Calculating sum of geometric series (mod m)

Question

I have a series

S = i^(m) + i^(2m) + ...............  + i^(km)  (mod m)   

0 <= i < m, k may be very large (up to 100,000,000),  m <= 300000
         


        

Answer 1

This is the algorithm for a similar problem I encountered

You probably know that one can calculate the power of a number in logarithmic time. You can also do so for calculating the sum of the geometric series. Since it holds that

1 + a + a^2 + ... + a^(2*n+1) = (1 + a) * (1 + (a^2) + (a^2)^2 + ... + (a^2)^n),

you can recursively calculate the geometric series on the right hand to get the result.

This way you do not need division, so you can take the remainder of the sum (and of intermediate results) modulo any number you want.