# Question

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

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.

Example 1:

Input: a = 2, b = [3]
Output: 8


Example 2:

Input: a = 2, b = [1,0]
Output: 1024


Example 3:

Input: a = 1, b = [4,3,3,8,5,2]
Output: 1


Constraints:

• 1 <= a <= 231 - 1
• 1 <= b.length <= 2000
• 0 <= b[i] <= 9
• b does not contain leading zeros.

# 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

• 
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;
}
}
}

############

class Solution {
private static final int MOD = 1337;

public int superPow(int a, int[] b) {
int ans = 1;
for (int i = b.length - 1; i >= 0; --i) {
ans = (int) ((long) ans * quickPowAndMod(a, b[i]) % MOD);
a = quickPowAndMod(a, 10);
}
return ans;
}

private int quickPowAndMod(int a, int b) {
int ans = 1;
while (b > 0) {
if ((b & 1) == 1) {
ans = (ans * (a % MOD)) % MOD;
}
b >>= 1;
a = (a % MOD) * (a % MOD) % MOD;
}
return ans;
}
}

• class Solution:
def superPow(self, a: int, b: List[int]) -> int:
MOD = 1337
ans = 1
for e in b[::-1]:
ans = ans * pow(a, e, MOD) % MOD
a = pow(a, 10, MOD)
return ans

############

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


• class Solution {
const int MOD = 1337;

public:
int superPow(int a, vector<int>& b) {
int ans = 1;
for (int i = b.size() - 1; i >= 0; --i) {
ans = (long) ans * quickPowAndMod(a, b[i]) % MOD;
a = quickPowAndMod(a, 10);
}
return ans;
}

int quickPowAndMod(int a, int b) {
int ans = 1;
while (b) {
if (b & 1) {
ans = (ans * (a % MOD)) % MOD;
}
b >>= 1;
a = ((a % MOD) * (a % MOD)) % MOD;
}
return ans;
}
};

• const mod = 1337

func superPow(a int, b []int) int {
ans := 1
for i := len(b) - 1; i >= 0; i-- {
ans = ans * quickPowAndMod(a, b[i]) % mod
a = quickPowAndMod(a, 10)
}
return ans
}

func quickPowAndMod(a, b int) int {
ans := 1
for b > 0 {
if b&1 > 0 {
ans = ans * a % mod
}
b >>= 1
a = ((a % mod) * (a % mod)) % mod
}
return ans
}