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,
- Find the a’th fibonacci number.
- Find the b’th fibonacci number.
- 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 memorizationint 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 bint 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 codeint 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 memoizationstatic 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 bstatic 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 codepublic 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 memoizationf = [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 bdef 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 codeif __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 memoizationstatic 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 bstatic 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 codestatic 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 memoizationlet 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 bfunction 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 codelet a = 3, b = 12;document.write(findLCMFibonacci(a, b));// This code is contributed by rag2127</script> |
Output:
144
Time Complexity: O(2n)
Auxiliary Space: O(MAX)
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