..hmm.. I am really curios now about your mistake
https://codeforces.net/contest/1465/submission/101887567
I tried this, ans is m initially. For every rook (i, i). Decrement ans. For every cycles formed by other rooks, increment ans.
Idk why it isn't correct
i think your cycle finding is wrong. It's a directed graph, so you need at least three modes of visited (untouched, active, completed).
I'm also getting 7 but getting wa on pretest 5.
Can C be done using number of cycles count?
Yes, initialise ans = m, for every cycle increment ans, for every rook at (i, i) decrement ans
Implemented this and got WA on test 5. How?
Try this test case — 1 5 4 1 2 2 3 4 1 5 5
Correct answer is 3. Many incorrect codes will output 4.
Thanks! My code is giving WA on this. I understood my mistake.
Implementation mistake. How to implement correctly?
I wrote a recursive solution Code
I also done it using number of cycles count and same i also got WA in pretest 5 . Maybe you also done same mistake . During checking cycles , some cycles are not exactly cycles. ( means 1 2 3 but 3 to 1 not possible ) But in last i changed and it got accepted .
my answer is 3 and i'm still WA on test 5
Same for me. Was it wrong answer on the 45th output? Did you find what was the error?
i'm WA on the 213rd output
OMG okay. I figured out my issue though. Initialization problem.
Yep, I passed the pretest. Don't know about system test though.
Yeah, any rook on the diagonal contributes 0 to the answer.
Any rook not on the diagonal contributes 1 to the answer.
Additionally, any cycle contributes 1 to the answer.
This is optimal because moving a rook already on the diagonal off the diagonal is a waste of moves. Any rook not on the diagonal needs at least 1 move to get on the diagonal. And any cycle of $$$n$$$ pieces needs $$$n+1$$$ to resolve (you need to move a piece to a free column/row to remove the cycle, put $$$n-1$$$ pieces on the diagonal, then move back the original piece to the diagonal, for a total of $$$n+1$$$ moves)
Can you please share your code?. Mine is bit lengthy and unreadable.
Not sure if you can see it now but it should be visible once system tests finish (and I hope mine passes :P link)
could anyone help me with pretest 4 of div2C ?
Aah! This $$$D2C$$$ today:
how to solve c??
ignore all points on diagonals. for every other point, connect every 2 sharing x or y.
for each connected component, check if at least 1 point can be moved to diagonal in 1 move.
if so, ans += no of points in the component. otherwise, ans += no of points in component + 1
did the exact same thing , got wrong answer on pretest 5 . any case i may be missing ?
Some hack tests:
5 4
1 2
2 1
4 5
5 4
Then the answer is 6.
For each rooks $$$(x, y)$$$, add an edge between $$$x$$$ and $$$y$$$.
For each connect component, if it is a tree, the answer of this component is the number of rooks. And if it is a cycle, the answer of this component is 1 + the number of rooks.
this approach also giving wrong answer
Oh, maybe you forget considering $$$(x, x)$$$ ? And this rook is ok, and needs no move.
okk thanks
can u explain the intuation behind it? pllzzzzzzzzz
When I tested this round, I firstly guess the answer is $$$m$$$ or $$$m+1$$$ (remove rooks at $$$(x,x)$$$).
The difference between this two case is that the set of x coordinates and the set of y coordinates. You can see that if these two sets are same, then the answer is $$$m+1$$$, otherwise $$$m$$$. Like 2 rooks $$$(1,2)$$$ and $$$(2,1)$$$, the answer is 3 because $$${1,2}={1,2}$$$. If the first rook $$$(1, 2)$$$ wants to go to $$$(1, 1)$$$, but $$$(1, 1)$$$ is banned by $$$(2, 1)$$$. And if the first rook wants to go to $$$(2, 2)$$$, $$$(2,2)$$$ is also banned by $$$(2, 1)$$$. Is it like a cycle?
And 2 rooks $$$(1, 3)$$$ and $$$(2, 1)$$$, the answer is 2. The first rook goes to $$$(3, 3)$$$, and $$$(2, 1)$$$ goes to $$$(2, 2)$$$, this may be a chain.
However, the above conclusion does not work only when the board can be split into many parts, like the above hack test. And then add an edge between x coordinate and y coordinate may be a trick with DSU. There is a similar problem 1013D - Chemical table. Finally, if you do this, you can get the above conclusion.
