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2066D1 - Club of Young Aircraft Builders (easy version)
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2066D2 - Club of Young Aircraft Builders (hard version)
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thank you for the quick editorial, questions were tricky.
I need to understand official approach, hoping to get some proofs of correctness
good contest , bad contest
The best contest i've ever particpated!
True
Reason?
please make editorial for problem D in English
How was it supposed to be deduced for A that 'n' is supposed to be a 'positive integer' only? The problem specified for 'n' to be an 'integer' which allows a larger set of valid pairs. More precisely for each currently accepted (x, y), (y, x) should also be a valid pair.
Or am I missing some detail?
oh yeah, correct, it says integer
There was an announcement stating that n is positive. Apart from that, there is nothing that says n should be positive.
oh man ,
many people who were solving it assuming negative and missed the announcement because they were looking at their notebook might have spent so much extra time.
Ok thanks, I fortunately read the announcement before submitting. It was just that the announcement seemed to imply that the fact was deducible from the problem statement itself, therefore I was confirming.
tricky but insteresting questions!
sevlll777 Problem D (Div2) or Problem A (Div1)'s Editorial is in Russian language only.. Please make it available in English !!
It is translated on Polygon. I guess some time is needed to pull it up from here, should be available soon
Interesting and exciting.Love it!
FST.
Is there a problem in the interactor of Div. 1 A (Div. 2 D) Object Identification? This submission 305637703 passes the sample locally but gets WA 1.
we are investigating it
I'd like to mention that I looked for format error for about 20 min due to the misleading information from the bugged interactor in this submission and the lack of interacting process for the sample.
Can anybody explain why this doesnt work for A?
There will be cases where $$$y\pmod{9} = 0$$$ and your code would check if $$$x\pmod{9}=-1$$$, when it should be checking if $$$x\pmod{9} = 8$$$. You can replace your code with
and it should work. Also you aren't printing a new line, which you should.
x=1,y=2; test case is failling. Consider num=100,num+1=101 then x=1 and y=2
See you have a integer number n let say it consist ABCDEF where A,B..F are digits
Thing about the two case:
Case 1: Where unit digit (here F belongs to [0,8]) n+1 make it belongs to [1,9] that mean the whole structure of number n would remains constant except the unit digit F so the sum would increase just by one x = A+B+C+D+E+F y = x + 1
Case 2: Where unit digit of n is 9 now n+1 will make unit digit 0 and carry 1 will be forward to E, again thing of two case E belongs to [0,8] or E == 9, if case 1 then simply it will add up 1 and stop other wise make digit E = 0 for n+1 and forward that carry to D.
so you got a sum x for n, for n+1 you are going to get some zeroes in place contiguous 9 from unit place (let say k no of 0 replacing k 9's) and finally adding 1 to very first non 9 digit.
so y = x- k.9 + 1
clearly visible (x + 1- y) must be divisible by 9 by taking care that (...) is >= 0.
That's it.... I hope it is helpful and easy to understand pardon my bad English
Here's a combinatorial solution for D1.
Let the number of planes launched from $$$i^{\text{th}}$$$ floor be $$$u_i$$$, note that $$$u_{n} = c$$$.
Now corresponding to this case in the original dp recursion solution we get the term $$$\prod_{i=1}^{n-1} \binom{c}{u_i}$$$
So the final answer is just sum over this expresion as $$$u_i$$$ ranges over all possible values constrained by $$$0\le u_i \le c$$$ and $$$\sum_{i=1}^{n-1} u_i = m-c$$$.
Now this is equivalent to a counting problem where we have $$$c(n-1)$$$ distinct objects divided into $$$n-1$$$ types with $$$c$$$ objects of every type and we want to choose a total of $$$m-c$$$ objects. The above expression occurs when we count by first deciding number of objects of each type and then choosing the objects per type. Instead we can directly choose the $$$m-c$$$ objects.
Hence the answer is
B was a really good problem. I wasn't able to submit the right code in time, however I still enjoyed taking my time solving the problem.
its awsome, Like your previous competitions, this one also had a lot of mathematics!
div1A >> div1B
Rating roll back when ? I want to be expert in problem solving
just curious, who or what is Devyatkino, and how is it related to the 2067C problem?
Devyat' = Nine in Russian
Problem D was really interesting theory-wise, pretty cool for an interactive problem to actually be accessible for regular users rather than being 2300+ rated
why in problem a i can't use this equation x-(y-1)? can anyone give me a test that makes my code wrong? https://codeforces.me/contest/2067/submission/305716615
A YouTube solution got leaked in the last 10 minutes of Problem E. I don't know much of Div. 1 but in Div. 2 more than 10 people in top 20 have been cheated having the same solution and it might get worse lower down in the standings. Some of the solutions are with comments and exactly same, I don't know how they escaped plagiarism but something needs to be done about these cheaters or else the fun of competitive programming will end soon.
Yeah completely agreed, Here I try my best to increase my rating and on other side my friend has higher rating because he is cheating.
why this not work for A? (x-(y-1))?
In C, if n = 6, answer is actually 9, not 8, right?
6+9*8=78
Solved ABC and had some idea about D. The problems were interesting and the observations were not very obvious. Overall a nice contest.
There are a lot of cheaters at the top of the leaderboard(Div 2) Please look into it..
After regrading, my submission to Div. 1 A now says "Wrong answer on pretest 4." If my rating is going to be calculated based on this, this feels unfair, since it would not be hard for me to fix my solution to the problem, if I knew this was the verdict. Also, it seems that some people had the opposite problem, where their solutions failed in contest but were correct. I think the correct decision here is to void the contest.
"It is not difficult to understand that it is optimal to take the leftmost zero in the subsequence" for problem Div2 E why taking left most zero is optimal..?Can anyone explain in detail?
sevlll777 Could you please tell the testcase where some of the rejudged solutions of Div 1.(A) have failed. Specifically testcase 37 on pretest 2.
testcase 37 on pretest 2 is:
Thanks
Brother, how bad was c :,(
Is there a way to solve the second problem using the frequency array?
305680148
Yes, of course here is how i solved it in the contest , do check it out bro. you take a frequency array and then check if it appears more than twice , redistribute the excess counts to the next number by increment operation. after that check if all frequencies are even then only it is possible.
I understand it now, Thank you very much bro