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Program to find LCM of two Fibonnaci Numbers

  • Difficulty Level : Basic
  • Last Updated : 02 Jun, 2021

Given here are two positive numbers a and b. The task is to print the least common multiple of a’th and b’th Fibonacci Numbers.
The first few Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……
Note that 0 is considered as 0’th Fibonacci Number.
Examples: 
 

Input : a = 3, b = 12
Output : 144

Input : a = 8, b = 37
Output : 507314157

 

Approach: The simple solution of the problem is, 
 

  1. Find the a’th fibonacci number.
  2. Find the b’th fibonacci number.
  3. Find their GCD, and with the help of the GCD find their LCM. The relation is LCM(a, b) = (a x b) / GCD(a, b) (Please refer here).

Below is the implementation of the above approach: 
 

C++




// C++ Program to find LCM of Fib(a)
// and Fib(b)
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1000;
 
// Create an array for memorization
int f[MAX] = { 0 };
 
// Function to return the n'th Fibonacci
// number using table f[].
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1) ? (n + 1) / 2 : n / 2;
 
    // Applying recursive formula
    // Note value n&1 is 1
    // if n is odd, else 0.
    f[n] = (n & 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1))
                   : (2 * fib(k - 1) + fib(k)) * fib(k);
 
    return f[n];
}
 
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the LCM of
// Fib(a) and Fib(a)
int findLCMFibonacci(int a, int b)
{
    return (fib(a) * fib(b)) / fib(gcd(a, b));
}
 
// Driver code
int main()
{
    int a = 3, b = 12;
 
    cout << findLCMFibonacci(a, b);
 
    return 0;
}

Java




// Java program to find LCM of Fib(a)
// and Fib(b)
import java.util.*;
 
class GFG
{
 
static int MAX = 1000;
 
// Create an array for memoization
static int[] f = new int[MAX];
 
// Function to return the n'th Fibonacci
// number using table f[].
static int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n] != 0)
        return f[n];
    int k = 0;
    if ((n & 1) != 0)
        k = (n + 1) / 2;
    else
        k = n / 2;
 
    // Applying recursive formula
    // Note value n&1 is 1
    // if n is odd, else 0.
    if((n & 1 ) != 0)
        f[n] = (fib(k) * fib(k) +
                fib(k - 1) * fib(k - 1));
    else
        f[n] = (2 * fib(k - 1) +
                    fib(k)) * fib(k);
 
    return f[n];
}
 
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the LCM of
// Fib(a) and Fib(a)
static int findLCMFibonacci(int a, int b)
{
    return (fib(a) * fib(b)) / fib(gcd(a, b));
}
 
// Driver code
public static void main(String args[])
{
    int a = 3, b = 12;
 
    System.out.println(findLCMFibonacci(a, b));
}
}
 
// This code is contributed by
// Surendra_Gangwar

Python3




# Python 3 Program to find LCM of
# Fib(a) and Fib(b)
MAX = 1000
 
# Create an array for memoization
f = [0] * MAX
 
# Function to return the n'th
# Fibonacci number using table f[].
def fib(n):
 
    # Base cases
    if (n == 0):
        return 0
    if (n == 1 or n == 2):
        f[n] = 1
        return f[n]
 
    # If fib(n) is already computed
    if (f[n]):
        return f[n]
 
    k = (n + 1) // 2 if (n & 1) else n // 2
 
    # Applying recursive formula
    # Note value n&1 is 1
    # if n is odd, else 0.
    if (n & 1):
        f[n] = (fib(k) * fib(k) +
                fib(k - 1) * fib(k - 1))
    else:
        f[n] = (2 * fib(k - 1) + fib(k)) * fib(k)
 
    return f[n]
 
# Function to return gcd of a and b
def gcd(a, b):
    if (a == 0):
        return b
 
    return gcd(b % a, a)
 
# Function to return the LCM of
# Fib(a) and Fib(a)
def findLCMFibonacci(a, b):
 
    return (fib(a) * fib(b)) // fib(gcd(a, b))
 
# Driver code
if __name__ == "__main__":
    a = 3
    b = 12
 
    print (findLCMFibonacci(a, b))
 
