卖萌的弱渣

I am stupid, I am hungry.

Super Pow

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

Example1:

a = 2 b = [3]

Result: 8

Example2:

a = 2 b = [1,0]

Result: 1024

Solution

  • Java
(Super-Pow.java) download
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public class Solution {
    public int superPow(int a, int[] b) {
        int ret=1;
        for(int i=b.length-1; i>=0; i--){
            ret = ret*pow(a,b[i],1337)%1337;
            a = pow(a,10,1337);
        }
        return ret;
    }
    // 计算a^b%c
    public int pow(int a, int b, int c){
        long ret = 1;
        long p = a;
        while(b>0){
            if((b&1)==1)
                ret = (ret*p)%c;
            p = (p*p)%c;
            b = b>>1;
        }
        return (int)(ret%c);
    }
}
  • c mod m = (a ⋅ b) mod m = [(a mod m) ⋅ (b mod m)] mod m
(Super-Pow.py) download
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class Solution(object):
    def superPow(self, a, b):
        """
        :type a: int
        :type b: List[int]
        :rtype: int
        """
        ans = 1
        mod = 1337
        for bi in b[::-1]:
            ans = ans*a**bi%mod
            a = a ** 10 % mod
        return ans