Can u explain how you are taking an edge between the two cells of the board?
If you can explain with your above example it would be really helpful.
Solved. Thanks
How to do D2C ?
I think Div 1 Problem E can turn into : the probability of choosing some number from the sg numbers that =0. Am I right?
If I am right , how to compute? Using basis?
It turns into "compute probability that the resulting Grundy number is $$$x$$$, for each $$$x$$$. You can write a classic recurrent probability formula $$$p(x) = \frac{(x = 0)}{N+1} + \sum_{i=1}^N \frac{1}{N+1} p(x \oplus g_i)$$$ and turn it into Gaussian elimination since the Grundy numbers of vertices are $$$0 \le g_i \lt 512$$$ (approx. $$$\sqrt{M}$$$).
I'm so sad, during the contest I (wrongly) concluded that the grundy values can only be 0 and 1, therefore we have just 2 states in our linear system, and Gaussian elimination isn't needed Q_Q
Followup question, which might be stupid: is it true that under modulo M, if you have $$$a = P_a*Q_a^{-1}$$$ and $$$b = P_b*Q_b^{-1}$$$, and you take $$$c = a * b $$$ $$$=$$$ $$$P_c*Q_c^{-1}$$$, that the $$$P_c$$$ and $$$Q_c$$$ are automatically coprime and you never need to do any GCD-ing?
I think it is but I'm not 100% sure, thanks.
You should notice that modulo $$$M$$$, a fraction $$$p/q$$$ is the same as $$$pa/qa$$$ for any integer $$$a$$$ coprime with $$$M$$$. We don't care if the numerator and denominator are coprime with each other, only that the denominator is coprime with $$$M$$$ — which makes sense, they represent the same thing. You'd only care about the exact values when printing fractions or dealing with some operations like "add square of denominator to numerator".
Of course the Grundy values aren't just 0 or 1, take a semicomplete DAG.
Thanks! I got AC
So the time complexity of the Gaussian elimination part is $$$O(M\sqrt M)$$$?
Yes,it can pass in $$$O(M\sqrt M)$$$. Though I found this dp transformation in the contest,but I didn't realize the size of dp status is up to $$$\sqrt M$$$(Maybe I'm not familiar with sg numbers),and I focus in finding the probability of "Xor equal to zero".In fact I have never seen the problem which need Gaussian elimination to update dp. It's a good problem ,thanks!
Do you know those annoying implementation problems that have like 100 cases? Well, C is the perfect example of it. :))
C is a graph problem.
can you tell your approach
I think it can also be done using arrays (to find cycles).
Yep storing y coordinate for every x coordinate and then looping till you get the initial x or some x which has no y. If you get your initial x you found a loop otherwise you didn't. Keep marking the visited vertices.
How? :/
What's the solution for Div2C/Div1A ? I did it by counting the number of cycles formed in the graph when each rook position (x,y) is connected to (x,x) and (y,y), and then added it to the number of rooks not on the main diagonal.
I'm sure there must be an easy solution.
mine is also the same I don't know why it's getting the wrong answer
For each $$$(i,i)$$$, connect the rook in row $$$i$$$ with the rook in column $$$i$$$ with an edge. Discard self-loops. The resulting graph is a union of cycles and paths, the answer is number of vertices + number of cycles.
can explain the intuition behind it??
Take a path. You can move one rook from one end to the diagonal, shortening the path. For a cycle, you can't do that, so you need at least one extra operation with an unused row/column ($$$M < N$$$ so it's always possible), and using that operation turns this cycle into a path.
Worst Div2B ever. I thought I'd get TLE with simple brute force approach but then I looked at the number of submissions.
how does that make Div2B worse if lot of people solved it fast and you didn't
Because a lot of people (including me) solved it by looking at speed at which people were getting AC.
How to solve B?
Keep increasing n until you find the one satisfying the required property.
is this a joke?
Only after you get AC.
I need to look at the constraints more
There exists at least one valid answer every $$$9!$$$ which equals $$$362880$$$. To be more precise, LCM of $$$1, 2, 3, 4, 5, 6, 7, 8, 9$$$ is $$$2520$$$, so at least one valid answer exists every $$$2520$$$. That's why you can do a linear search.
I wish I had thought this. I wrote the correct ans and them keep worry for next 2 hrs hope system test won't break my solution.