# This code is contributed by ita_c

C#




// C# program to find LCM of Fib(a)
// and Fib(b)
using System;
 
class GFG
{
 
static int MAX = 1000;
 
// Create an array for memoization
static int[] f = new int[MAX];
 
// Function to return the n'th Fibonacci
// number using table f[].
static int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n] != 0)
        return f[n];
    int k = 0;
    if ((n & 1) != 0)
        k = (n + 1) / 2;
    else
        k = n / 2;
 
    // Applying recursive formula
    // Note value n&1 is 1
    // if n is odd, else 0.
    if((n & 1 ) != 0)
        f[n] = (fib(k) * fib(k) +
                fib(k - 1) * fib(k - 1));
    else
        f[n] = (2 * fib(k - 1) +
                    fib(k)) * fib(k);
 
    return f[n];
}
 
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the LCM of
// Fib(a) and Fib(a)
static int findLCMFibonacci(int a, int b)
{
    return (fib(a) * fib(b)) / fib(gcd(a, b));
}
 
// Driver code
static void Main()
{
    int a = 3, b = 12;
 
    Console.WriteLine(findLCMFibonacci(a, b));
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP Program to find LCM of Fib(a)
// and Fib(b)
 
$GLOBALS['MAX'] = 1000;
 
// Create an array for memoization
$GLOBALS['f'] = array();
 
for($i = 0; $i < $GLOBALS['MAX']; $i++)
    $GLOBALS['f'][$i] = 0;
 
// Function to return the n'th Fibonacci
// number using table f[].
function fib($n)
{
    // Base cases
    if ($n == 0)
        return 0;
    if ($n == 1 || $n == 2)
        return ($GLOBALS['f'][$n] = 1);
 
    // If fib(n) is already computed
    if ($GLOBALS['f'][$n])
        return $GLOBALS['f'][$n];
 
    $k = ($n & 1) ? ($n + 1) / 2 : $n / 2;
 
    // Applying recursive formula
    // Note value n&1 is 1
    // if n is odd, else 0.
    $GLOBALS['f'][$n] = ($n & 1) ?
                        (fib($k) * fib($k) +
                         fib($k - 1) * fib($k - 1)) :
                        (2 * fib($k - 1) + fib($k)) * fib($k);
 
    return $GLOBALS['f'][$n];
}
 
// Function to return gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
 
    return gcd($b % $a, $a);
}
 
// Function to return the LCM of
// Fib(a) and Fib(a)
function findLCMFibonacci($a, $b)
{
    return (fib($a) * fib($b)) /
            fib(gcd($a, $b));
}
 
// Driver code
$a = 3;
$b = 12;
 
echo findLCMFibonacci($a, $b);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
// Javascript program to find LCM of Fib(a)
// and Fib(b)
 
let MAX = 1000;
 
// Create an array for memoization
let f = new Array(MAX);
for(let i=0;i<MAX;i++)
{
    f[i]=0;
}
 
// Function to return the n'th Fibonacci
// number using table f[].
function fib(n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
   
    // If fib(n) is already computed
    if (f[n] != 0)
        return f[n];
    let k = 0;
    if ((n & 1) != 0)
        k = (n + 1) / 2;
    else
        k = n / 2;
   
    // Applying recursive formula
    // Note value n&1 is 1
    // if n is odd, else 0.
    if((n & 1 ) != 0)
        f[n] = (fib(k) * fib(k) +
                fib(k - 1) * fib(k - 1));
    else
        f[n] = (2 * fib(k - 1) +
                    fib(k)) * fib(k);
   
    return f[n];
}
 
// Function to return gcd of a and b
function gcd(a,b)
{
    if (a == 0)
        return b;
   
    return gcd(b % a, a);
}
 
// Function to return the LCM of
// Fib(a) and Fib(a)
function findLCMFibonacci(a,b)
{
    return (fib(a) * fib(b)) / fib(gcd(a, b));
}
 
// Driver code
let a = 3, b = 12;
document.write(findLCMFibonacci(a, b));
 
 
// This code is contributed by rag2127
</script>
Output: 
144

 




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