Program to find GCD of floating point numbers

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 b
double 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 b
def 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 = 1.20
b = 22.5
print('{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 b
function 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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