int work()
{
    if(n==1)return 1;    
    int two=B(2)-B(5);
    int three=0,seven=0,nine=0;
    int nn=n;
    while(nn)
    {
        int n1=nn;
        while(n1)
        {
            int d=n1/10;
            three=(three+d)&3;
            seven=(seven+d)&3;
            nine=(nine+d)&3;
            d=n1%10;
            if(d>=3)three++;
            if(d>=7)seven++;
            if(d>=9)nine++;
            n1/=2;
        }
        nn/=5;
    }
    two%=4;
    three%=4;
    seven%=4;
    nine%=4;
    return m2[two]*m3[three]*m7[seven]*m9[nine]%10;
}

• Re:easy code (642) qylsw 2010-07-09 19:17:54