FBI's blog

By FBI, history, 5 hours ago, In English

2033A - Sakurako and Kosuke

Idea: FBI

Tutorial
Solution

2033B - Sakurako and Water

Idea: FBI

Tutorial
Solution

2033C - Sakurako's Field Trip

Idea: Vladosiya

Tutorial
Solution

2033D - Kousuke's Assignment

Idea: FBI

Tutorial
Solution

2033E - Sakurako, Kosuke, and the Permutation

Idea: FBI

Tutorial
Solution

2033F - Kosuke's Sloth

Idea: FBI

Tutorial
Solution

2033G - Sakurako and Chefir

Idea: Vladosiya

Tutorial
Solution
  • Vote: I like it
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4 hours ago, # |
  Vote: I like it 0 Vote: I do not like it

awesome contest!

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4 hours ago, # |
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Bro I sent F two seconds before the end of the round and it got Compilation Error in C++17 but when I sent it with C++20 1 minute after the end of the round it got OK. I'm literally unluckiest participant of this round.

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3 hours ago, # |
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my rating change is showing up as 0 wtf? is this possible?

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    3 hours ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Yeah, this is possible. This just means that:
    1) your performance wasnt' good enough for your rating to increase!
    2) your performance wasnt' bad enough for your rating to decrease!

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

I noticed that except D, all problems that involve "Kosuke" refer to "Kosuke" as "Kosuke" but D refers to "Kosuke" as "Ko'U'suke".
This means nothing of course, just a funny observation.

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2 hours ago, # |
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upper bound for the first appearance of $$$0$$$ on problem F is actually bounded by 2 * k.

https://jonkagstrom.com/articles/upper_bound_of_fibonacci_entry_points.pdf

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51 minute(s) ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I feel like proving that we don't miss any numbers divisible by k would be a nice touch. Proof: Let Fi be the first number divisible by k. Fm exists such that m>i, and m!=ni, n in the set of integers. Let us prove that it is not divisible by k. gcd(Fi,Fm) = F(gcd(i,m)). The gcd must be <= i since, i is less than m. Also, the gcd is not equal to i since m is not divisible by i. Thus, F(gcd(i,m)) is an element of the set of Fz, 0<=z<i. By premise, these numbers are not divisible by k, thus since the gcd of Fm and Fi is less than k, and Fi is divisible by k, Fm is not divisible by k. QED.

p.s: idk latex but someone can make this into latex and i will edit if anyone feels like it

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23 minutes ago, # |
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Implementation for problem F...