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372. Super Pow

DifficultyMedium

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

Explanation:

The array b = [3] represents the exponent 3. Therefore, a3 = 23 = 8, so the result is 8.

Example 2:

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

Output: 1024

Explanation:

The array b = [1, 0] represents the exponent 10. Therefore, a10 = 210 = 1024, so the result is 1024.

Example 3:

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

Output: 1

Explanation:

The array b = [4, 3, 3, 8, 5, 2] represents a positive exponent. Since a = 1, any positive power of 1 is 1. Therefore, the result is 1.

 

Constraints:

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

Solutions

Solution 1

Thinking

\(a\) is raised to a huge decimal \(b\) given as digits; return \(a^b\bmod 1337\). Building \(b\) as one integer is awkward. Write \(b=\cdots+e_i\cdot 10^i\).

From low digit to high: multiply by \(a^{e}\), then replace the base with \(a^{10}\). Modular fast pow keeps each step \(O(\log e)\).

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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

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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 = (int) 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;
    }
}

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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;
    }
};

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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
}

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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;
}

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