Modify the program so that the user enters both fractions at the same time, sepa

Question

Modify the program so that the user enters both fractions at the same time, separated by a plus sign:

Enter two fractions separated by a plus sign: 1/2+1/4

The sum is 3/4

#include
#include

int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom;

printf("Enter first fraction: ");
scanf("%d/%d", &num1, &denom1);

printf("Enter second fraction: ");
scanf("%d/%d", &num2, &denom2);

result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
printf("The sum is %d/%d ", result_num, result_denom);

return 0;
}

Explanation / Answer

I think something like the following would suffice:
#include <stdio.h>

int gcd(int a, int b) /* euclidian gcd */
{
if(b == 0) return a;
return gcd(b, a % b);
}

int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom, d; /* added d for divisor */

printf("Enter two fractions separated by a plus sign: ");
scanf("%d/%d+%d/%d", &num1, &denom1, &num2, &denom2);

result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
d = gcd(result_num, result_denom);
printf("The sum is %d/%d ", result_num/d, result_denom/d); /* divide by divisor */

return 0;
}

So whats happening? Well first, the user can now input the desired input of a/b+c/d

Secondly, given those values, it would have calculated (ad+b*c)/(b*d) ... that isn't exactly what we want from the instructions. For if it was 1/2+1/4, the output would have been 6/8.

So we introduce the procedure "GCD" which basically follows Euclid's rule for finding the greater common divisor of two numbers a and b. Basically put, if the output was caculated to be 6/8, gcd of 6 and 8 would find that both are divisible by 2, and so dividing 6 and 8 by 2 would give 3 and 4, as the instruction desired.

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