# Question

Formatted question description: https://leetcode.ca/all/372.html

 372	Super Pow

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;
}
}
}

• Todo

• class Solution(object):
def superPow(self, a, b):
"""
:type a: int
:type b: List[int]
:rtype: int
"""
ret = 1
k = 1
for num in reversed(b):
ret *= a ** (num) % 1337
a = a ** 10 % 1337
return ret % 1337