Comments

The contest is now also available in the gym.

Hi! The problems are available for upsolving in the gym

orz

On T1duSIOI 2021 predictions, 5 years ago
0

orz

+61

McDic orz

On samchen123Study Algorithms!, 6 years ago
+10

numb so rude

On samchen123Study Algorithms!, 6 years ago
+8

Good luck sir!

Thank you sir

We need a stefdasca x tmwilliamlin168 crossover.

For each node you can precompute jumps with lengths powers of 2 and answer queries using binary lifting.

+166

Orz

+25

Div 1 orz

+29

Use kmp

This is known as maximum matching in a bipartite graph and can be solved using Hopcroft Karp

+28

We hang out on the same lake.

Awesome, thanks for sharing

+13

Great!

I assume it is because your comp breaks the sort. Here is the modified correct version: https://mirror.codeforces.com/contest/1185/submission/55805231

+3

You should calculate exponent modulo phi(MOD) which is MOD-1 in this case.

+22

It's just a Controlled Not gate. https://en.wikipedia.org/wiki/Controlled_NOT_gate

+22

lol Are you taking quantum computing classes for the Olympiad?

It seems that the round is planned to begin at 12:00 Moscow time, contrary to what we were announced in the last e-mail. "December 16 at 16:00 Moscow time."

Only russian statements too.

+55

Author had solution for G, but this tutorial is too small to contain it.

0

There's an appearance issue in C tutorial

English tutorial for E?

On GreenGrapeCodeforces Round #505, 8 years ago
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Expected output?

On GreenGrapeCodeforces Round #505, 8 years ago
+10

Hack for D : 5 2 3 5 7 210 Answer : NO

What about long long main() ?

What the problem meant is: If you replace every character in s with f(s[i]) you should get string t. Meaning that "aab" and "abb" are not isomorphic since running s through your function would create string "bba" and not "abb".

Can someone explain their solution for E? I was thinking about something involving cliques but couldn't figure it out.

I finally understand. Thank you so much.

Could anyone explain me the solution to div2 D. I can't seem to understand why this always works. For 4 4 1 2 3 wouldn't the output be 4 1 2 3?