Every 2520th number has this property, so just brute force for each number from N to the next multiple of 2520. You could have also run a simple while loop, but that gives no proof to why it won't TLE.
For anyone wondering how 2520, it is the lcm of 1 2 3 4 5 6 7 8 9.
standings are available to everyone so you can see and guess whatever you want.Though i proved myself .
"standings are available to everyone so you can see and guess whatever you want" ==> And when whatever = solution, that makes a problem worse.
I'm damn sure 90% of the people solved B without even thinking of its complexity ,more like they just guessed it and you dont't expect it from DIV2B level problem.
Btw were you 100% sure that your code will get AC and not get TLE verdict?
yes i was 100% sure . I calculate 9! and used similar to pigeonhole principle to prove it .
Well I was sure, but again who knows about such a strict time complexity.
I checked we would get an answer in max 2520 iteration since 5*7*8*9 would include all the numbers in it. But again instead of using 2520 I used 3000 which sadly made me cost -167 rating :(
Same bruh.... it was only after I saw the brute force implementations that I realized that only a maximum of 2520 numbers needed to be checked. :") I should've moved on to C. :"(
The lcm of numbers from 1to9 is 2520 so it does not take more than 2520 for each test case if one does by simply increasing n and checking its fair or not
I was worried as well, but i've tried generating several cases and noticed that the number of iterations would always be pretty small, hence a brute force would pass.
LCM of first 9 natural numbers is 2520. So brute force will work.
Solved A, B , D in last 10 minutes it was thrilling experience.
Found Div2D to be not that difficult. Don't know why so fewer solutions.
Because Div-2 C seems more interesting.
I started 1hr late, solved A and B in <20mins and then lost hope after C and then went to play chess. Does C require some advanced data structure or algorithm?
I don't think it requires some advanced shit. I think it's just something simple that we did not quite figure out.
why do you use 1e9+7 in D and 998... in E? congratulations, you have trolled me. I hope your problemsetting career has ended today.
Congratulations for not copypasting modulos.
There should have been an example testcase for D which checks for the correct modulo.
Some testers noticed that but we were short on time and I didn't want to change anything. Definitely a thing to consider when problemsetting.
You can't really give big tests in samples.
Actually you can, as a downloadable file.
i was low on time after coding D so i forgot to recheck the modulo...
You're not supposed to "recheck". You're supposed to copy-paste. If you did that, you'd have more time too.
I should've used Zashikhin prime. I'm deeply sorry for your disappointment which I believe is immeasurable.
When will be system tests?
Why templates are evil Karma lol
unfortunately, this is another case because I haven't used any templates
Do you also ignore other parts of statements when solve? I wouldn't write it in this way if you weren't so aggressive. You should try to be more polite
Tell it to Errichto (comment)
And if I politely asked, "why do you use 1e9+7 in D and 998... in E?", would you apologize?
You impolitely hoped that the authors' problem setting career has ended today. I agree that it would be better to use the same modulo in both problems. I could apologize before someone else who is not so toxic
Ok thank you for the answer. Sorry for being impolite.
And I'm sorry for such inconvenience. Next time I'll pay more attention to it
No worries, man. Just funny to see you complaining about modulos.
Umm...It's a woman
i'm a man
My solution to D1C passed the 21 pretests but I found a NullPointerException hack right after contest ended :'(
Second day in a row with E<D<C XD
I found C pretty easy. The last 2 signs obviously have to be + and -, but then for any string, we can always proceed from the last operation and "merge" a sign into the next sign if they're different, so that lets us turn any sequence of — to one -, any sequence of + (except the last, which is fine here) to +, and when we're left with +-+-+, that can be reduced to + as well. The rest is some careful implementation.
Can anyone prove why brute force works in B? I calculated such numbers upto 10, 000 only to find a pattern!
LCM of all digits from 1-9 is 2520. So max 2520 iterations
Numbers that divide all possible digits occur at most every 9! numbers, which is 300k
In DIV2 C did we have to count the no. of cycles?
Hi! This context was not designed like the previous contexts because until the second question the questions were designed at the ideal level, but from the third question onwards the level of questions was very high and this was really a big weakness !!
In DIV2 C did we have to count the no. of cycles?
Yes https://codeforces.net/blog/entry/85746?#comment-735295
Okkk, Thanks
Hi . This contest was not designed like the previous contests because until the second question the questions were designed at the ideal level, but from the third question onwards the level of questions was very high and this was really a big weakness
I read there are solutions to problem E that don't utilize the fact that the nimber is at most $$$\sqrt{M}$$$, and works in $$$O(N\log{N})$$$. How do those solutions work?
