Given two positive integers a and n, calculate an modulo 1337.
Example 1:
Input: a = 2, n = 3 Output: 8
Example 2:
Input: a = 2, n = 10 Output: 1024
Example 3:
Input: a = 1, n = 1000 Output: 1
Example 4:
Input: a = 2147483647, n = 200 Output: 1198
Constraints:
1 <= a, n <= 231 - 1
The core challenge of this problem is to compute large powers efficiently. Direct computation of an
can result in extremely large numbers, which are impractical to handle directly due to time and space constraints. The significance of this problem lies in its applications in cryptography, computer graphics, and numerical methods.
Potential pitfalls include integer overflow and inefficient algorithms that do not scale well with large inputs.
To solve this problem, we can use the method of Exponentiation by Squaring, which is an efficient way to compute large powers modulo a number. This method reduces the time complexity significantly compared to the naive approach.
The naive solution involves multiplying a
by itself n
times, which has a time complexity of O(n)
. This approach is not feasible for large values of n
.
Exponentiation by Squaring reduces the time complexity to O(log n)
. The idea is to break down the power calculation into smaller parts using the properties of exponents:
n
is even, an = (a2)n/2
n
is odd, an = a * an-1
Here is a step-by-step breakdown of the Exponentiation by Squaring algorithm:
n
is greater than 0:
n
is odd, multiply the result by a
and reduce n
by 1.a
and halve n
.
#include <iostream>
using namespace std;
// Function to perform modular exponentiation
int powerMod(int a, int n, int mod) {
int result = 1;
a = a % mod; // Update a if it is more than or equal to mod
while (n > 0) {
// If n is odd, multiply a with result
if (n % 2 == 1) {
result = (result * a) % mod;
}
// n must be even now
n = n >> 1; // n = n / 2
a = (a * a) % mod; // Change a to a^2
}
return result;
}
int main() {
int a, n;
cout << "Enter a and n: ";
cin >> a >> n;
cout << "Result: " << powerMod(a, n, 1337) << endl;
return 0;
}
The time complexity of the Exponentiation by Squaring algorithm is O(log n)
because we halve the exponent in each step. The space complexity is O(1)
as we use a constant amount of extra space.
Consider the following edge cases:
a = 1
and any n
: The result is always 1.n = 0
: The result is 1 for any a
(by definition of exponentiation).a
and n
: The algorithm handles these efficiently due to its logarithmic time complexity.To test the solution comprehensively, consider a variety of test cases:
a = 2, n = 3
a = 1, n = 1000
a = 2147483647, n = 200
Use a testing framework like Google Test for automated testing.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to efficiently compute large powers modulo a number using the Exponentiation by Squaring method. This approach significantly reduces the time complexity and handles large inputs effectively. Understanding and solving such problems is crucial for applications in various fields, including cryptography and numerical methods.