如何高效计算N！使用定点大数库？

T2 = { (2N+1).(2N+3).(2N+5)...(4N-1) }.(2^N)/(N!)

// edit beg:
// Sorry, forget to copy sorted list of all primes up to max n here it is
// end of table is marked with 0
// Primes are in DWORDs so they only 4Byte per number
// so the table is very small compared with lookup table for the same max n!
// and also primes are needed for many other routines in bignum
// can compute n! for n <= max prime in table
DWORD _arithmetics_primes[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,0};
// edit end.

longnum fact(const DWORD &amp;x)
    {
    if (x<=4)
        {
        if (x==4) return 24;
        if (x==3) return  6;
        if (x==2) return  2;
        if (x==1) return  1;
        if (x==0) return  1;
        }
    int N4,N2,p,i,j,e; longnum c,pp;
    N4=(x>>2)<<2;
    N2=N4>>1;
    c=fact(N2); c*=c;                 // c=((2N)!)^2;
    for (i=0;;i++)                    // c*= T2
        {
        p=_arithmetics_primes[i];
        if (!p) break;
        if (p>N4) break;
        for (e=0,j=N4;j;e+=j&amp;1,j/=p);
        if (e)                        // c*=p^e
            {
            if (p==2) c<<=e;
            else for (pp=p;;)
                {
                if (int(e&amp;1)) c*=pp;
                e>>=1; if (!e) break;
                pp*=pp;
                }
            }
        }
    for (i=N4+1;i<=x;i++) { c*=i; } c.round();
    return c;
    }