CF 913A - Modular Exponentiation

Rating: 900
Tags: implementation, math
Solve time: 6m 16s
Verified: yes

Solution

Problem Understanding

The problem asks us to compute the remainder when an integer m is divided by $2^n$, where n and m are positive integers. In other words, we are given the size of a power-of-two modulus and a dividend, and we must find what is left after dividing the dividend by that modulus. The input n specifies the number of times we multiply 2 by itself to obtain the divisor, while m is the number to be reduced modulo that divisor. The output is a single integer between 0 and $2^n - 1$, representing the remainder.

The constraints are moderate but have subtle implications. Both n and m can be as large as 10^8. Directly computing $2^n$ is impractical, because storing or operating with numbers of roughly $10^{30,000,000}$ digits is impossible in memory. This rules out any approach that literally constructs $2^n$. The algorithm must therefore rely on a property of powers of two, rather than computing them explicitly.

Edge cases arise where m is smaller than $2^n$, in which case the remainder is simply m, or when m is exactly divisible by $2^n$, yielding a remainder of zero. For example, if n = 5 and m = 32, $2^5 = 32$ and the remainder is 0. A careless implementation that tries to compute $2^n$ literally would fail or overflow in such cases.

Approaches

The brute-force approach would attempt to compute $2^n$ and then perform m modulo $2^n$. This is correct in principle, but with n up to 10^8, the number $2^n$ is astronomically large and cannot be represented in standard integer types. Attempting to construct it directly would require exponential time and memory, far exceeding the problem limits.

The key insight is that $2^n$ has a very simple binary representation: it is a 1 followed by n zeros in binary. Computing m modulo $2^n$ is equivalent to taking the last n bits of m. In practice, this is exactly what the bitwise AND operation with $2^n - 1$ achieves, because $2^n - 1$ in binary is n ones. This observation reduces the problem to a single fast operation that works even for the largest constraints, and it avoids constructing massive numbers.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(n)	Too slow, impossible for large n
Bitwise Modulo	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the integers n and m. These specify the exponent of 2 and the number to reduce, respectively.
Compute $mask = 2^n - 1$. This creates a number with n ones in binary. The reason is that any number modulo a power of two is equivalent to its last n bits, which this mask isolates.
Compute the remainder as $result = m & mask$. The bitwise AND operation keeps exactly the last n bits of m, discarding the higher bits. This operation directly produces $m \bmod 2^n$ without any large-number arithmetic.
Print the result. This completes the calculation efficiently.

Why it works: the invariant is that for any integer m, the remainder of division by $2^n$ depends solely on the lowest n bits of m. The AND operation with $2^n - 1$ preserves those bits while zeroing out the higher-order bits, which have no effect on the remainder. This guarantees correctness regardless of how large m or n is.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
m = int(input())

# 2^n - 1 is a mask with n ones in binary
mask = (1 << n) - 1

# remainder of m divided by 2^n
result = m & mask

print(result)

The code reads input using fast I/O, computes the mask using a left-shift to avoid computing large powers explicitly, and then applies a bitwise AND to extract the last n bits. Using 1 << n is both fast and safe in Python because Python integers are unbounded. The final print outputs the remainder as required.

Worked Examples

Sample 1: n = 4, m = 42

Step	Value	Explanation
n	4	read from input
m	42	read from input
mask	15	$2^4 - 1 = 16 - 1 = 15$
result	10	42 & 15 = 101010 & 1111 = 1010 (binary) = 10

This shows that the last 4 bits of 42 are 1010, which gives the remainder 10.

Sample 2: n = 1, m = 58

Step	Value	Explanation
n	1	read from input
m	58	read from input
mask	1	$2^1 - 1 = 1$
result	0	58 & 1 = 111010 & 1 = 0

This demonstrates the edge case where m is divisible by $2^n$, yielding remainder 0.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	single bitwise AND operation
Space	O(1)	only three integers stored (n, m, mask)

The time complexity does not depend on the magnitude of n or m, so it easily fits within the 1-second limit. Space usage is minimal, well within the 256 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    m = int(input())
    mask = (1 << n) - 1
    return str(m & mask)

# provided samples
assert run("4\n42\n") == "10", "sample 1"
assert run("1\n58\n") == "0", "sample 2"

# custom cases
assert run("5\n32\n") == "0", "m equals 2^n"
assert run("5\n31\n") == "31", "m just below 2^n"
assert run("8\n300\n") == "44", "medium n and m"
assert run("30\n1073741825\n") == "1", "large n with overflow-like value"
assert run("10\n1023\n") == "1023", "m exactly 2^n - 1"

Test input	Expected output	What it validates
5\n32\n	0	remainder zero when m = 2^n
5\n31\n	31	remainder equal to m when m < 2^n
8\n300\n	44	typical case, checks correct masking
30\n1073741825\n	1	large n, tests correctness with big numbers
10\n1023\n	1023	m = 2^n - 1 edge case

Edge Cases

When m is smaller than $2^n$, the algorithm correctly returns m itself. For example, n = 5, m = 3 results in mask = 31, and 3 & 31 = 3. When m equals $2^n$, such as n = 5, m = 32, the mask is 31, and 32 & 31 = 0. Large n such as 30 and m = 1073741825 still works because Python supports arbitrary precision integers, and the bitwise AND isolates the correct last 30 bits. These cases confirm the algorithm handles all boundaries correctly.