The greatest common divisor (GCD) of two or more numbers, which are not all zero, is the largest positive number that divides each of the numbers.
Example:
Input : 0.3, 0.9 Output : 0.3 Input : 0.48, 0.108 Output : 0.012
The simplest approach to solve this problem is :
a=1.20
b=22.5
Expressing each of the numbers without decimals as the product of primes we get:
120
2250
Now, H.C.F. of 120 and 2250 = 2*3*5=30
Therefore,the H.C.F. of 1.20 and 22.5=0.30
(taking 2 decimal places)
We can do this using the Euclidean algorithm. This algorithm indicates that if the smaller number is subtracted from a bigger number, GCD of two numbers doesn’t change.
C++
// CPP code for finding the GCD of two floating// numbers.#include <bits/stdc++.h>using namespace std;// Recursive function to return gcd of a and bdouble gcd(double a, double b){ if (a < b) return gcd(b, a); // base case if (fabs(b) < 0.001) return a; else return (gcd(b, a - floor(a / b) * b));}// Driver Function.int main(){ double a = 1.20, b = 22.5; cout << gcd(a, b); return 0;} |
Java
// JAVA code for finding the GCD of // two floating numbers.import java.io.*;class GFG { // Recursive function to return gcd // of a and b static double gcd(double a, double b) { if (a < b) return gcd(b, a); // base case if (Math.abs(b) < 0.001) return a; else return (gcd(b, a - Math.floor(a / b) * b)); } // Driver Function. public static void main(String args[]) { double a = 1.20, b = 22.5; System.out.printf("%.1f" ,gcd(a, b)); }}/*This code is contributed by Nikita Tiwari.*/ |
Python
# Python code for finding the GCD of# two floating numbers.import math# Recursive function to return gcd # of a and bdef gcd(a,b) : if (a < b) : return gcd(b, a) # base case if (abs(b) < 0.001) : return a else : return (gcd(b, a - math.floor(a / b) * b)) # Driver Function.a = 1.20b = 22.5print('{0:.1f}'.format(gcd(a, b)))# This code is contributed by Nikita Tiwari. |
C#
// C# code for finding the GCD of // two floating numbers.using System;class GFG { // Recursive function to return gcd // of a and b static float gcd(double a, double b) { if (a < b) return gcd(b, a); // base case if (Math.Abs(b) < 0.001) return (float)a; else return (float)(gcd(b, a - Math.Floor(a / b) * b)); } // Driver Function. public static void Main() { double a = 1.20, b = 22.5; Console.WriteLine(gcd(a, b)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP code for finding the GCD// of two floating numbers.// Recursive function to // return gcd of a and bfunction gcd($a, $b){ if ($a < $b) return gcd($b, $a); // base case if (abs($b) < 0.001) return $a; else return (gcd($b, $a - floor($a / $b) * $b));}// Driver Code$a = 1.20;$b = 22.5;echo gcd($a, $b);// This code is contributed// by aj_36?> |
Javascript
<script>// javascript code for finding the GCD of // two floating numbers. // Recursive function to return gcd // of a and b function gcd(a , b) { if (a < b) return gcd(b, a); // base case if (Math.abs(b) < 0.001) return a; else return (gcd(b, a - Math.floor(a / b) * b)); } // Driver Function. var a = 1.20, b = 22.5; document.write( gcd(a, b).toFixed(1));// This code is contributed by aashish1995 </script> |
Output:
0.3
Time Complexity: O(log n)
Auxiliary Space: O(log n)
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