Suppose that all Grundy numbers are in the range $$$[0, 2^d)$$$. Then, let $$$c_i$$$ be the number of vertices with Grundy number $$$i$$$, by definition, all $$$c_i$$$'s are nonnegative integer and their sum is $$$n$$$. Let $$$p_{k, i}$$$ be the probability that after $$$k$$$ steps of the first type the current Grundy number is $$$i$$$ (including the fact that $$$k$$$ steps may not even happen, so $$$p_{k, i}$$$ go to zero as $$$k$$$ increases). Then, we have a very simple formula: $$$p_{0, 0} = 1$$$, $$$p_{0, i} = 0$$$ for $$$i \neq 0$$$ and $$$p_{k + 1, i} = \sum\limits_{j=0}^{2^d - 1} \dfrac{p_{k, i} c_{i \oplus j}}{n + 1}$$$.
In other words, $$$p_{k + 1} = (p_k \cdot c) / (n + 1)$$$, where by $$$\cdot$$$ I mean the Hadamard product of vectors. Because $$$p_0 = 1$$$, we get $$$p_k = c^k / (n + 1)^k$$$ (again, $$$c^k$$$ here means the $$$k$$$-th Hadamard power of $$$c$$$). Hence, the probability to end with Grundy number $$$0$$$ is the $$$0$$$-th coefficient of $$$\dfrac{1}{n+1} \sum\limits_{k=0}^{+\infty} \dfrac{c^k}{(n+1)^k}$$$. Indeed, to finish, we need to make $$$k$$$ steps of the first type (summands on the right) and then a step of the second type (multiplier on the left) for some $$$k$$$.
Now, we can find the Hadamard transform $$$H(c)$$$ of $$$c$$$. Because Hadamard transform is linear
($$$H(\alpha f + \beta g) = \alpha H(f) + \beta H(g)$$$, if $$$\alpha$$$ and $$$\beta$$$ are numbers) and multiplicative ($$$H(fg) = H(f) H(g)$$$), we can get the coefficients of $$$H \left (\dfrac{1}{n+1} \sum\limits_{k=0}^{+\infty} \dfrac{c^k}{(n+1)^k} \right)$$$ just by replacing each coefficient $$$x$$$ of $$$H(c)$$$ by $$$\dfrac{1}{n+1} \sum\limits_{k=0}^{+\infty} \dfrac{x^k}{(n+1)^k} = \dfrac{1}{n+1} \cdot \dfrac{1}{1 - x/(n+1)}$$$.
Finally, just apply inverse Hadamard transform to $$$H \left (\dfrac{1}{n+1} \sum\limits_{k=0}^{+\infty} \dfrac{c^k}{(n+1)^k} \right)$$$ that we just found. The probability of Bob's win is the $$$0$$$-th coefficent of the result.
There are no issues with division by $$$0$$$ here: because $$$c_i$$$'s are nonnegative integers with sum $$$n$$$, the coefficients of $$$H(c)$$$ are integers from the range $$$[-n, +n]$$$. Hence, $$$1 - x/(n+1)$$$ can't be $$$0$$$ modulo $$$998244353$$$, because otherwise $$$n + 1$$$ and $$$x$$$ are equal modulo $$$998244353$$$, meaning that $$$998244353 \leqslant 2n + 1$$$. In fact, this solution proves that there are no bad tests (tests, where the denominator of the answer is divisible by $$$998244353$$$): there are no bad tests exactly because there are no issues with division by $$$0$$$ in this solution.
This solution works in $$$O(n + d \cdot 2^d)$$$, if the input is just $$$n$$$ numbers from the range $$$[0, 2^d)$$$.
It's not an improvement on complexity but you can in fact save yourself the inverse transform, since the sum of all coefficients in the forward transform is $$$2^d$$$ times the $$$0$$$-th coefficient in the initial vector (every other coefficient cancels out).
This is easier. Consider the string as sequence of numbers
2^(s[i]-'a')
. Last two numbers always will wid fixed signs:s[n-1]
is positive ands[n-2]
is negative. Other numbers can be with any signs. There are2^(n-2)
ways to distribute signs, but it's possible to find the desired distribution by the binary search principle. Sort the array ofn-2
numbers by descending, then iterate over it. Initially all signs are positive. For each number find out, whether you have to change the sign of this number. This is necessary only if and only if the sum of alln
numbers after changing remains greater or equal tot
. If after iterating this sum is equal tot
answer isYes
Why don't you like my approach???
