as a participant, even if you didnt test, you wouldn't be able to win the t-shirt LLoooooLL

vro this isn't good and bro is advertising it everywhere

no you can't

Bruh, only 2

Ignore this comment, I misunderstood.

I found B a bit tricky !! but didn't use random....

you can prove there must be an answer at arround 30 or 40 numbers.

Problem B:

If you have two even numbers $$$x \lt y$$$ in the list, $$$y \mod x$$$ is even so it's done.

Assume all numbers are odd. If you find two adjacent numbers $$$x \lt y$$$ such that $$$2x \gt y$$$, you're done because $$$y \mod x = y-x$$$ which is even.

Note that if $$$2x \gt y$$$ never occurs, then each number must be at least twice the previous one. This sequence would quickly exceed the $$$10^9$$$ limit, so you only need to brute force the first ~35 numbers. Because of this, even if you write the $$$O(n^2)$$$ solution it will quickly find the answer.

https://codeforces.me/blog/entry/148062?#comment-1323661

how did you come up with that ? did you solve some problems like that before ? curious to know

wait for system testing my friend

Each number x has y mod x=2 for atleast half of the numbers greater than x.Thus,if n>30,answer will always be YES,so you can simply write brute fruce.

That's pretty cool. I was thinking of doing the same thing after frustrating 2 hours, but I had no proof for this either. Apparently, one should use pen and paper while solving problems. >:(

sorry, but exponentials/logarithms is as basic as math can get though

Yes B was horrible, D was just implantation base think in reverse direction of string

please share idea for D

breaks early

actually it's O(n*log_n)

check pairs among the first min(n,32) elements: for each pair x<y, if y % x is even print them; otherwise print -1. Using 32 elements is enough because longer sequences without such a pair would grow beyond 1e9

UPD: Got it, thanks

size of n for -1 answer is bounded by a<=10^9

1. If array has at least 2 even numbers, any pair of evens should work. Easy to prove.

2. otherwise check adjacent pairs for the answer. If none of those pairs works which implies a[i + 1] is at least 2 times bigger than a[i], now brute force kicks in.

If there’s a 1 with any x, or if there are two even numbers, then the answer is trivial.

So we only care about the odd x. Now, loop over all odd x. If you ever find two consecutive odds where (x[i+1] % x[i]) is even, you’re done. the easiest way when you will be done is the following: x[i+1] = q*x[i] + r. where q = floor(x[i+1]/x[i]) and q = 1

If that never happens, we know tat q > 1 and which means X[i+1] >= 2*x[i]. Voilà, the odd numbers double each step, so there are only O(log) of them. Just brute-force those.

Wow, you have 1 << 5 submissions for the problem C. :(

I think if you get TLE, the jury doesn't compare the output.