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Given two positive integers a and a large positive integer array b representing an integer b containing digits from b[0] to b[n-1] (i.e., b is expressed as an array). Calculate a raised to the power b mod 1337.

Essentially, the task is to compute (a^{b} \mod 1337), where (b) is a large number represented by an integer array.

Clarifying Questions

  1. Is a always a positive integer?
    • Yes, a is always a positive integer.
  2. What is the range of elements in the array b?
    • The elements of b are digits, so each element is between 0 and 9.
  3. Can the array b be large (e.g., tens of thousands of elements)?
    • Yes, the array b can be very large, which is why directly computing the power is not feasible.
  4. What should be the output if a or b is zero?
    • If a is 0 and b is not zero, the answer should be 0 (i.e., 0^b = 0 for any b != 0). If b is zero, a ^ b should be 1 (i.e., a^0 = 1 for any a != 0).

Strategy

To solve this problem efficiently:

  1. Modular Arithmetic:
    • Use properties of modular arithmetic to simplify the calculation, especially due to a large b.
    • One of the properties we will use is ((a \cdot b) \mod c = [(a \mod c) \cdot (b \mod c)] \mod c).
  2. Exponentiation by Squaring:
    • Use this method to efficiently calculate large powers modulo a number.
  3. Handling the Large Exponent:
    • Given b is an array, e.g., [1, 2, 3], it represents the number 123.
    • We will use the property (a^{(b \cdot 10 + c)} \equiv a^b \cdot a^c \mod \text{mod}) to incrementally compute the result.
  4. Python Function:
    • Implement the solution where we decompose (a^b \mod 1337) using the above properties.

Code

Let’s implement the solution:

def superPow(a, b):
    MOD = 1337

    # Helper function to compute (x ^ y) % mod using iterative squaring
    def power_mod(x, y, mod):
        result = 1
        x = x % mod  # Update x if it is more than or equal to mod
        while y > 0:
            if y % 2 == 1:  # If y is odd, multiply x with result
                result = (result * x) % mod
            y = y >> 1  # y = y // 2
            x = (x * x) % mod  # Change x to x^2
        return result

    # Reduce problem using the properties of modulus
    a %= MOD
    result = 1

    for digit in b:
        result = power_mod(result, 10, MOD) * power_mod(a, digit, MOD) % MOD

    return result

# Example usage:
a = 2
b = [1, 0]
print(superPow(a, b))  # Output: 1024

Time Complexity

The time complexity of this algorithm is O(n * log(10)), where n is the number of digits in the array b and log(10) arises from the power_mod function.

Explanation

  1. power_mod Function:
    • Computes ((x^y) \mod \text{mod}) using exponentiation by squaring, an efficient method for large powers.
  2. Iterate Over Array b:
    • For each digit in b, adjust the current power result using the previous result taken to the power of 10 (because each new digit represents the next power of 10).

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