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)
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