how to work with fractions in c ? (help neded);

jabka · 11-17-2006, 11:53 PM

Howdy..

i got hw and ine asinment is like that :
the user enter two numbers (a and b) and the prog will show the fraqtion in his smallest form :
Exm : 4 6
the ouptut will be 2/3
so what i tried was :
//a,b,c are int the if(b<a) is requeired
if (b>a){
c= b / a;
a=a/c;
printf("%d/%d ",a,c);
}
it fine for 2/4 and 5/25 but for more complex form :
4 6
i don't know what to do

pda_h4x0r · 11-18-2006, 12:58 AM

Code:

#define Min(x,y) ((x)<=(y)?(x):(y))
#define Abs(x) ((x)<0?-(x):(x))

void main(int argc, char** argv) {

// assume a and b have been entered
int min = Min(Abs(a), Abs(b));
while(min--) {
if(a%min == 0 && b%min == 0) {
a /= min; b /= min;
}
}
printf("%d/%d", a, b);

}

At least, that's how I'd do it.

xhi · 11-18-2006, 09:46 AM

Quote:

Originally Posted by pda_h4x0r

Code:

#define Min(x,y) ((x)<=(y)?(x):(y))
#define Abs(x) ((x)<0?-(x):(x))

void main(int argc, char** argv) {

// assume a and b have been entered
int min = Min(Abs(a), Abs(b));
while(min--) {
if(a%min == 0 && b%min == 0) {
a /= min; b /= min;
}
}
printf("%d/%d", a, b);

}

At least, that's how I'd do it.

void main() ? bahh.. sorry i just threw up a little. main always should return int.

jabka · 11-18-2006, 11:10 AM

np .. thnx for the help ..

acid_kewpie · 11-19-2006, 06:59 AM

oh good grief, how does this kid learn anything if he's spoon fed his own coursework answers??? members should *NOT* ask homework questions, and other members should *NOT* provde full answers as and when they do get posted.

pda_h4x0r · 11-19-2006, 10:56 PM

LOL, but that's how WE learn to code

Joking, of course.

Quigi · 11-20-2006, 12:55 PM

pda_h4x0r: Inefficient for big numbers.

Acid_kewpie is right, it's Jabka's homework. But maybe a tiny clue would be OK? From middle school you remember that you divide numerator and denominator by their greatest common denominator. A few thousand years ago a Greek proposed a nifty GCD algorithm; I think the guy's name was Euclid.

pda_h4x0r · 11-20-2006, 03:06 PM

Yes, I am aware that it is HORRIBLY inefficient for large numbers (in fact, horribly inefficient in general--my algorithm is O(n) and Euclid's is O(log n)). However, the algorithm demonstrates a simplified, conceptual implementation, intended to show Jabka the concepts of the solution. This code snippet is analogous to the Bubble Sort algorithm: easy to understand for newbies but without a doubt one of the WORST algorithms of its kind, in terms of performance.