Problem Overview

Codeforces 913A - Modular Exponentiation is a straightforward yet insightful problem that checks your understanding of powers of two, modulo arithmetic, and edge case handling. You are given two integers n and m, and your task is to compute m % (2^n) efficiently.

Although it seems simple at first glance, this problem is excellent for understanding how exponential operations can quickly overflow and how to prevent it using clever constraints.

C++ Solution

Here’s the full C++ solution to this problem:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    long long n, m;
    cin >> n >> m;
    if (n > 27) {
        cout << m << endl;
    } else {
        int x = pow(2, n);
        x = m % x;
        cout << x << endl;
    }
    return 0;
}

Detailed Explanation of the Approach

Understanding the Problem

We are given two integers n and m. Our goal is to compute:

m % (2^n)

But there's a catch: the values of n and m can be quite large. So we need to ensure that our operations are both time-efficient and memory-safe.

Key Insight

The maximum value of n can be up to 1e9, which means 2^n would become astronomically large. However, the constraints of the problem implicitly suggest a safe upper limit.

By inspecting the sample test cases and doing a bit of analysis, we realize that if n exceeds 27, the value of 2^n becomes larger than any possible value of m within int or even standard input range. Since m % x == m when x > m, we can safely return m in such cases.

This avoids unnecessary computation of massive powers of 2.

 Step-by-Step Breakdown

Let’s take an example:

Input:

n = 3, m = 12

We need to compute:

12 % (2^3) = 12 % 8 = 4

Let’s walk through the logic in the code:

  1. Input reading:
    We take n and m as input.

  2. Check if n > 27:
    If yes, return m because 2^n > m.

  3. Otherwise:
    Calculate 2^n using pow(2, n).
    Then return m % (2^n).

Dry Run Example

Let’s dry run with a few values to verify correctness.

Example 1:

Input:

n = 5, m = 30

Processing:

2^5 = 32  
30 % 32 = 30

Output:

30

Example 2:

Input:

n = 28, m = 100

Since n > 27, we directly return m:

Output:

100

Time and Space Complexity

Time Complexity:

  • Reading input takes constant time: O(1)
  • If n <= 27, we compute pow(2, n) and take a modulo — both are constant-time operations.
  • So the total time complexity is O(1).

Space Complexity:

  • We only use a few variables, so space complexity is also O(1).

Common Mistakes to Avoid

  1. Not Checking the Overflow Edge Case:
    If you attempt to compute 2^n for n > 30, you may encounter overflow even with long long. That's why the special case n > 27 is important.

  2. Using pow() for Integer Exponentiation:
    pow() returns a double, not an integer. So be cautious — cast it appropriately or use bit shifting (1 << n) for small n.

  3. Forgetting Modulo Priority:
    Ensure you apply the modulo only after computing the correct power, not during intermediate steps (unless optimizing).

 Alternate Approaches

Instead of using pow(2, n), you can also use bit shifting:

int x = 1 << n;

This only works when n <= 30, because shifting by more than 30 bits may cause undefined behavior in 32-bit systems. It's a faster and safer alternative when you're sure of the bounds.

 Related Codeforces Problems

If you found this problem helpful and want to practice similar problems, check these out:

These problems involve modular arithmetic, powers, and efficient number operations.

 Conclusion

Codeforces 913A is a great example of how a simple mathematical operation can be turned into an efficient and clever solution by observing constraints and optimizing edge cases.

You learned:

  • When to use direct logic vs. optimization
  • Importance of edge case handling (like overflow)
  • How to safely compute powers of 2
  • Efficiency through condition short-circuiting

Solving such problems helps sharpen your understanding of number theory and optimization — two core pillars in competitive programming.

If you enjoyed this problem, make sure to bookmark it and explore the related problems above. Keep practicing, and you'll keep improving!