Revision: 34585
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at October 25, 2010 23:58 by skiabox
Initial Code
#include <stdbool.h>
#include <stdio.h>
//gcf function - return gcd of two numbers
int gcd(int n, int m)
{
int gcd, remainder;
while (n != 0)
{
remainder = m % n;
m = n;
n = remainder;
}
gcd = m;
return gcd;
}//end gcd function
int main (int argc, const char * argv[]) {
// insert code here...
//--declarations
int number1, number2;
int newNumber1, newNumber2;
//--get user input
printf("Enter a fraction: ");
scanf("%d/%d", &number1, &number2);
//--calculations
//find the gcd of numerator and denominator
//then divide both the numerator and denominator by the GCD
newNumber1 = number1 / gcd(number1, number2);
newNumber2 = number2 / gcd(number1, number2);
//--results
printf("In lowest terms: %d/%d", newNumber1, newNumber2);
}
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Initial Description
To reduce a fraction to lowest terms, first we compute the GCD(greatest common divisor) of the numerator and denominator. Then we divide both the numerator and denominator by the GCD.
Initial Title
Reduce a fraction to lowest terms
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Initial Language
C