C++ Long Division - c++

Question

Whilst working on a personal project of mine, I came across a need to divide two very large arbitrary numbers (each number having roughly 100 digits).
So i wrote out the very basic code for division (i.e., answer = a/b, where a and b are imputed by the user)and quickly discovered that it only has a precision of 16 digits! It may be obvious at this point that Im not a coder!
So i searched the internet and found a code that, as far as i can tell, uses the traditional method of long division by making a string(but too be honest im not sure as im quite confused by it). But upon running the code it gives out some incorrect answers and wont work at all if a>b.
Im not even sure if there's a better way to solve this problem than the method in the code below!? Maybe there's a simpler code??
So basically i need help to write a code, in C++, to divide two very large numbers.
Any help or suggestions are greatly appreciated!
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std; //avoids having to use std:: with cout/cin
int main (int argc, char **argv)
{
string dividend, divisor, difference, a, b, s, tempstring = ""; // a and b used to store dividend and divisor.
int quotient, inta, intb, diff, tempint = 0;
char d;
quotient = 0;
cout << "Enter the dividend? "; //larger number (on top)
cin >> a;
cout << "Enter the divisor? "; //smaller number (on bottom)
cin >> b;
//making the strings the same length by adding 0's to the beggining of string.
while (a.length() < b.length()) a = '0'+a; // a has less digits than b add 0's
while (b.length() < a.length()) b = '0'+b; // b has less digits than a add 0's
inta = a[0]-'0'; // getting first digit in both strings
intb = b[0]-'0';
//if a<b print remainder out (a) and return 0
if (inta < intb)
{
cout << "Quotient: 0 " << endl << "Remainder: " << a << endl;
}
else
{
a = '0'+a;
b = '0'+b;
diff = intb;
//s = b;
// while ( s >= b )
do
{
for (int i = a.length()-1; i>=0; i--) // do subtraction until end of string
{
inta = a[i]-'0'; // converting ascii to int, used for munipulation
intb = b[i]-'0';
if (inta < intb) // borrow if needed
{
a[i-1]--; //borrow from next digit
a[i] += 10;
}
diff = a[i] - b[i];
char d = diff+'0';
s = d + s; //this + is appending two strings, not performing addition.
}
quotient++;
a = s;
// strcpy (a, s);
}
while (s >= b); // fails after dividing 3 x's
cout << "s string: " << s << endl;
cout << "a string: " << a << endl;
cout << "Quotient: " << quotient << endl;
//cout << "Remainder: " << s << endl;
}
system ("pause");
return 0;
cin.get(); // allows the user to enter variable without instantly ending the program
cin.get(); // allows the user to enter variable without instantly ending the program
}

Answer 1

There are much better methods than that. This subtractive method is arbitrarily slow for large dividends and small divisors. The canonical method is given as Algorithm D in Knuth, D.E., The Art of Computer Programming, volume 2, but I'm sure you will find it online. I'd be astonished if it wasn't in Wikipedia somewhere.