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On lyftThe Lyft Level 5 Challenge, 47 hours ago

That can't be your friend. A friend will not post such a repulsive joke. A classmate I guess?

On lyftThe Lyft Level 5 Challenge, 47 hours ago

I have to point out that you have a VERY AGGRESSIVE avatar. No matter who "ypy" and "dyf" is, It is very impolite to draw a piece of green shit on other's head and write other's name on the avatar. DO NOT use this. Change it as fast as you can! And the comment is awful. It seems like someone that uses your account illegally and change your avatar, and post that piece of bloody comment.

On lyftThe Lyft Level 5 Challenge, 9 days ago

This algorithm is great! It has helped me alot in this problem!

This algorithm is great! It has helped me alot in this problem!

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On lyftThe Lyft Level 5 Challenge, 11 days ago

We are processing scoreboard and contacting people to confirm they can come to onsite round. We will do our best to keep everyone updated.

Among those who confirmed the possibility of onsite participation, we will invite the top-30 to the Final based on the results in the Elimination Round.

IT WON'T BE RATED!!! -------->

"We invite top-30 participants who don't need sponsored travel to our location. If you get good score in round-1, want to be invited to onsite, and can manage travel costs, don't hesitate to send a personal message and we'll definitely consider you when processing scoreboard."

This was what they said here. So, you don't even really need to live there. You just need to be there at the time of the contest.

And then comes a time you have to do something like this XD

long long int sq = sqrt(n);
while(sq*sq > n)
while(sq*sq <= n)

BTW, should you permanently live in Bay Area or you can just be here for the time of contest?

Thanks for your reply, it now runs in 77ms.

The sieving part is correct, the part to blame is actually:


Sadly, this is .

Alternatives are

  1. s += '0'
  2. use std::stringstream
  3. std::string s(n, '0')

The best one in this particular case is 3.

I solved it with DP , states are <CurrentNodeValue , Turn > , Current State is to pick the optimal choice depends on transitions (For example :- currentTurn for Alice , Alice must search for any odd moves to Win and Bob must search for even moves to Win )

In problem C , can any one tell me what's wrong in my approach ? why does it give wrong answer on Test 23 ?

Test 16 is a random test, where each number is a product of two primes slightly over 109. Pretest 10 is the same random test, but with a different seed. So there should not be anything special about it.

I solved with Fermat factorization (ofc after eliminating p^2, p^3, p^4 and small prime numbers)

Ok i got it, Thanks brother :)

How do the organizers know who lives in the Bay Area?

If one of the player wins for a given position, then the other one loses. Because there is always one player who wins and the other one loses. I don't see why would you need a second dimension.

Unless your state refers to which player starts playing, but I still don't understand why it would be needed since it is specified in the statement that Alice starts playing.

What corner case does test 16 for problem D have? My code passed pretests but gave WA for that one.

When i though about C problem as a DP problem on DAG, i defined the state such that

dp[i][j]=1 if the player j can win when the game starts at position i, else it equals 0.

I found some submissions only use dp[i] and nothing else changed in the rest of the solution, so what is the idea behind the possibility of removing the second dimension in my defined state?

Thanks in advance :)

My C got TLE, it ran in 1450ms with time limit being 1s, can anybody explain why is it so? It is sieve-like implementation so the time complexity should be O(n*log(n)).

Hi, majk, could you help?

Submission : 43960105

In reality, the uses around 17000 queries.

How well did we need to do to get an intership?

Oh, thanks! I've just find out that the order doesn't matter.

Why did you cut off solution from solution? First one is already tedious enough to write...

Wow I barely reached yellow! Purples are half div2 — half div1, so now I'm completely div 1. Also it seems now I'm a nerd: :)

Made redundant recursive calls and hence it caused TLE. Now I get AC! >_<

I instead multiplied it by (k+1)^2 as there is no other occurrence of those p and q. Why are you multiplying with ((k+2)/(K+1))^2 ?

After running into such stuff once or twice you'll probably start doing things like "let's check +-10 in each direction, just to be sure", even if you don't really need that :)

Nice Contest , Quick editorial,Fast sys testing. Thanks Codeforces!!!!

Thank you! Together with comment by KAN, it cleared my doubts.

Until this moment I realized how weak the pretests of D was. Still, my bad after all, so no huge complaints. Great problemset overall, to be honest (even I myself took away my very chance to be in top100).

If only the constraints of D was 1e18,I've been a candidate :((

The input is 3 15 15 21 (the first integer on the line is the number of prime factors of ai).

Precision... how I love it. Changing from int(c ** (1/3)) to int(c ** (1/3) + 0.0001) changed WA11 on AC

Hi majk, arsijo,

There is something wrong with the checker for task D. For example this submission is getting WA on test 9:

I assume that the test case is: 3 30 30 42

My program locally returns 24 and the checker says it returns 45. I do not see any reason why there should be any difference. Could you take a look into this?


AFAIK the TL on any judge executable (validator, checker, etc) is 10 or 20 seconds. I suppose it can be raised, but is it needed? Also, we'll have to have the same format for all system tests, so we need to factorize them too, that might take long time.

You would get a "Judgement Failed" verdict with a comment that validator exceeded the time limit. Which is what I was getting in Polygon when preparing the problem.

Someone with a deeper knowledge of the platform might be able to tell you more details.

You need to flush the output using fflush(stdout) or cout.flush() in C++;. Anyways you'll fail pretest 1.

"You can't do that without factorising the input, and that factorisation cannot be done fast enough."

So what exactly will go wrong if you'll try to factorize it? Will it be something like "checker TL exceeded", since you have to proceed that input within (let's say) 1 second, or is it "we can do that, but it will load CF servers too much" (sounds questionable due to small number of hacks). I don't think it is "you'll have to wait for the verdict for too long" either, since it isn't instant anyway. Could you please add a few details about what the issue is, for those who don't know how CF works? I'm just curious about what is the technical limitation behind it.

On lyftThe Lyft Level 5 Challenge, 12 days ago

It is not clear how you will distinguish local from not local participants (there was no option during registration).

Could you give an idea of what can be considered as a "good score"?

On vsbLevel 5 Engineer Story, 13 days ago
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