Блог пользователя FBI

Автор FBI, история, 7 часов назад, По-английски

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
  • Проголосовать: нравится
  • +26
  • Проголосовать: не нравится

»
6 часов назад, # |
  Проголосовать: нравится -7 Проголосовать: не нравится

awesome contest!

»
6 часов назад, # |
  Проголосовать: нравится -6 Проголосовать: не нравится

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.

»
5 часов назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

my rating change is showing up as 0 wtf? is this possible?

  • »
    »
    5 часов назад, # ^ |
    Rev. 2   Проголосовать: нравится +4 Проголосовать: не нравится

    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!

»
5 часов назад, # |
  Проголосовать: нравится +3 Проголосовать: не нравится

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.

»
4 часа назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

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

»
3 часа назад, # |
Rev. 2   Проголосовать: нравится 0 Проголосовать: не нравится

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

»
2 часа назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Implementation for problem F...

»
62 минуты назад, # |
  Проголосовать: нравится +1 Проголосовать: не нравится

There are some things that I have discussed and discovered after the contest for understanding F. The following points are in context of a fibonacci sequence mod $$$k$$$, where $$$k \geq 2$$$. Pisano period refers to the period of such a sequence.

  1. The period is bounded by $$$6k$$$.

  2. The zeros of fibonacci mod $$$k$$$ are evenly spaced as proved here thanks to Dominater069 and observed to first occur no later than $$$2k$$$ as mentioned here.

  3. The Pisano period is either 1, 2 or 4 times the period of its zero. It is mentioned in wikipedia that "The number of occurrences of 0 per cycle is 1, 2, or 4." and since we have established that the all the zeros are evenly spaced, it implies that the Pisano period is either 1, 2 or 4 times the period of the zeros.

»
13 минут назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

4th question can also be done by using map.

storing the prefix sum in map and making a counter variable . if current element sum matches map then increment the counter and delete the map. do this iteration for whole array.

at the end cout counter.