simplifiing fractions part 2
thanks everyone for your help, but i have a second question.... is there a way to simplify the fractions using loops? because that is about where i am in my programming. and i dont want to get ahead of my self. any assitance would be much appreciated thanks again.
Code:int gcd(int a, int b) { int r; while (b != 0) { r = a % b; a = b; b = r; } return a; }
this isn't the ideal solution, but it'll give you an idea of how to do it.
I have compiled this on MSVC++ 6.
hope this helpsCode:#include <iostream.h> #include <math.h> int main(void) { int i = 0; // we are assuming that numerator >= denominator int numerator = 9; int denominator = 12; int simplified = 0; int newNumerator = numerator; int newDenominator = denominator; while(!simplified) { int i = 0; for(i = 2; i < newNumerator; i++) { // if they are both divisible by the number if(newNumerator % i == 0 && newDenominator % i ==0) { // divide both by the number newNumerator /= i; newDenominator /= i; // reset i to make sure that the exit condition is met properly i = 0; break; } } // if we've made it all the way through the loop, we have no more factors if(i == newNumerator) { simplified++; } } cout << "The simplifed version of " << numerator << "/" << denominator; cout << " is " << newNumerator << "/" << newDenominator << endl; return(0); }
U.
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.
the greatest common divisor wasn't the answer he was looking for. didn't he want to simplify the fraction??
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.
Once you have the gcd all that's need to simplify is
Code:g = gcd(numer, denom) (numer / g) ----------- (denom / g)
i can't believe that i didn't see that the greatest common divisor is the way ...... i'm going to hang my head in shame after that!
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.
