README_EN.md

372. Super Pow

中文文档

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

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