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Shayan's blog

Codeforces Round 959 by NEAR — Video Tutorial
By Shayan, 2 months ago, In English

1994A — Diverse Game

1994B — Fun Game

1994C — Hungry Games

1994D — Funny Game

1994E — Wooden Game

1994F — Stardew Valley

1994G — Minecraft

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2 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Was literally waiting for this!

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2 months ago, # |
  Vote: I like it +3 Vote: I do not like it

waiting...

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2 months ago, # |
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@shayan please sir can you explain this : PROBLEM D: we are using f[] array to make the node dead and lets say we get node z and node y' for operation x then we make node y' as dead .How are we ensuring that down the line for any other x ,node y' will be not be critical node(only node forming pair with some other node divisible by operation number )??? or in simple words please prove choosing any node among two would be fine .

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    2 months ago, # ^ |
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    So if you look closely, actually the Pegion-hole principle ensures that. At every step down the line, let's say we have (x) alive nodes and at that point we will be requiring (x-1) edges. So there will surely be a pair that will be having the same modulo value (as per Pegion-hole principle). Hence it does not matter which edge we had removed previously as we are sure to find a pair at every step!

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      2 months ago, # ^ |
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      Thanks for the answering!!! But I got this answer after thinking like this see for any operation x ,I will have x+1 numbers and their remainder would be in range from 0 to x-1 and that means there will always be at least two numbers with same remainder and that's the pigeon hole principle.

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2 months ago, # |
  Vote: I like it +3 Vote: I do not like it

waited for this , thanks brother.

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2 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Chinese video editorial by me

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5 weeks ago, # |
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there is no video for C

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