Fast way to manually mod a number
Posted
by Nikolai Mushegian
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Published on 2009-06-12T17:32:13Z
Indexed on
2010/05/09
21:08 UTC
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I need to be able to calculate (a^b) % c for very large values of a and b (which individually are pushing limit and which cause overflow errors when you try to calculate a^b). For small enough numbers, using the identity (a^b)%c = (a%c)^b%c works, but if c is too large this doesn't really help. I wrote a loop to do the mod operation manually, one a at a time:
private static long no_Overflow_Mod(ulong num_base, ulong num_exponent, ulong mod)
{
long answer = 1;
for (int x = 0; x < num_exponent; x++)
{
answer = (answer * num_base) % mod;
}
return answer;
}
but this takes a very long time. Is there any simple and fast way to do this operation without actually having to take a to the power of b AND without using time-consuming loops? If all else fails, I can make a bool array to represent a huge data type and figure out how to do this with bitwise operators, but there has to be a better way.
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