reducing a fraction in c

Question

#include <stdio.h>
#include <stdlib.h>
int main(){
    system("color f0");
    int a,b,c,d,e,f,g,h,z;
    printf("First numerator:");
    scanf("%d",&a);
    printf("First denominator:");
    scanf("%d",&b);
    printf("Second numerator:");
    scanf("%d",&c);
    printf("Second denominator:");
    scanf("%d",&d);

    a=a*d;
    c=c*b;
    e=a+c;
    f=b*d;
    printf("Simple form:%d/%d\n",e,f);
    return 0;
}

This is my code i want to reduce that simple fraction to the lowest possible but without using maths library

thats an assignment i have to do the homework but i am unable to do so we cant use the math library — Sanee Reimoo, Nov 15 '16 at 17:25
@SaneeReimoo OK, how would you do it *with* the math library? — Eugene Sh., Nov 15 '16 at 17:26
**Don't panic**. We can still help. I'll make a psuedo code algorithm for you :) — SIGSTACKFAULT, Nov 15 '16 at 17:27
pseudo code will help. i heard about euclidean too but i am unable to get it. Waiting for pseudo code — Sanee Reimoo, Nov 15 '16 at 17:29
`a/b` is reduced to `(a/gcd(a,b)) / (b/gcd(a,b))` . Now look for the `gcd` algorithm (Euclidean). There are plenty around there. — Eugene Sh., Nov 15 '16 at 17:32

MarianD · Accepted Answer · 2019-01-04T19:58:25.477

You did some weird things with your code:

First, you ask a user for two nominators and two denominators.

So your lines

printf("Second numerator:");
scanf("%d",&c);
printf("Second denominator:");
scanf("%d",&d);

are superfluous and you may delete them.

Second, you lines

a=a*d;
c=c*b;
e=a+c;
f=b*d;

are horrible - a reader (and you) will be lost in your amount of 1-letter names.

So why don't give a variable for a nominator the name nominator and for a variable for a denominator the denominator? And so on?

So replace your whole code with this one:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int numerator, denominator, smaller, i;

    system("color f0");

    printf("Numerator  : ");
    scanf("%d",&numerator);

    printf("Denominator: ");
    scanf("%d",&denominator);

    printf("\nOriginal fraction: %d/%d\n", numerator, denominator);

    // Method: We get the smaller number from numerator and denominator
    // and will try all numbers from it decreasing by 1 for common divisor

    smaller = numerator < denominator ? numerator : denominator;

    for (i = smaller; i > 1; --i)
        if (numerator % i == 0 && denominator % i ==0)
        {
            numerator   /= i;
            denominator /= i;
            break;
        }

    printf("Reduced fraction : %d/%d\n", numerator, denominator);
    return 0;
}

Marian u r rude but awesome thnx for ur help. :D – Sanee Reimoo Nov 15 '16 at 18:25 — Sanee Reimoo, Nov 15 '16 at 18:25

score 0 · Answer 2 · answered Nov 15 '16 at 17:36

0

Stack overflow != /r/homeworkhelp

A psuedocode algorithm:

get a fraction in the form of a/b

while a/b is not in lowest terms:
    find a common divisor, k
    divide a by k  
    divide b by k
end while

to check if something is lowest terms:

if A == B:
    [not lowest terms]
if A is a multiple of B:
    [not lowest terms]
if there is a common divisor: // this catches the first two.
    [not lowest terms]
else:
    [lowest terms]

I'm going to leave the divisor-finder to you, otherwise it would be too easy.

answered Nov 15 '16 at 17:36

SIGSTACKFAULT

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I'm *pretty* sure this works. Yell at me if it doesn't, I'll fix it. – SIGSTACKFAULT Nov 15 '16 at 17:37
i will get back to u people in a bit. hope for the best – Sanee Reimoo Nov 15 '16 at 17:38
r=e%f; if (r == 0) printf("Reduced form:%d/%d",e/e,f/e); else e=f; f=r; r=e%f; printf("Reduced form:%d/%d",e/r,f/r); return 0; this is what i have done but its not reducing correctly – Sanee Reimoo Nov 15 '16 at 17:47

reducing a fraction in c

2 Answers2