You are answering for Div2E whereas the above discussion is for Div1E.
This problems have same letters, I didn’t understand. But 9 contribution points cannot be returned
contribution points are also absolutely irrelevant
It's still a shame that the minus dirts my profile. It's not beautiful!
In past I thought that contribution is how much money are donated by him
(I still bad know English, I'm Russian. Fixed me if.)
I was very close to this solution during the contest but I didn't know about the linearity property of Hadamard Transform, which is why I couldn't proceed further. I have read a few articles on FWHT(including serbanalogy's blog) but never saw this property anywhere, can you suggest some source where I can learn more Hadamard Transform and its properties?
Thanks a lot.
You participated in contest ? I can't see the contest in your contest section .
I hate D1D...
Can anyone help explain the idea of D2D? Thanks!
In today's Div2 ~ C problem: // after contest
My code's link : https://codeforces.net/contest/1465/submission/101896978
As far as I think my complexity is O(3*m) => O(n) But I got TLE can anyone explain me why it got TLE
You can use heading little shorter .I thought you were giving solution .
It's because inner for loop can run 'n' times in worst case thus m*n complexity .
xD
It's more about Div2: D
Am I the only one solve D2C without modeling it as graph and write very long and stupid code 101886029?
For Problem D (div.2) Greedy Solution WA test case 9: Dude wtf was with that D problem? I figured I would start off the round by solving D first since it would be 6 problems D should have been solvable. Anyway I read it and I figured if I went greedy it would work since for any test case that I could come up with greedy solution worked. When I say greedy what I did was from left to right when meeting a '?' I know the '1's and '0's before (two counters) and after (this can be done with a sum array from right to left) so I can figure which one is better to put. Can anyone please give me a test case where this doesn't work? I was getting WA in test 9 with greedy solution. Also for the cases where I would get the same cost I went recursively and tried both solutions to see which would give the best answer. Again I did not get TLE but a WA in test case 9. So please somebody give me a test case now that the contest is over. Also let's say greedy doesn't work, what kind of a DP solution would work, like I just need a complexity? Though I don't think a DP solution would work either. This problem must have the need of some really sophisticated data-structure/algorithm combination like Treaps with matrix exponentiation or some. But if you are gonna say 6 problems then make D at least solvable. Look at how many people solved C and how many solved D it's insane. Oh I can see why now... it's because they mostly used orange and red testers. Dude like if you are orange you know every algorithm known to man and every technique you have it already memorized. Competitive programming started ageing like chess smh.
For Problem D (div.2) Greedy Solution WA test case 9: Dude wtf was with that D problem? I figured I would start off the round by solving D first since it would be 6 problems D should have been solvable. Anyway I read it and I figured if I went greedy it would work since for any test case that I could come up with greedy solution worked. When I say greedy what I did was from left to right when meeting a '?' I know the '1's and '0's before (two counters) and after (this can be done with a sum array from right to left) so I can figure which one is better to put. Can anyone please give me a test case where this doesn't work? I was getting WA in test 9 with greedy solution. Also for the cases where I would get the same cost I went recursively and tried both solutions to see which would give the best answer. Again I did not get TLE but a WA in test case 9. So please somebody give me a test case now that the contest is over. Also let's say greedy doesn't work, what kind of a DP solution would work, like I just need a complexity? Though I don't think a DP solution would work either. This problem must have the need of some really sophisticated data-structure/algorithm combination like Treaps with matrix exponentiation or some. But if you are gonna say 6 problems then make D at least solvable. Look at how many people solved C and how many solved D it's insane. Oh I can see why now... it's because they mostly used orange and red testers. Dude like if you are orange you know every algorithm known to man and every technique you have it already memorized. Competitive programming started ageing like chess smh.
If
x>y
, all?
will filled by one of sequences like11...100...0
, from000..00
to111..11
. In the other case, iterate over all sequences like00...011...1
my code for div2 C is ans=m, decrease if x==y, increase if a cycle is found. my solution solves all test cases suggested in the comment section. and still WA on test case 5. help pls :) Edit: found my mistake, got AC.
