Last Digit of a Large Fibonacci Number fast algorithm

last digit of a large fibonacci number python
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I'm trying to solve Fibonacci using java, but my code takes so long with big numbers.

Problem Description Task. Given an integer 𝑛, find the last digit of the 𝑛th Fibonacci number 𝐹𝑛 (that is, 𝐹𝑛 mod 10).

Input Format. The input consists of a single integer 𝑛.

Constraints. 0 ≤ 𝑛 ≤ 10⁷.

Output Format. Output the last digit of 𝐹𝑛.

My code:

public class FibonacciLastDigit {

private static int getFibonacciLastDigitNaive(int n) {
    if (n <= 1) {
        return n;
    }
    BigInteger first = BigInteger.ZERO;
    BigInteger second = BigInteger.ONE;
    BigInteger temp;

    for (int i = 1; i < n; i++) {
        temp = first.add(second);
        first = second;
        second = temp;
    }
    return second.mod(BigInteger.TEN).intValue();
}

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    System.out.println(getFibonacciLastDigitNaive(n));
}}

My code fails if input = 613455 it takes 30 seconds to get value and max allowed time is 1.5 second.

I had to use big Integer because long isn't enough.

Your implementation is indeed naive, because you're asked to get the last digit of the Fibonacci number not the actual Fibonacci number itself. You only need to keep the track of the last digit, other digits don't matter.

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    System.out.println(getFibonacciLastDigit(n));
}

private static int getFibonacciLastDigit(int n) {
    if (n <= 1) {
        return n;
    }
    int first = 0;
    int second = 1;
    int temp;

    for (int i = 1; i < n; i++) {
        temp = (first + second) % 10;
        first = second;
        second = temp;
    }
    return second;
}

Program to find last digit of n'th Fibonnaci Number, Given a number 'n', write a function that prints the last digit of n'th ('n' can also be a Simple approach is to calculate the n'th Fibonacci number and printing the last digit. Fibonacci numbers grow exponentially fast. n'th Fibonacci number, there is a direct algorithm to just calculate its last digit (that is, large-numbers. Fibonacci numbers grow exponentially fast. For example, the 200’th Fibonacci number equals 280571172992510140037611932413038677189525. And F (1000) does not fit into the standard C++ int type. To overcome this difficulty, instead of calculating n’th Fibonacci number, there is a direct algorithm to just calculate its last digit (that is, F (n) mod 10).

There is a cycle in the last digit of the Fibonacci numbers. It repeats for every 60 numbers. So just build a table of the last digit of the first 60 numbers, then do a modulo 60 operation on the input and a table lookup.

You may see the cycle in any online (or offline) table of Fibonacci numbers. One link at the bottom.

For building the table, for each calculated number you may subtract 10 if it’s over 9 since you only need the last digit, and the last digit of each number only depends on the last digit of the two previous numbers. You can use int math (you neither need long nor BigInteger).

Link: The first 300 Fibonacci numbers, factored

Last Digit of a Large Fibonacci Number fast algorithm, Your implementation is indeed naive, because you're asked to get the last digit of the Fibonacci number not the actual Fibonacci number itself. The Last Digit of a Large Fibonacci Number — aadimator Explanation: Well there’s not much to say, as it is not a very tricky problem. I just implemented the algorithm that they suggested in

Your implementation is indeed naive, because you're asked to get the last digit of the Fibonacci number not the actual Fibonacci number itself. You only need to keep the track of the last digit, other digits don't matter. Code in py must be

lookup = {0:0,1:1}
N = int(input())
for i in range(2,N+1):
       lookup[i] = (lookup[i-1] + lookup[i-1])%10
print(lookup[N])

Calculating the last digits of large fibonacci numbers, Calculating the last digits of large fibonacci numbers way, the "standard" way, and in the end a sub-linear algorithm for calculating the nth Fibonacci number that allows for calculating huge Fibonacci numbers very quickly. 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

Solution:Last Digit of a Large Fibonacci Number – Chen Hong, Problem: Last Digit of a Large Fibonacci Number Problem Recall that Fibonacci numbers grow exponentially fast. The starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Problem: The Last Digit of a Large Fibonacci Number Python: Max time used: 0.17/5.00, max memory used: 8699904/536870912 Java: Max time used: 0.19/1.50, max memory used: 28651520/536870912

Last digits of Fibonacci numbers, Cool topic. It does seem erratic, but on a larger scale, some simple straight lines appear. https://repl.it/@prof_pantaloni/cycle-length  Last Digit of a Sum of Fibonacci Numbers; Last Digit of a Partial Sum of Fibonacci Numbers; Last Digit of the Sum of Squares of Fibonacci Numbers; Week-3 (pdf) Money Change; Maximum Value of the Loot (Fractional Knapsack) Maximum Advertisement Revenue (Maximum Dot Product) Collecting Signatures (Covering Segments by Points) Maximum Number of

Python: Calculate the last digit of a large Fibonacci Number with less , Uses python3 # Compute the Last Digit of a Large Fibonacci Number def This is one of the simplest answers,i'm sure, there is a faster algorithm. Here is what I  Please design and implement your own algorithms to pass the course. Week 1- Programming Challenges . Sum of Two Digits; Maximum Pairwise Product; Week 2- Algorithmic Warm-up . Fibonacci Number; Last Digit of a Large Fibonacci Number; Greatest Common Divisor; Least Common Multiple; Fibonacci Number Again; Last Digit of the Sum of Fibonacci Numbers

Comments
  • I don't get it.. constraint says n<=107? Why enter 613455 then?
  • Hint: modulo addition.
  • I don't know why this happens if I stop him from entering more than 107 test case fails, so i had to remove this constraint
  • Yes it is meant to fail, isn't it? That means there won't be any input higher than 107. Also, you don't need big numbers. @meowgoesthedog's hint is excellent.
  • I expect "107" should be 10⁷. I made an edit to fix the question.
  • This is also less than optimal, since there is an algorithm that takes O(log(n)) steps in addition to the algorithm in the other answer.
  • If this is anything else than a comp sci exercise (and even then), then this should be the only acceptable answer. Because this algorithm pretty much guarantees constant computational complexity.