Last M digits of expression A^N

Dave Angel davea at ieee.org
Sat Feb 6 23:59:44 CET 2010


monkeys paw wrote:
> <div class="moz-text-flowed" style="font-family: -moz-fixed">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
>
Because the N= N/2 operation in the loop means that the number of loops 
is equal to the number of  bits in N, or the log-base-2 of N.

DaveA



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