Question
Formatted question description: https://leetcode.ca/all/372.html
372 Super Pow
Your task is to calculate: a^b mod 1337
where a is a positive integer and b is an extremely large positive integer given in the form of an array.
Example 1:
Input: a = 2, b = [3]
Output: 8
Example 2:
Input: a = 2, b = [1,0]
Output: 1024
Algorithm
Related maths
a^b % 1337 = (a%1337)^b % 1337
xy % 1337 = ((x%1337) * (y%1337)) % 1337
Reduced by half, the difference is that the remainder of 1337 is added. Since the given exponent b is a representation of a one-dimensional array, it must be very inconvenient to deal with it if we halve it.
So we use bitwise processing, such as 2^23 = (2^2)^10 * 2^3, so we can start from the highest bit of b, calculate a result and store it in res, and then go to the next bit , The tenth power of res is multiplied by the power of a and the remainder is 1337.
Code
Java
public class Super_Pow {
class Solution {
// 2^23 = (2^2)^10 * 2^3
public int superPow(int a, int[] b) {
int res = 1;
for (int i = 0; i < b.length; ++i) {
res = pow(res, 10) * pow(a, b[i]) % 1337;
}
return res;
}
int pow(int x, int n) {
if (n == 0) return 1;
if (n == 1) return x % 1337;
return pow(x % 1337, n / 2) * pow(x % 1337, n - n / 2) % 1337;
}
}
}