How do you get C programming to reduce a fraction?

Answer 1

#include "stdio.h"

int gcd(int a, int b);

void reduce(int* numerator, int* denominator);

int main(int argc, char* argv[]) {

int a, b;

fscanf(stdin, "%d/%d", &a, &b);

reduce(&a, &b);

fprintf(stdout, "%d/%d", a, b);

return 0;

}

int gcd(int a, int b) {

int c;

while (b) {

c = a % b;

a = b;

b = c;

}

return a;

}

void reduce(int* numerator, int* denominator) {

int g = gcd(*numerator, *denominator);

*numerator /= g;

*denominator /= g;

}

You get the greatest common divisor between the numerator and denominator and divide them by it.

Reduce uses integer pointers so that changes the numerator and denominator to their reduce form.

Answer 2

There are probably better or easier ways, but the following works.

#include <iostream>

using namespace std;

void Simplify( int * Num, int * Den )

{

int H = 0;

bool b = true;

while( b )

{

b = !b;

for( H = 10; H > 1; --H )

{

if( b = ((*Num) %H 0 ))

{

*Num /= H;

*Den /= H;

break;

}

}

}

}

int main()

{

int N = 0;

int D = 0;

printf( "Input the numerator: " );

scanf( "%u", &N );

printf( "Input the denominator: " );

scanf( "%u", &D );

int n = N;

int d = D;

Simplify( &n, &d );

printf( "%u/%u simplifies to %u/%u\n", N, D, n, d );

return( 0 );

}