unfortunately not true for mac os 15+ (sequoia) right now (as of 22/10/2025)

Any submission results in a rated contest. It won't be rated if you did not make a single submission tho, even you registered beforehand

now you've got to make another div3

meow :3

good job young padawan

when's the editorial come out?

i did like $$$2^n ⋅ (n+m)$$$ but with 5 additional iterations so the constant is x5, but it still got AC lol, i can't even prove why it's correct

Maybe it's not, i guess the tests just aren't strong enough

how did you solve F if you couldn't solve E, xD

E is like even easier than D i think

cats do help as well

fr

my cope for that i'm not yellow yet is that i don't do contests as much, but actually it's a skill issue

in codeforces it's rating that matters, much like in the real world, who has more wealth/looks nicer gets more respect

basically ratism

rainboy

Nominated by moonpieXXIV

I use usually assume it's 1.5e8 ops a second, works for me

Though it's much better to memorize max time complexities for various values of $$$n, q$$$ (number of queries for example) etc.:

$$$n \leq 10^5$$$ means you can go with $$$O(n$$$ $$$log^2),$$$ $$$O(n\sqrt{n})$$$ and so on. For $$$n \leq 10^6$$$, usually only solutions with $$$O(n$$$ $$$log)$$$ pass. There are exceptions, $$$O(n$$$ $$$log^2)$$$ might pass as well, if you feel like you can't optimize further — try it, you might get lucky :)

no they don't, might as well be even more interested, icpc, job opportunities and such

you did, good job :)

안녕, i am a (beta) tester

cause you can solve the problem using 2-sat and many other methods

solution with 2-SAT

of course not

well basically the problem is to count the numbers from $$$1$$$ to $$$n$$$ that have odd number of divisors.

if you think about divisors of a number, you can see that for every divisor $$$d$$$ of a number $$$x$$$, there's its let's say mirror divisor $$$x/d$$$. So you may guess that every number actually should have even number of divisors (if for every $$$d$$$ you have $$$x/d$$$ you should end up with even number of divisors right?). But it's not the case. In which cases a number has odd number of divisors then? When $$$d = x/d$$$, since you only count number of unique divisors (you don't count the same number twice).

Now the problem is to find number of such $$$i$$$ that there's some divisor $$$d = i/d$$$. Simple math and you see that $$$d^2 = i$$$, so $$$i$$$ should be perfect square. You can count them manually, since the square root of the biggest perfect square $$$i$$$ cannot be bigger than $$$d^2 = i \leq n, d \leq \sqrt{n}$$$ (complexity $$$O(\sqrt{n})$$$)

And finally you can notice that integer $$$\sqrt{n}$$$ is literally the biggest such number $$$d$$$, and every $$$d$$$ less than $$$\sqrt{n}$$$ works as well (if $$$d^2 \leq n$$$, then obviously $$$(d-1)^2 \leq n$$$ if $$$d \gt 0$$$). So the answer is $$$\sqrt{n}$$$

I don't see the reason why it is so disliked

I solved this problem like the second I read the statement, it's all about practice I guess then afterall. Solve more problems like this and you'll get the idea how to think to solve this kind of problems.

:like:

In CP you can just try and guess that some idea (maybe greedy) is correct. Code brute force approach about that idea and submit it. It's indeed correct if you got TLE (which is equal to not wrong answer) or AC :)

In order to really prove that some idea is true you have to study math OR solve bunch of problems and learn from them.

And I did look at your profile, I see you solved many A and B problems. Learn to solve them quick (virtual contests). Once you got that (you'd probably already reach high pupil) start to solve C and D.

huh?

just solved a problem like that (C++20 64)

totally agree with lis05

interesting note: you can even leave "type" space blank! like just main() {.., only matter of preference, choose whatever you want

glad to help :3

no, there's no way

if you are talking about the number of tests themselves, then it's basically $$$X-1$$$ if you got "Wrong answer on test $$$X$$$"

nice clickbate

p1 is maximum sum path, and p3 is minimum, which means we can take any other path since its sum will always be less than p1 and greater than p3 (or equal)

fifty-seven registered 46 hours ago ( 0 _ 0 )

cp-algorithms is an english version of e-maxx.ru, and it is a little bit better because it is being updated frequently

it works fine for me?

maybe something's wrong with your internet connection.

Btw, that is korean online judge, I am currently in Korea so may be that is the reason it works for me and doesn't for ya

good job :)

but I believe there are already some other resources with the same functionality (https://cfviz.netlify.app/ for instance)

you didn't?

apparently he didn't even participate in the contest, no need to call him cheater for a bug in system

you want me to be your bae? :)

you won't believe but fast ABCs are closely enough for CM

with the right training you can reach even CM in a year

they don't believe in you i guess

reach IM and prove them wrong

good luck, no kidding :)

only once in a year, at the beginning of the new year

Round was cool, especially problem D and E. Hope in future will be rounds like this!

Thanks for vovuh for the nice div3 round.

Just go to https://paste.ofcode.org or https://ideone.com (usually I use paste.ofcode.org 'cause it more easy to use) and paste your code, then you can just send to someone the link.

Like this -> https://paste.ofcode.org/329jEhW7par3TKGBpYxpsNK or just link the word -> click.

Hope it helps u.

Very nice round, problems was really good to solve and there was no bugs!

One of the interesting rounds I've ever solved.

Thank you McDic ^3^