C++ (x86_64-specific) - 220000

(easily portable to x86-32, porting to other architectures implies a significant speed loss)

Here's my own micro-implementation of long arithmetic.
The calculation itself takes 10 seconds, and while the output is in easily printable form (see the operator<< overload), it takes some more time to print it.

#include <vector>
#include <iostream>
#include <stdint.h>
#include <ctime>

typedef uint64_t digit;
typedef std::vector<digit> number;

std::ostream &operator<<(std::ostream &s, const number &x)
{
    std::vector<char> o;
    size_t size = x.size() * 21;
    o.resize(size);
    size_t lud = 0;
    for(number::const_reverse_iterator i = x.rbegin(), end = x.rend(); i != end; i++)
    {
        digit carry = 0;
        int j;
        for(j = 0; j <= lud || carry; j++)
        {
            digit r = o[j] * (1LL << 32) + carry;
            o[j] = r % 10;
            carry = r / 10;
        }
        lud = j;
        carry = 0;
        for(j = 0; j <= lud || carry; j++)
        {
            digit r = o[j] * (1LL << 32) + carry;
            o[j] = r % 10;
            carry = r / 10;
        }
        lud = j;
        carry = *i;
        for(j = 0; carry; j++)
        {
            digit r = o[j] + (carry % 10);
            carry /= 10;
            carry += r / 10;
            o[j] = r % 10;
        }
        if(j > lud)
            lud = j;
    }
    for(int j = lud; j--;)
        s.put(o[j] + '0');
    return s;
}

inline digit dmul(digit x, digit y, digit &carry)
{
    asm("mulq %2" : "+a"(x), "=d"(carry) : "r"(y));
    return x;
}
inline digit dadd(digit x, digit y, digit &carry)
{
    asm("movq $0, %1; addq %2, %0; adcq %1, %1" : "+r"(x), "=r"(carry), "+r"(y));
    return x;
}

void multiply(number &x, digit y)
{
    x.resize(x.size() + 21);
    digit carry = 0;
    for(number::iterator i = x.begin(), end = x.end(); i != end; i++)
    {
        digit nc, res = dmul(*i, y, nc);
        *i = dadd(res, carry, carry);
        carry += nc;
    }
    size_t sz = x.size();
    for(number::const_reverse_iterator i = x.rbegin(), end = x.rend(); i != end; i++)
    {
        if(*i)
            break;
        sz--;
    }
    x.resize(sz);
}

int main()
{
    const int r = 220000;
    clock_t start = clock();
    number n;
    n.push_back(1);
    for(digit a = 2; a <= r; a++)
        multiply(n, a);
    std::cout << "Took: " << (clock() - start)/((double)CLOCKS_PER_SEC) << std::endl;
    std::cout << n << std::endl;
}