Here are a few title options, based on the content of your article: Focus on Efficiency: * How to Calculate (a^b)%MOD Efficiently for Large Exponents * Optimizing (a^b)%MOD Calculations: A Log(b) Ti

Calculating (a^b)%MOD with Large Exponents

In computer programming, the problem of calculating (a^b)%MOD arises when we need to find the remainder when raising a number 'a' to a large exponent 'b', modulo a fixed constant 'MOD'. This is a common task in various cryptographic applications and mathematical computations.

Log(b) Time Complexity Method

A naive approach to this problem is to use the built-in pow() function in C , which calculates a to the power of b using the multiplication algorithm. However, this method becomes inefficient when 'b' is large, as it takes O(b) time.

Euler's Theorem

A more efficient approach involves using Euler's theorem, which states that for any integer 'a' and a prime modulus 'p', a^p mod p = a^(p-1) mod p. By extension, this can be generalized to any positive integer 'MOD' using Euler's totient function φ(MOD).

Euler's Totient Function

Euler's totient function counts the number of positive integers less than 'MOD' that are coprime to 'MOD'. It can be efficiently computed using the prime factorization of 'MOD'.

Calculating (a^b)%MOD with Large Exponents

Combining Euler's theorem and the Euler's totient function, we can calculate (a^b)%MOD for large exponents efficiently.

1. Calculate the totient function φ(MOD).
2. Calculate (a ^ φ(MOD)%MOD).
3. Calculate (a ^ (b % φ(MOD)) %MOD).

This approach reduces the time complexity to O(log(φ(MOD))) and makes it possible to handle exponents that cannot fit in a "long long" data type.

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