Can anyone explain div2D .Thanks
Consider all
?
positions. Ifx>y
, in the optimally case all the positions with?
filled by sequence like11...100...0
. Iterate over all possible sequences beginning a sequence with all zeroes. Next sequence is100..0
then1100..0
and etc. until sequence with all ones.In order to maintain the answer for each sequence, precalc for each position from 0 to n, number of
0
and number of1
to the left and to the right of this position. While iterating the sequences you recalc the numbers of0
and1
to the left
after each position with?
If
x<=y
, iterate over all sequences from the other side:0..000
,0..001
,0..011
, ...,1..111
and recalc the numbers of0
and1
to the right
after each position with?
why some people are getting TLE in B ?
I saw some of them use unordered map, i believe it has more constant factor(not sure), therefore failing the 2 sec time limit.
That must have made them MAD since there are less submissions in C...
Time Limit is 2 seconds in B
Thanks! Updated.
Why was there so weak pretests in B?
If you mean that actually brute force isn't acceptable, then I will tell you that it can be proved it's acceptable
pretests are not weak, the time limit is too strict :(
1000 TLEs for problem B :))
I think you should make the test set to more simple and rejudge. 1000 points to such is rationable.
A strategy to divide n by each digit just once, but adding a small log(n) factor gives TLE. come on!!
It was more than 1000 :(
And I am one of those people. This contest ruined my sleep. XD. Gonna practice some questions.
I got TLE in Div2 B. I used the obvious iterative approach i.e. keep incrementing value until a value is encountered which satisfies the condition.
The only possible reason for TLE I reckon is I used maps to store frequency ( occurrence ) of each digit.
Why such strict time-limit? :| This is really frustrating, even if time limit was INTENDED to be strict, why were the pretests weak? Why so much intricate optimizations were expected in div2B? 101865684 P.S. I am losing 160 points today!
Yeah that's the reason. I used set and got TLE too :/
Yes. anyone who has added a log factor has a TLE. I dont see the point of such tight limits.
I also think it is not reasonalbe to fail. Because we only want to store 9 digits in unordered_map, there is even not much collision for hashing. That strict time limit is meaningless.
Agreed. anta.baka I am not being rude but I'd genuinely like to know, from experienced coders and problem setters about their views on the FSTs of problem B.
I personally didn't expect FSTs. T * LCM * log sounds good in theory.
Sorry it is T*LCM*18*LOG SO 1000*2520*18. Anything extra to that will be costly. 18 can not be ignored as it is too close to TL
Python submissions are getting TLE.
i think problem B is such a bad problem for a contest.
Please Check problem B , it is bruteforce i agree but looking at c++ submissions they are getting accepted with fast i/o. Java solutions should be given consideration in such problems.
I was supposed to get +73, but after system tests for B failed it's -13. Seriously, guys, strong pretests please. 1000 people failed B system tests.
I was supposed to get +73 (and get my all-time high), but after system tests for B failed -23. Such a pity that 1000 other people failed system tests. A disgrace.
Hey, how were you able to determine these +73 and -23 numbers. Can you please tell where I can find them.
Here
Thank you
download the extension on chrome of Codeforces predictor or use codeforces live bot . BTW Todays contest was horrible for me , I solved C just after 2 minutes of contest ending and thus it is counted as upsolved (which is not wrong), but after this I Got a TLE in B as well , Due to a small log factor , log factor of 10 :(
Worst Problem B I ever came across. Problem B is not equally good for all the available programming languages
Agreed 100%. And I've seen quite a few Bs in my life.
NO DIV2 B was simple simplehere is my solution
No one said it’s hard. I just said it’s bad because Python solutions couldn’t pass and pretests were weak.
Interesting to see that all but one Python 3 solutions TLEd in system testing for Div 2 B.
The one that passed test 11 (system), used an LRU cache. Test 11 was all the same number. 101880982
That is not acceptable, problem-writers.
optimizeForce not codeforce
There should be a re-evaluation for B anta.baka neckbotov
Yes, Python solutions with the same logic as C++ solutions should pass.
Come on. There are a lots of problems that if you want to get AC with python, time limit should be 30 second.
People with same logic for B, someone getting TLE, someone getting AC. I've seen adhocforces, mathforces, but now optimizeforces ?
True, also almost all solutions in python failed B! B needs re-evaluation!!!
