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
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