Given two integers Num1 and Num2 as input. These two integers can be represented as fractions Num1/Num2. The goal is to reduce this fraction to its simplest form.
We will calculate the greatest common divisor of these two numbers.
Divide both numbers by their greatest common divisor.
Set these two variables to the quotient after division.
The simplest fraction will be Num1/Num2.
Input - Num1=22 Num2=10
Output - Num1 = 11 Num2 = 5
The simplest fraction is: 11/5
Explanation- The greatest common divisor of 22 and 10 is 2.
22/2=11, 10/2=5
The simplest fraction is 11/5
Input- Num1=36 Num2=40
Output- Num1 = 9 Num2 = 10
The simplest fraction is: 9/10
Explanation - 36 and The greatest common divisor of 40 is 4.
40/4=10, 36/4=9
The simplest fraction is 9/10
In In this method, we first use a recursive method to calculate the greatest common divisor of the input numbers. Divide two numbers by their greatest common divisor and get the quotients, which will be part of the simplest fraction.
Get the input variables Num1 and Num2.
The function findGCD(int a, int b) accepts num1 and num2 and returns the greatest common divisor of the two.
If b is 0, return a, otherwise return findGCD(b,a%b).
The function lowestFraction(int num1, int num2) accepts two numbers as input and prints the simplest fraction.
Use the variable denom to represent the greatest common divisor.
Set num1=num1/denom and num2=num2/denom.
Print num1 and num2.
Print the simplest fraction as num1/num2.
#include <bits/stdc++.h> using namespace std; int findGCD(int a, int b) { if (b == 0) return a; return findGCD(b, a % b); } void lowestFraction(int num1, int num2){ int denom; denom = findGCD(num1,num2); num1/=denom; num2/=denom; cout<< "Num1 = " << num1<<endl; cout<< "Num2 = " << num2<<endl; cout<< "Lowest Fraction : "<<num1<<"/"<<num2; } int main(){ int Num1 = 14; int Num2 = 8; lowestFraction(Num1,Num2); return 0; }
If we run the above code it will generate the following output
Num1 = 7 Num2 = 4 Lowest Fraction : 7/4
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