# 372. Super Pow

## Description

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.

## Solutions

• class Solution {
private final int mod = 1337;

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

private long qpow(long a, int n) {
long ans = 1;
for (; n > 0; n >>= 1) {
if ((n & 1) == 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return ans;
}
}

• class Solution {
public:
int superPow(int a, vector<int>& b) {
using ll = long long;
const int mod = 1337;
ll ans = 1;
auto qpow = [&](ll a, int n) {
ll ans = 1;
for (; n; n >>= 1) {
if (n & 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return (int) ans;
};
for (int i = b.size() - 1; ~i; --i) {
ans = ans * qpow(a, b[i]) % mod;
a = qpow(a, 10);
}
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


• func superPow(a int, b []int) int {
const mod int = 1337
ans := 1
qpow := func(a, n int) int {
ans := 1
for ; n > 0; n >>= 1 {
if n&1 == 1 {
ans = ans * a % mod
}
a = a * a % mod
}
return ans
}
for i := len(b) - 1; i >= 0; i-- {
ans = ans * qpow(a, b[i]) % mod
a = qpow(a, 10)
}
return ans
}

• function superPow(a: number, b: number[]): number {
let ans = 1;
const mod = 1337;
const qpow = (a: number, n: number): number => {
let ans = 1;
for (; n; n >>= 1) {
if (n & 1) {
ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
}
a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
}
return ans;
};
for (let i = b.length - 1; ~i; --i) {
ans = Number((BigInt(ans) * BigInt(qpow(a, b[i]))) % BigInt(mod));
a = qpow(a, 10);
}
return ans;
}