how to calculate a^b%mod where b is very much larger than mod
can we do b = b%mod???
e.g. mod = 1000000007
a = 2 b = 1134903170 after this can we do b = 1134903170%mod
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how to calculate a^b%mod where b is very much larger than mod
can we do b = b%mod???
e.g. mod = 1000000007
a = 2 b = 1134903170 after this can we do b = 1134903170%mod
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You can do b = b % phi(mod)
(euler totient function)
Well, first of all, for your example you can always just do fast exponentiation, in O(log b) multiplications. If b is even, then ab = (ab / 2)2, otherwise ab = ab - 1·a. Though I never implement it recursively, but iteratively (C++):
The modding is covered in the
mul
function.But to answer your question, yes you can reduce b. If b is greater than φ(mod), then you can reduce b to b%φ(mod) + φ(mod). If you knew that a is relatively prime to mod then you wouldn't need the + φ(mod), but in the general case it wouldn't work without + φ(mod) (for example, a = 2, mod = 220, b = φ(mod) = 219, the solution is 0). φ(x) is the Euler totient function.