Last M digits of expression A^N

[monkeys paw]

mukesh tiwari 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.

How do you arrive at log N as the performance number?

> 
> 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