Raising a number to a huge exponent

Question

I am given the number 3 and a variable \'n\', that can be as high as 1 000 000 000 (a billion). I have to print the answer of 3^n modulo 100003. I tried the followi

Answer 1

This is to augment Kaidul's answer.

100003 is a prime number, which immediately casts in the Fermat's Little Theorem: any number raised to a prime power is congruent to itself modulo that prime. It means that you don't need to raise to n'th power. A n % 100002 power suffices.

Edit: example.

Say, n is 200008, which is 100002 * 2 + 6. Now,

3 ^ 200007 =
3 ^ (100002 + 100002 + 6) = 
3 ^ 100002 * 3 ^ 100002 * 3 ^ 6

FLT claims that (3 ^ 100002) % 100003 == 1, and the last line above, modulo 100003, reduces to 3 ^ 6. In general, for a prime p,

(k ^ n) % p == k ^ (n % p)

Of course, it only speeds the computation if the exponent n is greater than p. As per your request (exponent 100, modulo 100003) there is nothing to reduce. Go straight to the Kaidul's approach.

Answer 2

Take advantage of property of modular arithmetic

(a × b) modulo M == ((a module M) × (b modulo M)) modulo M

By using above multiplication rule

(a^n) modulo M 
= (a × a × a × a ... × a) modulo M 
= ((a module M) × (a modulo M) × (a modulo M) ... × (a modulo M)) modulo M

Calculate the result by divide and conquer approach. The recurrence relation will be:

f(x, n) = 0                     if n == 0

f(x, n) = (f(x, n / 2))^2       if n is even
f(x, n) = (f(x, n / 2))^2 * x   if n is odd

Here is the C++ implementation:

int powerUtil(int base, int exp, int mod) {
    if(exp == 0) return 1;
    int ret = powerUtil(base, exp / 2, mod) % mod;
    ret = 1LL * ret * ret % mod;
    if(exp & 1) {
        ret = 1LL * ret * base % mod;
    }
    return ret;
}

double power(int base, int exp, int mod) {
    if(exp < 0) {
        if(base == 0) return DBL_MAX; // undefined
        return 1 / (double) powerUtil(base, -exp, mod);
    }
    return powerUtil(base, exp, mod);
}