# Adding a series of fractions

 Patricia Ghann · January 6, 2015 Hello Bucky, Happy New Year. Please I want to write a program in c++ to add a series of fractions and am finding it difficult so I need your help please. I want to add for example 1/3+3/5+5/7+7/9+9/11+...............+97/99.

## Replies

- page 1

 Patricia Ghann · January 6, 2015 This is what I have come out with so far.#includeusing namespace std;Hello, what seems to be the problem? This program builds and runs alright, but its not doing what I want. I need help have to submit by Thursday. Thanks.#includeusing namespace std;int main(){int num = 1;int den = 3;float fraction(num/den);int numberOfFractionEntered = 0;int sum = 0;cout<< "Enter first fraction or -1 to quit" <> fraction;while (fraction < 1/3){sum = sum + fraction;numberOfFractionEntered++;cout<< "Enter next fraction or -1 to quit" <>fraction;}cout<using namespace std;int main(){int num = 0;int den = 0;float fraction;fraction =float(num)/float(den);int numberOfFractionEntered = 0;double sum = 0;The court<< "Enter first fraction or -1 to quit" <> fraction;while (fraction !=-1){//sum = sum + fraction;//numberOfFractionEntered++;cin>>num;cout<> den;cout<>fraction;sum = sum + fraction;numberOfFractionEntered;}cout< fractionResult.num = fraction1.num*fraction2.den + fraction2.num*fraction1.den> fractionResult.den = fraction1,den*fraction2.denan example: if you want to calculate 1/3 + 3/5:> fractionResult.num = 1*5+3*3 = 5+9 = 14> fractionResult.den = 3*5 = 15resulting in a fraction result struct representing 14/15, save the value and add on the next input fraction Number Double07 · January 7, 2015 It would be easier if you requested the user to input the numerator, then the denominator.  I tried the fraction input method using your code, but I'm not sure what is going on there myself.  It's more complicated, it can probably be done, but I don't know how to do it.  Also, both requests (cout statements of numerator and denominator)  need to be inside the while loop so it knows to end when it user inputs the -1 value. c student · January 7, 2015 if it's not too late, here is my functioning code written in C since i dont code in c++ however should be easy to translate across.  i have added comments to show psuedo code for better understanding:``#include #include // define a struct for a fraction type to hold// both numerator and denominatortypedef struct {    int num;    int den;} fraction;fraction fracAdd (fraction, fraction);int main (void) {    // request amount of fractions    puts ("Please enter the amount of fractions");    // acquire number of fractions to be calculated    int amount = getchar()-48;    // declare array of fraction types according to the previous line    fraction frac[amount];    // declare a resultant fraction    fraction result;    // loop to enter in each fraction    for (int n = 0; n < amount; n++) {        // request numerator for fraction n        printf ("Please enter in the numerator for fraction %d\n", n+1);        // input into fraction array n for numerator member        scanf ("%d", &frac[n].num);        // request denominator for fraction n        printf ("Please enter in the denominator for fraction %d\n", n+1);        // input into fraction array n for denominator member        scanf ("%d", &frac[n].den); if (frac[n].den == 0) { fprintf (stderr, "Zero is not a valid denominator value"); exit (1); }        // this is just to place the first fraction into result        if (n == 0) {            result = frac[n];        // then for the rest of the fractions, add them onto the result        } else {            result = fracAdd (result, frac[n]);        }    }    // print result with numerator and denominator    printf ("Result: %d/%d\n", result.num, result.den);    return 0;}// function to add fractionsfraction fracAdd (fraction a, fraction b) {    // declare resultant fraction    fraction result;    // adding numerators    result.num = a.num*b.den+b.num*a.den;    // adding common denominator    result.den = a.den*b.den;    // return resultant    return result;}``good lucknote: does not simplify fractions.  if you wish to simplify the fractions, just make a function to calculate gcd (i think).
• 1
• 2

## C++

124,375 followers