如何将2个大数相乘？

Question

i created a small console app that multiply 2 long int number. i don't know where my problem is. this app work fine until the number of digits is 3.

but if number of digit was bigger than 3 , the app's output is wrong. :(

please show me where my problem is that i solve it.

here is my code:

int digits (int n)
{
    int counter = 0;
    while (n > 0)
    {
        n/=10;
        counter++;
    }
    return counter;
}

long longMultiply(long a, long b)
{
    const int S = 3;
    int w,x,y,z;
    int n = max(digits(a),digits(b));

    if(a == 0 || b ==0) {
        return 0;
    } else if (n <= S) {
        return a*b;
    } else {
        int m = (n/2);

        //first number
        x = a/(10^m);
        y = a%(10^m);

        //second number
        w = b/(10^m);
        z = b%(10^m);

        return (longMultiply(x,w)*(10^(2*m)) + (longMultiply(x,z) + longMultiply(w,y)))*(10^m) + longMultiply(y,z) ;

    }
}

int main() {
    //digits(12345);
    cout << longMultiply(100,100);
    return 0;
}

Answer 1

10^m is not the m-th power of 10, in fact this is 10 xor'ed by m

You can use the pow function from cmath library instead ( http://www.cplusplus.com/reference/cmath/pow/ ), but it works on floating-point numbers.

Alternatively to get 10^m, you could simply multiply 1 m times by 10.

int m = (n/2);
long tenToM = 1;
for (int i=0; i<m; i++)
    tenToM *= 10;
long tenToTwoM = tenToM * tenToM;

and then instead of 10^m use tenToM and instead of 10^(2*m) use tenToTwoM

Answer 2

It seems like your problem is in the logic of the else portion. It works up to 3 digits because it is simply outputting the product when that fails it runs your else block which I am not sure I understand. What exactly is setting m = n/2 trying to do?

Answer 3

If the product is less than or equal to 10 ^ 18 ; you can simply use

  long long product = a * b ;

If a or b are greater than the range of long long ; one can simply take one as long long and another as string . Suppose a > 10^18 and b < 10^18 . The below code is valid only when b * 9 is in the range of long long .

       string a ; long long b ;
       cin >> a >> b ;
       reverse ( a.begin() , a.end() ) ;
       string prod ;
       long long temp ,carry ;
        temp = carry = 0 ;
       for ( i = 0 ; i < a.length() ; i++ ){
          temp =  (a[i] - '0') * b + carry ;
          prod += ( temp % 10 ) + '0' ;
          carry = temp / 10  ;
       }

        while ( carry != 0 ){
        prod += ( carry % 10 ) + '0' ;
        carry /= 10 ;
        }
        reverse ( prod.begin() , prod.end() ) ;
        cout << prod ; // this string contains the required product .

However if both are very big you can consider using a third party Big Integer Library. For external Big Integer Library you can consider using BOOST BigInteger Library , which is quite fast and highly tested.