On Fri, Feb 5, 2010 at 3:14 PM, mukesh tiwari <mukeshtiwari.ii...@gmail.com> wrote: > Hello everyone. I am kind of new to python so pardon me if i sound > stupid. > I have to find out the last M digits of expression.One thing i can do > is (A**N)%M but my A and N are too large (10^100) and M is less than > 10^5. The other approach was repeated squaring and taking mod of > expression. Is there any other way to do this in python more faster > than log N. > > def power(A,N,M): > ret=1 > while(N): > if(N%2!=0):ret=(ret*A)%M > A=(A*A)%M > N=N//2 > return ret > -- > http://mail.python.org/mailman/listinfo/python-list >
http://docs.python.org/3.1/library/decimal.html#decimal.Context.power -- Gerald Britton -- http://mail.python.org/mailman/listinfo/python-list
