Actually Serval is from Japari Park. :P

You'll make it, just believe in yourself.

Wow 8 years old girl

As an Eight year old

Unfortunately, the problems are 2085 but not TECHNOPOLIS.

caused by 2085

2085 nerd

orz

youre already blue.

1100 1101 1110 1111??

Hope to see you as Candidate Master today...

Note that the contests vary in difficulty. It can be Div.1, Div.1+2, Div.2, Div.3, Div.4, where Div.1 is the most difficult and Div.4 is the easiest. Normally Div.3 and Div.4 are for school pupils, Div.2 for university students and Div.1 for pros. Actually, you can participate in any.

Practice with the problems from previous contests, register for the contest, prepare mentally for a sprint, start in time and do your best!

[off-topic] you are so nice man. you been 1600 rated by 4 contest nice. What's your first account? or it is first?

Answering to myself: it was actually available during the contest, so likely just an error?..

whoa man... C was kind of hardest for me today

I come to talk with u again

WLOG, let x < y. Then, add to x and y such that x only has a single bit. Then, if x and y still share a bit, do the same thing to y. You can easily enough see that x and y will not share bits after this due to x < y in the first place.

Well, first if $$$x = y$$$, there is no answer, clearly.

Otherwise, let's say $$$x \gt y$$$ (if not, we can swap), you want to make $$$(x + k) & (y + k) = 0$$$, which gives an idea of making $$$x + k$$$ having one set bit, which is $$$2^p$$$ ($$$2^p \ge x$$$). When we have $$$x + k = 2^p$$$, $$$y + k$$$ cannot reach $$$2^p$$$ since $$$x \gt y$$$, so they won't have common set bit.

So result is $$$2^p - x$$$.

2^30 — max(x, y)

Make the maximum number power of 2 and if number same then -1

One of xor identity: x + y = x^ y when there's no bit of x and y are equals.

So you need to find a k to flood max(x,y) to 1 bit to the msb(most-significant-bit) only, the lesser number cannot reach there and there you have it.

the formular:

k = (1 << bit_width) - max(x, y)

I accidentally AC, then realized this after AC rofl.

My solution

I find the fact a + b = a ^ b <=> a & b = 0 pretty obvious, if you just observe the bitmasks.

Well now in future problems u'll know, as it is pretty standard result used in multiple problems.

First, it is well known; Secondly, it is not hard to observe that $$$(a+b)=(a\oplus b)\Leftrightarrow a\operatorname{\&}b = 0$$$. You just need to look at the binary representation.

I didn't know that beforehand, derived it mid-contest

the easiest way to understand is to try and add two numbers by hand in binary, then try to xor them in binary. What you'll find is xor will equal addition if there is no "carrying" of the bit.

note that a+b=(a&b)<<1+(a^b) so now you know that (x+k)&(y+k) must be 0 note that if x!=y one is larger, so if you add diff between larger one and 1<<(msb+1) to both of them, one will have all 0s in the original places and one in the larger place, other wont have that 1 in the larger place, meaning & of the two is 0, meaning its euqal a+b to a^b

gg

$$$2^{50}-\max(x,y)$$$

use 'well known' equality a + b = (a & b) << 1 + a ^ b

then you will find (x + k) + (y + k) = (x + k) ^ (y + k) iff (x + k) & (y + k) = 0

wlog let x < y, let X = x + k, and d = y — x. then X & (X + d) = 0. then it's fairly easy to find X

I submitted dp solution only to a TLE in pretest 3. How to D? Cute and simple problem but just aaaghhhhh!!!!

me too... At the last minute of the contest I realized that this problem could be solved using greedy algorithms with heap. But I don't have enough time to solve it >_<

I also thought about dp for D and stuck for a long time, then attempted F1 but failed too...