Is there a good way to optimize the multiplication of two BigNums?

Question

I have a class BigNum:

struct BigNum{
    vector <int> digits;
    
    BigNum(vector <int> data){
        for(int item : data){d.push_back(item);}
    }
    
    int get_digit(size_t index){
        return (index >= d.size() ? 0 : d[index]);
    }
};

and I'm trying to write code to multiply two BigNums. Currently, I've been using the traditional method of multiplication, which is multiplying the first number by each digit of the other and adding it to a running total. Here's my code:

BigNum add(BigNum a, BigNum b){ // traditional adding: goes digit by digit and keeps a "carry" variable
    vector <int> ret;

    int carry = 0;
    for(size_t i = 0; i < max(a.digits.size(), b.digits.size()); ++i){
        int curr = a.get_digit(i) + b.get_digit(i) + carry;

        ret.push_back(curr%10);
        carry = curr/10;
    }

    // leftover from carrying values
    while(carry != 0){
        ret.push_back(carry%10);
        carry /= 10;
    }

    return BigNum(ret);
}

BigNum mult(BigNum a, BigNum b){
    BigNum ret({0});

    for(size_t i = 0; i < a.d.size(); ++i){
        vector <int> row(i, 0); // account for the zeroes at the end of each row

        int carry = 0;
        for(size_t j = 0; j < b.d.size(); ++j){
            int curr = a.d[i] * b.d[j] + carry;

            row.push_back(curr%10);
            carry = curr/10;
        }

        while(carry != 0){ // leftover from carrying
            row.push_back(carry%10);
            carry /= 10;
        }

        ret = add(ret, BigNum(row)); // add the current row to our running sum
    }

    return ret;
}

This code still works pretty slowly; it takes around a minute to calculate the factorial of 1000. Is there a better way to multiply two BigNums? If not, is there a better way to represent large numbers that will speed up this code?

Answer 1

If you use a different base, say 2^16 instead of 10, the multiplication will be much faster.

But getting to print in decimal will be longer.