CF 102697060 - Maximum Number

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Solve time: 45s
Verified: yes

Solution

Problem Understanding

The task asks for the largest signed integer that can be represented using exactly (N) bits. For a signed integer representation, one bit is reserved for the sign, leaving (N-1) bits for the magnitude of the largest positive value. The maximum is consequently

[ 2^{N-1}-1. ]

The input consists of a single integer (N), where (1 \le N \le 64). The output is the corresponding maximum signed integer. The official problem statement gives the same formula and specifically warns that values above 32 bits require an integer type capable of representing larger values.

The constraint is deliberately small because the mathematical operation itself is constant time. Even though (N) can reach 64, we never need to iterate over the possible integers or simulate individual bits. An algorithm proportional to (N) would already be more than sufficient, while anything exponential in (N) is unnecessary and becomes impossible at (N=64).

The main edge cases come from the smallest bit widths and from large values. For (N=1), there is one sign bit and no magnitude bits, so the largest representable signed integer is (2^0-1=0). A careless implementation that assumes there must be at least one magnitude bit could incorrectly produce 1.

For (N=2), the answer is (2^1-1=1). The complete input and output are:

An implementation that accidentally computes (2^N-1) would return 3, because it would be treating all bits as magnitude bits and ignoring the signed representation.

Large (N) is another practical edge case. For example, with (N=64), the answer is

[ 2^{63}-1=9223372036854775807. ]

Using floating-point exponentiation can lose integer precision, and using a fixed-width signed 32-bit integer would overflow. Python's integers are arbitrary precision, so the direct integer expression is safe.

Approaches

The most literal brute-force approach would generate every signed integer representable with (N) bits, test whether it is within the representation's positive range, and keep the largest one. There are (2^N) different bit patterns in an (N)-bit representation, so this requires (O(2^N)) candidates in the worst case. At (N=64), that is (2^{64}), or about (1.84\times10^{19}), operations. Such an approach is completely infeasible.

The brute force works conceptually because every possible representation is considered, so eventually the largest valid positive value is found. The problem is that the representation has an enormous number of possibilities even though the answer has a simple mathematical form.

The key observation is that signed integers use one bit for the sign. To maximize a positive signed integer, the sign bit must be zero. Every remaining (N-1) bit should be one, because changing any of those bits from zero to one increases the value while keeping the sign positive. Thus the optimal bit pattern is a zero followed by (N-1) ones.

A binary number consisting of (N-1) ones has value (2^{N-1}-1). We therefore do not need to construct or inspect any candidate numbers. One exponentiation followed by one subtraction gives the answer directly.

The brute-force search is replaced by the observation that all positive values are ordered by their magnitude bits. Setting every available magnitude bit to one is always optimal, which reduces the problem from exponential enumeration to a constant number of arithmetic operations.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	(O(2^N))	(O(1))	Too slow
Optimal	(O(1))	(O(1))	Accepted

Algorithm Walkthrough

Read the number of bits (N). The input contains only one value, so no test-case loop is needed.
Compute (2^{N-1}). There are (N-1) magnitude bits because one of the (N) bits is the sign bit.
Subtract one from the result. The value (2^{N-1}-1) is exactly the binary number consisting of (N-1) ones, which is the largest positive value possible.
Print the resulting integer. Python's arbitrary-precision integer type handles the maximum case (N=64) without overflow.

Why it works

For a positive signed (N)-bit integer, the sign bit is fixed to zero. The remaining (N-1) bits contribute positive powers of two, so making any one of them equal to one can only increase the represented value. The maximum is reached when every magnitude bit is one. The resulting binary value is (1+2+4+\dots+2^{N-2}), which equals (2^{N-1}-1). Hence the algorithm always prints exactly the largest signed integer representable with (N) bits.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
print((1 << (n - 1)) - 1)

The expression 1 << (n - 1) computes (2^{N-1}) using an integer bit shift. This avoids floating-point arithmetic entirely, which matters because floating-point numbers cannot represent every large integer exactly. The official statement explicitly warns about floating-point precision for this problem.

The subtraction is performed after the shift, matching the mathematical formula exactly. The order is significant: (1 << n) - 1 would incorrectly treat all (N) bits as magnitude bits and produce (2^N-1).

There is no overflow concern in Python because its integers automatically grow to whatever size is necessary. In this problem the largest intermediate value is only (2^{63}), which is well within Python's practical integer range.

Worked Examples

The official sample uses (N=32).

(N)	Sign bits	Magnitude bits	(2^{N-1})	Answer
32	1	31	2147483648	2147483647

The computation produces (2^{31}-1=2147483647), which matches the sample output. This demonstrates the intended interpretation of one sign bit and confirms that the formula works for a familiar 32-bit signed integer.

For a second example, consider (N=64).

(N)	Sign bits	Magnitude bits	(2^{N-1})	Answer
64	1	63	9223372036854775808	9223372036854775807

The answer is (9223372036854775807), the largest positive value of a signed 64-bit integer. This case exercises the upper boundary and shows why an implementation must not use 32-bit arithmetic or floating-point exponentiation.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(1))	One bit shift and one subtraction are performed.
Space	(O(1))	Only the input value and resulting integer are stored.

The constraint (N\le64) makes the arithmetic trivial. Even without relying on the small bound, the algorithm performs a fixed number of operations and never enumerates possible bit patterns, so it easily fits the 1 second time limit and 256 MB memory limit specified by the problem.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())
    print((1 << (n - 1)) - 1)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()
    output = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout

    return output

# Provided sample
assert run("32\n") == "2147483647\n", "sample 1"

# Minimum-size input
assert run("1\n") == "0\n", "one-bit signed integer"

# Small boundary case
assert run("2\n") == "1\n", "two-bit signed integer"

# All magnitude bits are set
assert run("8\n") == "127\n", "eight-bit signed integer"

# Maximum allowed input
assert run("64\n") == "9223372036854775807\n", "maximum bit width"

Test input	Expected output	What it validates
`32`	`2147483647`	Official sample and standard 32-bit boundary
`1`	`0`	Minimum input and zero available magnitude bits
`2`	`1`	Smallest case where a positive value exists
`8`	`127`	Correct placement of all magnitude bits
`64`	`9223372036854775807`	Maximum input and large-integer handling

Edge Cases

For (N=1), the input is:

The algorithm evaluates (1 << 0) - 1, giving (1-1=0). There are no magnitude bits, so zero really is the largest signed value. The output is:

For (N=2), the input is:

The algorithm computes (1 << 1) - 1 = 2-1 = 1. The available positive representation is binary 01, while 11 represents a negative value under signed two's-complement interpretation. The output is:

For the maximum value (N=64), the input is:

The algorithm computes (2^{63}-1), producing:

9223372036854775807

No overflow occurs because Python integers have arbitrary precision. The calculation also avoids floating-point precision problems by using an integer bit shift.

The most common off-by-one mistake is using (2^N-1) instead of (2^{N-1}-1). For example, with (N=8), that incorrect formula gives 255, but an 8-bit signed integer can represent positive values only up to 127. The correct calculation uses seven magnitude bits, giving (2^7-1=127).