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;
            }
        }
    }
    
  • 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
    
    

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