I've seen such solutions in Python itself, where both have same logic, but one is giving TLE and the other is giving AC. The only difference was:- one was using list, one was using string.
But it is not optimization . How can you know your code will get FST after getting pretest passed in 125 ms where limit is 2000 ms . Why will u optimise ur solution in this case?
Yes you are right, I would say that the Testcases were not appropriate.
For example see this two submuissions. AC TLE
After having 20+ testers, it is very sad to see there are 1000+ system test fails on problem B (mostly for python3). It is advisable to keep at least one python coder as your tester
Why do I get TLE on main test cases for problem B? What the hell man! I don't think this was supposed to happen
Edit: my solution is cpp !!
Worst round ever! Make it unrated
[deleted]
With this problem B, I came from cyan to almost gray in one week and the next contest is in 10 days :(
start a 9-day vigorous practice session, hope you'll bring out the fire in good bye 2020!
My rating prediction just went from +100 to -100 only because of storing digits xD
mine from +30 to -30 :( for the same.
I can't understand people who use python for competitive programming. If you use python for competitive programming, so you won't become good in it.
Ever heard of pajengod bruh?
no
Most people do competitive programming to learn a language and to get better at data structures and algorithms. I use Python because I can, what is the meaning of "If you use python for competitive programming, so you won't become good in it"? What does a language have to do with algorithms and logic?
got tle because of storing digits...
same I used set to store the digits and TLED on test case 11 but pretest were not much strong.
This Contest went Tough for me but Contests like these are real tester of our skills like patience , how good we are in making comeback after slow start ,implementation . Overall it was a great contest and we will surely gain a lot of things (keeping aside ratings ) for long run
Why Ternary search gave AC for problem D div2/B div1.
Why Div 2 B rejudge on improved time limits would be fair.
A large number of submissions times out on system test (excl. C++) which are of optimal time complexities.
Had it been a case of TLE on pre-tests, (or even close to time limit), one have the opportunity to re-implement in faster languages. (I saw 93ms and thought it'd be fine)
Languages like Python is not an odd choice for the problem, as it particularly helps iterating over digits without divide and modulo over 10s.
I totally agree with you I was a bit sceptical at first but when I saw that it passed all the pretests so efficiently I didn't think it needed any optimisation at all. Suppose I'd have got something above maybe 1 seconds I would've tried to optimise it or would've thought about my approach again. But those silly pretests kept us clueless. In such cases why to even bother giving the pretests. Just don't tell us anything and let us match the answers just from the example given in the problem T-T
Can someone tell why i am getting wrong answer in test case 3 in system testing for div2 B my code :- 101899785
I think it's better if you explain your reasoning/logic as well.
I have counted individually digits for 2,4,8 --> max of three for 6 ---> 6 for 3,9 ---> max of two for 5 or 7 ---> 5 or 7
then checked if it exits and divides the no. , if it divides then break else increment
Your code is a bit hard to follow. Try not to name variables as 'ttt','se' etc. Why do you check if each digit of 'n'(given number) is divisible by every digit?
I hope, you will not do unfair with us by making this rated. Atleast there should be increase of TL and a rejudge.
DIV2 B was just bruteforce ,and if they just want bruteforce to pass then why so strict time limit. Using map gives TLE while withot map AC. I know , Its my mistake that I should have taken the complexity of map implementation into account, but Whats the point of making those submissions System Fail. Anyways, Got to know today, "Use map very Carefully"!
Strict time limit and poor pretests , this is unfair ... and you have titled div2B as "Fair Number"
Pretty good contest. First two questions boosted confidence. Rest questions were really good. Good job setters :)
Good contest when 1000 people failed system tests for B? Think again.
Maybe we should not blame the author, because that test case also not in pretest. They also don't expect that. But hope they can give a reasonable response
1) I never blamed the authors in my response? 2) I never said this was a bad contest in my response; I simply said it’s undeserving of being called “good” when >1000 participants failed system tests. Only one Python code passed system tests. 3) In Div1E and Div2B submitting the exact same code again sometimes give AC. The time limit was far too tight.
Atleast ,they could have designed a single pretest to check that time limit over submissions during the contest rather than making more than 1k submission fail in just a bruteforce problem!
Or maybe a pretest that takes more execution time like more than 1 second. So that people could get a hint that it might fail.
Yeah thats the points I was trying to make. Thanks For understanding.
