Last M digits of expression A^N
mensanator at aol.com
Fri Feb 5 21:28:37 CET 2010
On Feb 5, 2:18 pm, Mark Dickinson <dicki... at gmail.com> wrote:
> On Feb 5, 8:14 pm, mukesh tiwari <mukeshtiwari.ii... at 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
> The built-in pow function does exactly this,
It doesn't do 'exactly' that. M is the number of digits.
If you want 17 digits, the third number should 10**17.
Using 17 directly will give you a 2-digit answer as shown.
Try this instead:
> if you give it three
> >>> pow(12345, 67891, 17)
> >>> 12345 ** 67891 % 17
> (Though this won't work for negative N.)
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