 Huge Fibonacci Number modulo m — aadimator

# Huge Fibonacci Number modulo m

This problem was taken from the Coursera Data Structures and Algorithms Specialization, specifically from the Algorithmic Toolbox Course, Week 2, that I’ve recently completed. If you are taking that course or plan to take that course, please don’t look ahead at the solution as it will be against the Honor Code and won’t do you any good.

#### Problem Introduction

The Fibonacci numbers are defined as follows: F(0) = 0, F(1) = 1, and
F(i) = F(i−1) + F(i−2) for i ≥ 2.

#### Problem Description

Task: Given two integers n and m, output F(n) mod m (that is, the remainder of F(n) when divided by m).
Input Format: The input consists of two integers n and m given on the same line (separated by a space).
Constraints: 1 ≤ n ≤ 10^18 , 2 ≤ m ≤ 10^5 .
Output Format: Output F(n) mod m
Time Limits: C: 1 sec, C++: 1 sec, Java: 1.5 sec, Python: 5 sec. C#: 1.5 sec, Haskell: 2 sec, JavaScript: 3 sec, Ruby: 3 sec, Scala: 3 sec.
Memory Limit: 512 Mb

#### Sample

Input:
281621358815590 30524
Output:
11963

#### What To Do

In this problem, the given number n may be really huge. Hence an algorithm looping for n iterations will not fit into one second for sure. Therefore we need to avoid such a loop.
To get an idea how to solve this problem without going through all F(i) for i from 0 to n, let’s see what happens when m is small — say, m = 2 or m = 3.

Take a detailed look at this table. Do you see? Both these sequences are periodic! For m = 2, the period is 011 and has length 3, while for m = 3 the period is 01120221 and has length 8. Therefore, to compute, say,
F(2015) mod 3 we just need to find the remainder of 2015 when divided by 8. Since 2015 = 251·8+7, we conclude that F(2015) mod 3 = F 7 mod 3 = 1.

This is true in general: for any integer m ≥ 2, the sequence F(n) mod m is periodic. The period always starts with 01 and is known as Pisano period.

### My Solution:

#### Explanation:

I couldn’t find a suitable definition for Pisano period so that I could make a naive algorithm. Well, it turns out, I was looking at the wrong place. The whole time, it was right in front of my eyes but I couldn’t see it. So, without further ado, here’s the definition. This was a very tricky one. I had to search a lot and read quite a few posts in the Course forum to understand the algorithm.

The algorithm to compute the get_fibonacci_huge was given in the What To Do section. “Therefore, to compute, say, F(2015) mod 3 we just need to find the remainder of 2015 when divided by 8. Since 2015 = 251·8+7, we conclude that F(2015) mod 3 = F 7 mod 3 = 1.” We just have to generalize it.

First we need to find the remainder of F(n) divided by the length of Paisano period given m. To find the length of Paisano period for m, simply find the remainder of F(0) mod m to F(m*m) mod m and stop as soon as you encounter 01, as they indicate that the next iteration is being started, and return the length up to that point.

Now, you only need to find the F(remainder), which is going to be a lot less than F(n) and simply return it.

If you can think of any other way to improve this algorithm or if you could point out something that you think I did wrong, you’re more than welcome to reach out to me by responding to this and pointing out the mistakes. If you like this solution, please hit the “Recommend” button below, it’ll mean a lot to me. Thanks.