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hi codeforces admin

Your solution 149167827 for the problem 1651D significantly coincides with solutions WHAT_ARE_YOU--A_COP/149167827, testing_123332222/149168444.

here's what happened in round 1651 ( Mar/10/2022 ), 4 years ago,

iirc it was my initial days on codef, I may have forwarded my soln to my friend to check the validtity of soln who use to have an alt id for this purpose,

think iirc Its safe to say I was not aware of this rule that time, and since I put in lot of fair work into this while enjoying post that event,

recently noticed about this skipped contest which I was really not aware existed,

ask — do you think its possible to revert this contest as I really don't like it

if not, can share some alternatives for me ?

thanks, let me know if additional info is required
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can someone pls help with checking why this is tle,

problem E,

https://mirror.codeforces.com/contest/2127/submission/336913209
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can anyone be kind enough to share where can I find problems like E where we work around tree structures to count answers, really facing hard time around these problems.

thanks a lot

think combinatorics + tree filter works ?
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nice, good to know,

+1

feels good to see an edu round, high hopes
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- people will realise you don't have to be red to create problems that can't be solved by reds

Do you honestly think that's even possible?
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I think you got it wrong aksLolCoding,

if we check for each number with each magic, whats the max reduction at each step, then probably, this obs looks correct to me
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problem — divisors analysis —

think, there should be more explanation on how this is done — have to carefully divide the exponent by 2

case1 — all divisor count is even

can simply find the square root of number by considering only half count,

case2 — there is atleast one divisor with odd count —

as this divisor occurs in half the total divisors,

considering only half count means dividing total divisors by 2,

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I didn't get the first obs.,

can you please elaborate?
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yeah, I do know that, just wanted to know if someone solved this by just first obs,

from each point, we move to three possible cells(up, down, right), using bfs we find the ans for last cell
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hi, about problem F, do we really this observation to come up with a solution,

"Secondly, we don't need the UP move until the last column."

I think we can use simple bfs to reach the end
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why do we have to make a profit, can't we just break even(no profit no loss).

In other words, is it true that for any integer n , Sasha can make bets so that for any outcome that does not contradict the rules described above, at some moment of time he will have at least n coins.
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why can't we just use line sweep?
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ohh, I see, thanks.
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hey, why are you returning 2 here?

if(i==n-1) { if(ai==0)return 1; else return 2; }
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nice problems.
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thanks a lot.
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can't we use lower bound on vector of pairs of type long long int? it's not working for me.
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nice handle.