Super Pow Solution in C++

Problem Description

Welcome to the world of Super Pow, where numbers are raised to the power of other numbers, and you might just need a calculator that can handle more than your average math homework! The problem is simple: given an integer a and an array of integers b, you need to compute a^{b[0] × 10^{k-1} + b[1] × 10^{k-2} + ... + b[k-1]} mod 1337. Sounds easy, right? Just like trying to find your keys in a messy room—it’s all about the right approach!

Imagine you’re trying to impress your friends with your math skills at a party, and you decide to show off your ability to calculate large powers. But wait! You realize that your brain can only handle so much, and you need a clever way to break it down. That’s where our code comes in!

Code Solution

class Solution {
 public:
  int superPow(int a, vector& b) {
    int ans = 1;

    a %= kMod;
    for (const int i : b)
      ans = modPow(ans, 10) * modPow(a, i) % kMod;

    return ans;
  }

 private:
  static constexpr int kMod = 1337;

  long modPow(long x, long n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * modPow(x % kMod, (n - 1)) % kMod;
    return modPow(x * x % kMod, (n / 2)) % kMod;
  }
};
    

Approach

The approach taken in this solution is to break down the problem using modular arithmetic and exponentiation by squaring. The modPow function efficiently computes powers while keeping the results manageable by taking modulo 1337. The main function iterates through the array b, updating the result by combining the powers of a and the current value in b. This way, we avoid the pitfalls of handling excessively large numbers.

Time and Space Complexity

Time Complexity: O(k log n), where k is the length of the array b and n is the maximum value in b. This is due to the logarithmic nature of the exponentiation by squaring.

Space Complexity: O(1) since we are using a constant amount of space for variables.

Real-World Example

Imagine you’re a magician at a birthday party, and you want to perform a trick where you multiply your age by a power of 10, but you can only do it modulo 1337. You’d need to break down the calculation into manageable parts, just like our code does! Instead of trying to compute 30^{100} directly (which is a number so large it could scare off the kids), you cleverly use the superPow method to keep it fun and manageable.