Program to add two fractions

Add two fraction a/b and c/d and print answer in simplest form.
Examples :

Input:  1/2 + 3/2
Output: 2/1

Input:  1/3 + 3/9
Output: 2/3

Input:  1/5 + 3/15
Output: 2/5

Algorithm to add two fractions

• Find a common denominator by finding the LCM (Least Common Multiple) of the two denominators.
• Change the fractions to have the same denominator and add both terms.
• Reduce the final fraction obtained into its simpler form by dividing both numerator and denominator by their largest common factor.

C++

// C++ program to add 2 fractions
#include<bits/stdc++.h>
using namespace std;

// 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 convert the obtained fraction
// into it's simplest form
void lowest(int &den3, int &num3)
{
    // Finding gcd of both terms
    int common_factor = gcd(num3,den3);

    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3/common_factor;
    num3 = num3/common_factor;
}

//Function to add two fractions
void addFraction(int num1, int den1, int num2,
                 int den2, int &num3, int &den3)
{
    // Finding gcd of den1 and den2
    den3 = gcd(den1,den2);

    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1*den2) / den3;

    // Changing the fractions to have same denominator
    // Numerator of the final fraction obtained
    num3 = (num1)*(den3/den1) + (num2)*(den3/den2);

    // Calling function to convert final fraction
    // into it's simplest form
    lowest(den3,num3);
}

// Driver program
int main()
{
    int num1=1, den1=500, num2=2, den2=1500, den3, num3;
    addFraction(num1, den1, num2, den2, num3, den3);
    printf("%d/%d + %d/%d is equal to %d/%d\n", num1, den1,
                                   num2, den2, num3, den3);
    return 0;
}

Java

// Java program to add 2 fractions

class GFG{
// 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 convert the obtained fraction
// into it's simplest form
static void lowest(int den3, int num3)
{
    // Finding gcd of both terms
    int common_factor = gcd(num3,den3);

    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3/common_factor;
    num3 = num3/common_factor;
    System.out.println(num3+"/"+den3);
}

//Function to add two fractions
static void addFraction(int num1, int den1,
                        int num2, int den2)
{
    // Finding gcd of den1 and den2
    int den3 = gcd(den1,den2);

    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1*den2) / den3;

    // Changing the fractions to have same denominator
    // Numerator of the final fraction obtained
    int num3 = (num1)*(den3/den1) + (num2)*(den3/den2);

    // Calling function to convert final fraction
    // into it's simplest form
    lowest(den3,num3);
}

// Driver program
public static void main(String[] args)
{
    int num1=1, den1=500, num2=2, den2=1500;
    System.out.print(num1+"/"+den1+" + "+num2+"/"+den2+" is equal to ");
    addFraction(num1, den1, num2, den2);
}
}
// This code is contributed by mits

Python3

# Python3 program to add 2 fractions

# Function to return gcd of a and b
def gcd(a, b):
    if (a == 0):
        return b;
    return gcd(b % a, a);

# Function to convert the obtained
# fraction into it's simplest form
def lowest(den3, num3):

    # Finding gcd of both terms
    common_factor = gcd(num3, den3);

    # Converting both terms
    # into simpler terms by
    # dividing them by common factor
    den3 = int(den3 / common_factor);
    num3 = int(num3 / common_factor);
    print(num3, "/", den3);

# Function to add two fractions
def addFraction(num1, den1, num2, den2):

    # Finding gcd of den1 and den2
    den3 = gcd(den1, den2);

    # Denominator of final
    # fraction obtained finding
    # LCM of den1 and den2
    # LCM * GCD = a * b
    den3 = (den1 * den2) / den3;

    # Changing the fractions to
    # have same denominator Numerator
    # of the final fraction obtained
    num3 = ((num1) * (den3 / den1) +
            (num2) * (den3 / den2));

    # Calling function to convert
    # final fraction into it's
    # simplest form
    lowest(den3, num3);

# Driver Code
num1 = 1; den1 = 500;
num2 = 2; den2 = 1500;

print(num1, "/", den1, " + ", num2, "/",
      den2, " is equal to ", end = "");
addFraction(num1, den1, num2, den2);

