The given inputs are fractions, namely a/b and c/d, where a, b, c and d can be any integer value except 0. The task is to combine these two fractions are added to produce their final sum.
The fraction is represented by −
Input-: 1/4 + 2/12 Output-: 5/12 Since both the fractions denominators are unequal so to make them equal either GCD or LCM can be calculated. So in this case by multiplying the denominator which is 4 by 3 we can make them equal (1 * 3) / (4 * 3) = 3 / 12 Add both the terms: 3 / 12 + 2 / 12 = 5 / 12 Input-: 1/4 + 2/4 Output-: 3/4 Since both the terms have same denominator they can be directly added
In function int gcd(int a, int b) Step 1-> If a == 0 then, return b Step 2-> Return gcd(b%a, a) In function void smallest(int &den3, int &n3) Step 1-> Declare and initialize common_factor as gcd(n3,den3) Step 2-> Set den3 = den3/common_factor Step 3-> Set n3 = n3/common_factor In Function void add_frac(int n1, int den1, int n2, int den2, int &n3, int &den3) Step 1-> Set den3 = gcd(den1,den2) Step 2-> Set den3 = (den1*den2) / den3 Step 3-> Set n3 = (n1)*(den3/den1) + (n2)*(den3/den2) Step 4-> Call function smallest(den3,n3) In Function int main() Step 1-> Declare and initialize n1=1, den1=4, n2=2, den2=12, den3, n3 Step 2-> Call add_frac(n1, den1, n2, den2, n3, den3) Step 3-> Print the values of n1, den1, n2, den2, n3, den3
#include <stdio.h> int gcd(int a, int b) { if (a == 0) return b; return gcd(b%a, a); } void smallest(int &den3, int &n3) { // Finding gcd of both terms int common_factor = gcd(n3,den3); den3 = den3/common_factor; n3 = n3/common_factor; } void add_frac(int n1, int den1, int n2, int den2, int &n3, int &den3) { // to find the gcd of den1 and den2 den3 = gcd(den1,den2); // LCM * GCD = a * b den3 = (den1*den2) / den3; // Changing the inputs to have same denominator // Numerator of the final fraction obtained n3 = (n1)*(den3/den1) + (n2)*(den3/den2); smallest(den3,n3); } // Driver program int main() { int n1=1, den1=4, n2=2, den2=12, den3, n3; add_frac(n1, den1, n2, den2, n3, den3); printf("%d/%d + %d/%d = %d/%d</p><p>", n1, den1, n2, den2, n3, den3); return 0; }
1/4 + 2/12 = 5/12
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