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// 372. Super Pow
#include "leetcode.h"
/*
* your task is to calculate `a^b % 1337` where 'a' is a positive integer and
* 'b' is an extremely large positive integer given in the form of an array
*/
/*
* the number 1337 only has two divisors: 7 and 191 (excluding 1 and itself), so
* see if 'a' has a divisor of either of these. also note that 7 and 191 are
* prime numbers, meaning we can use euler's theorem, or really just fermat's
* little theorem if the mod 'n' is prime.
*/
int power(int x, int n, int mod) {
int res = 1;
for (x %= mod; n; x *= x % mod, n >>= 1)
if (n & 1)
res *= x % mod;
return res;
}
int superPow(int a, int *b, int bSize) {
if (!(a % 1337))
return 0;
int n = 0;
for (int i = 0; i < bSize; i++)
n = (n * 10 + b[i]) % 1140;
if (!n)
n += 1140;
return power(a, n, 1337);
}
int main() {
int b1[] = {3}, b2[] = {1, 0}, b3[] = {4, 3, 3, 8, 5, 2};
printf("%d\n", superPow(2, b1, ARRAY_SIZE(b1))); // expect: 8
printf("%d\n", superPow(2, b2, ARRAY_SIZE(b2))); // expect: 1024
printf("%d\n", superPow(1, b3, ARRAY_SIZE(b3))); // expect: 1
}
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