How to Efficiently Calculate (a^b)%MOD with Large Exponents?

Calculating (a^b)%MOD with Large Exponents

In this coding challenge, the task is to calculate the value of pow(a, b)%MOD, where the exponent b can be extremely large. While the conventional log(b) time complexity method is suitable for smaller values, it becomes impractical when b exceeds the capacity of long long data types in C .

However, a more efficient approach involves leveraging Euler's totient function, φ(MOD). Euler's theorem states that a^φ(MOD)≡1(mod MOD). This means that the power of a can be significantly reduced to a^(b % φ(MOD)).

Calculating φ(MOD) is itself a non-trivial task, but can be achieved using integer factorization methods. Once calculated, the exponent b can be replaced with b % φ(MOD) to dramatically reduce the computation time.

Further Refinements

In 2008, Schramm demonstrated that φ(b) can be obtained from the discrete Fourier transform of gcd(b, i), for i ranging from 1 to b. This eliminates the need for explicit factorization.

Additionally, Carmichael's function, λ(MOD), can be used to obtain the correct answer, especially when a and MOD share common factors.

Code Implementation

The following code snippet serves as an example in C :

<code class="cpp">#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) { return (b == 0) ? a : gcd(b, a % b); }

ll pmod(ll a, ll b, ll mod) {
    if (b == 0) return 1;
    if (b % 2 == 1) {
        return (a * pmod(a, b - 1, mod)) % mod;
    } else {
        ll tmp = pmod(a, b / 2, mod);
        return (tmp * tmp) % mod;
    }
}

int main() {
    ll a, b, mod;
    cin >> a >> b >> mod;
    cout << pmod(a, b % phi(mod), mod) << endl;
    return 0;
}</code>

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