# This code is contributed by mits

C#

// C# program to add 2 fractions

class GFG{
// 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 convert the obtained fraction
// into it's simplest form
static void lowest(int den3, int num3)
{
    // Finding gcd of both terms
    int common_factor = gcd(num3,den3);

    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3/common_factor;
    num3 = num3/common_factor;
    System.Console.WriteLine(num3+"/"+den3);
}

//Function to add two fractions
static void addFraction(int num1, int den1, int num2, int den2)
{
    // Finding gcd of den1 and den2
    int den3 = gcd(den1,den2);

    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1*den2) / den3;

    // Changing the fractions to have same denominator
    // Numerator of the final fraction obtained
    int num3 = (num1)*(den3/den1) + (num2)*(den3/den2);

    // Calling function to convert final fraction
    // into it's simplest form
    lowest(den3,num3);
}

// Driver program
public static void Main()
{
    int num1=1, den1=500, num2=2, den2=1500;
    System.Console.Write(num1+"/"+den1+" + "+num2+"/"+den2+" is equal to ");
    addFraction(num1, den1, num2, den2);
}
}
// This code is contributed by mits

PHP

<?php
// PHP program to add
// 2 fractions

// Function to return
// gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}

// Function to convert the
// obtained fraction into
// it's simplest form
function lowest(&$den3, &$num3)
{
    // Finding gcd of both terms
    $common_factor = gcd($num3, $den3);

    // Converting both terms
    // into simpler terms by
    // dividing them by common factor

    $den3 = (int)$den3 / $common_factor;
    $num3 = (int) $num3 / $common_factor;
}

// Function to add
// two fractions
function addFraction($num1, $den1, $num2,
                     $den2, &$num3, &$den3)
{
    // Finding gcd of den1 and den2
    $den3 = gcd($den1, $den2);

    // Denominator of final
    // fraction obtained finding
    // LCM of den1 and den2
    // LCM * GCD = a * b
    $den3 = ($den1 * $den2) / $den3;

    // Changing the fractions to
    // have same denominator Numerator
    // of the final fraction obtained
    $num3 = ($num1) * ($den3 / $den1) +
            ($num2) * ($den3 / $den2);

    // Calling function to convert
    // final fraction into it's
    // simplest form
    lowest($den3, $num3);
}

// Driver Code
$num1 = 1; $den1 = 500;
$num2 = 2; $den2 = 1500;
$den3; $num3;
addFraction($num1, $den1, $num2,
            $den2, $num3, $den3);
echo $num1, "/", $den1, " + ",
     $num2,"/", $den2, " is equal to ",
               $num3, "/", $den3, "\n";

// This code is contributed by aj_36
?>

Javascript

<script>

// Javascript program to add 2 fractions

// Function to return gcd of a and b

const gcd = (a, b) => {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to convert the
// obtained fraction into
// it's simplest form

const lowest = (den3, num3) => {
    // Finding gcd of both terms
    let common_factor = gcd(num3, den3);

    // Converting both terms
    // into simpler terms by
    // dividing them by common factor

    den3 = parseInt(den3 / common_factor);
    num3 = parseInt(num3 / common_factor);

    document.write(`${num3}/${den3}`)
}


// Function to add two fractions
const addFraction = (num1, den1, num2, den2) => {
    // Finding gcd of den1 and den2
    let den3 = gcd(den1, den2);

    // Denominator of final
    // fraction obtained finding
    // LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1 * den2) / den3;

    // Changing the fractions to
    // have same denominator Numerator
    // of the final fraction obtained
    let num3 = ((num1) * (den3 / den1) +
            (num2) * (den3 / den2));

    // Calling function to convert
    // final fraction into it's
    // simplest form
    lowest(den3, num3);
}

// Driver Code
let num1 = 1;
let den1 = 500;
let num2 = 2;
let den2 = 1500;

document.write(`${num1}/${den1} + ${num2}/${den2} is equal to `);

addFraction(num1, den1, num2, den2);

// This code is contributed by _saurabh_jaiswal

</script>

Output :

1/500 + 2/1500 is equal to 1/300

Time Complexity: O(log(min(a, b)), where a and b are two integers.

Auxiliary Space: O(1), no extra space required so it is a constant.

See below for doing the same using library functions. 
Ratio Manipulations in C++ | Set 1 (Arithmetic)
This article is contributed by Rahul Agrawal .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.