Add two fractions in C++ In this article we will discuss the program for add two fractions in C++. For this purpose we need to ask the user to enter two fractional values where each fraction must consist a Numerator and a Denominator.
Program to add two fractions Addtion of two fractions in C++ Concept For understanding this in a better way lets suppose an example :
Suppose, First fraction consist of 1 as numerator and 3 as denominator, and Second fraction consist of 3 as numerator and 9 as denominator.
(1 / 3) + (3 / 9) = (6 / 9) = (2 / 3)
Find LCM of 3 and 9 and the result will be 9. Multiply 3 in first fraction : (3 / 9) + (3 / 9) Add both fractions and then the result will be : (6 / 9) Now simplify it by finding the HCF of 6 and 9 and the result will be 3. So divide 6 and 9 by 3 and then the result will be : (2 / 3) This will be your simplified answer for the given problem. Algorithm Initialize variables of numerator and denominator Take user input of two fraction Find numerator using this condition (n1d2) +(d1n2 ) where n1,n2 are numerator and d1 and d2 are denominator Find denominator using this condition (d1*d2) for lcm Calculate GCD of a this new numerator and denominator Display a two value of this condition x / gcd, y/gcd Related Pages Permutations in which n people can occupy r seats in a classroom
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// GCD function int findGCD(int n1, int n2) { int gcd; for(int i=1; i <= n1 && i <= n2; i++) { if(n1%i==0 && n2%i==0) gcd = i; } return gcd; }
// Main Program int main() { int num1,den1;
//user input first fraction
cout << "Enter numerator and denominator of first number : "; cin >> num1 >> den1;
int num2,den2;
//user input second fraction
cout << "Enter numerator and denominator of second number: "; cin >> num2 >> den2;
//finding lcm of the denominators
int lcm = (den1*den2)/findGCD(den1,den2);
//finding the sum of the numbers
int sum=(num1*lcm/den1) + (num2*lcm/den2);
//normalizing numerator and denominator of result
int num3=sum/findGCD(sum,lcm);
lcm=lcm/findGCD(sum,lcm);
//printing output
cout<<num1<<"/"<<den1<<" + "<<num2<<"/"<<den2<<" = "<<num3<<"/"<<lcm;
return 0;
} Output Enter numerator and denominator of first number : 3 4 Enter numerator and denominator of second number: 5 6 3/4 + 5/6 = 19/12