It isn't that much about good or bad contest. I solved 3 questions and found all of them to be quite interesting. But the problem here is that pretests should be set more carefully. What is even the use of pretests then if isn't able to provide any useful feedback. Like maybe not fail such code in pretests but at least have large execution time (under the constraints). Although this was still a lesson for me :)
Can you tell me what's the point of using a map and how problem b is interesting?
What I found interesting is how the concept of LCM got used here to know that what is the max increment under which we will surely find our answer.
my exact same code for div2B gave Tle on system testing but now it is accepted why?
Dude your EXACTLY SAME CODE is not EXACT. Just see my recent submission I copied same code of yours and it's showing TLE. So, don't lie at least My submission(Same code as yours) . Your code
just see the solution which was accepted for 1996ms(this one is exactly same) not the one of 1560ms. I just tried it and it's accepted again. Edit : also you tried using c++17, try it by c++14 it will get accepted
It's really sad man but even your AC submission passed by 4 ms. lol
In Div1 E, I passed pretest on contest, but I failed the same pretest when systest running. After systest completed, I submitted exactly same code but it accepted!
I am really sad. If I were more lucky, I would pass Div1 E......
My submission for Div2-B on contest was TLE on system test , but after system test I submit same code and could get AC. The standard of AC become ambiguous.
i am also facing the same issue, the one i did during contest is still showing pretests passed and the one i submitted (same code) again , got accepted.
If you got TLE on B that's your fault for not analyzing the complexity of your solution, especially considering how many people that 'solved' it didn't know what was going on.
then wats the mistake of people using python??? is it fair
Yes? If your know your solution makes around $$$2500 \cdot 1000 \cdot 20 = 5 \cdot 10^7$$$ operations in the worst case you know that python will be too slow.
Even my solution failed when it was making around 5*10^7*3 (3 for log factor) and even if I remove that 3 , i.e. use unordered set it is still giving TLE , I used C++
ohhh i got it they should learn c++ to solve this problem!!! now its fair
Dude if you want to python which is such a damn slow language you should know that your just can't make that much operations
1397B - Power Sequence In this problem, many C++ solutions were considered WA whereas many python solutions were considered AC which used the same logic just because of the fact that the overflow limits of Python are larger than those of C++. So, I don't think your morale should be down by this
I also failed the tests for problem B. Which was because of a factor of log n. Right after I modified it to eliminate the log n factor it gave AC. Mine was in C++. Although I'm still kind of fine with what happened with C++ submissions. But it really sucks for those who gave the contest with Python.
Anyone did B with the use of string ?
I used string along with map for which now I feel stupid. XD
1464A - Peaceful Rooks, submission 101878185
Says a Master (T_T)
anta.baka If not for the C++ submissions. At least do something for the python submissions.
Nice contest
I wonder, why is there a huge gap between the number of participants who solved C and D? in two recent contests.
Мне пришло сообщение о копировании кода самого себя, тоисть я отправил один и тот же код и на Технокубок и на див2 раунд по нему, вот и все, я не хочу бан... "Ваше решение 101883259 по задаче 1411C значительным образом совпадает с решениями других участников и находится в группе одинаковых решений Betelgeyze/101883259, Betelgeyze/101883936." Это я писал, не понимаю в чем проблема
Hey, could anyone please tell me their approach for problem C? I thought about it for an hour or so but couldn't come up with a solution.
Some hints:
1) initially ans = m 2) if we have a cycle => ans++ 3) if x[i] == y[i] => ans--
my div2B solution is still stuck at pretest passed.. later i submitted exact same code and it got accepted. rating for same is also not given[user:MikeMirzayanov] anta.baka
Hi, it is fixed now. But why do you use bold for the whole message?
Maybe he wants the message to have the Headquarters title.
This contest makes me know that you should never give up halfway during a contest! Last night it was 0:30 a.m. here and I really wanted to quit the contest and sleep, but I didn't. And I managed to come up with the solution to C in the last 10 minutes, successfully implemented it in about 5 minutes and got AC! And finally I became international master ;)
I have come up solutions in last few minutes for many time,but usually I can't write the correct code in such a stressful situation.
Will there be any other contest before Good Bye 2020 or Not
Hello everyone, when will there be information about who made it to the Technocup final in the third qualifying round?
please change the colors normal MikeMirzayanov It is not fitting well for dark mode users for example :
please change the colors normal MikeMirzayanov It is not fitting well for dark mode users for example: ![ ]()