C++ Long Division -

whilst working on personal project of mine, came across need divide 2 large arbitrary numbers (each number having 100 digits). wrote out basic code division (i.e., answer = a/b, , b imputed user)and discovered has precision of 16 digits! may obvious @ point im not coder!

so searched internet , found code that, far can tell, uses traditional method of long division making string(but honest im not sure im quite confused it). upon running code gives out incorrect answers , wont work @ if a>b. im not sure if there's better way solve problem method in code below!? maybe there's simpler code??

so need write code, in c++, divide 2 large numbers. or suggestions appreciated!

#include <iostream>  #include <iomanip> #include <cmath>  using namespace std;  //avoids having use std:: cout/cin    int main (int argc, char **argv) {     string dividend, divisor, difference, a, b, s, tempstring = ""; // , b used store dividend , divisor.     int quotient, inta, intb, diff, tempint = 0;     char d;      quotient = 0;      cout << "enter dividend?  ";      //larger number (on top)     cin  >> a;     cout << "enter divisor?  ";           //smaller number (on bottom)     cin  >> b;      //making strings same length adding 0's beggining of string.     while (a.length() < b.length())  = '0'+a;          // has less digits b add 0's      while (b.length() < a.length())  b = '0'+b;          // b has less digits add 0's      inta = a[0]-'0';                        // getting first digit in both strings     intb = b[0]-'0';      //if a<b print remainder out (a) , return 0     if (inta < intb)     {         cout << "quotient: 0 " << endl << "remainder: " << << endl;     }       else     {         = '0'+a;         b = '0'+b;         diff = intb;         //s = b;        // while ( s >= b )                  {             (int = a.length()-1; i>=0; i--)            // subtraction until end of string             {             inta = a[i]-'0';                    // converting ascii int, used munipulation             intb = b[i]-'0';                if (inta < intb)                 // borrow if needed             {               a[i-1]--;                         //borrow next digit               a[i] += 10;             }             diff = a[i] - b[i];           char d = diff+'0';            s = d + s;                           //this + appending 2 strings, not performing addition.              }          quotient++;          = s;         // strcpy (a, s);         }           while (s >= b); // fails after dividing 3 x's      cout << "s string: " << s << endl;      cout << "a string: " << << endl;      cout << "quotient:  " << quotient << endl;      //cout << "remainder: " << s << endl;      }      system ("pause");     return 0;     cin.get(); // allows user enter variable without instantly ending program     cin.get(); // allows user enter variable without instantly ending program 

}

there better methods that. subtractive method arbitrarily slow large dividends , small divisors. canonical method given algorithm d in knuth, d.e., the art of computer programming, volume 2, i'm sure find online. i'd astonished if wasn't in wikipedia somewhere.