CF 105925L - qPhones Production Line

CF 105925L - qPhones Production Line

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Solve time: 40s
Verified: yes

Solution

Problem Understanding

We are given a classical memory size expressed in megabytes and asked how many qubits are needed in a quantum-style model so that it can represent every possible classical memory state of that size.

A classical device with $M$ megabytes has a certain number of bits, and therefore $2^{\text{(number of bits)}}$ possible memory configurations. The quantum device behaves differently in description but is being used here as a simulation tool: $k$ qubits can represent $2^k$ basis states simultaneously, which is equivalent to $2^k$ classical bit configurations.

The task reduces to matching capacities. If a classical device has $B$ bits, then we need the smallest $k$ such that $2^k \ge 2^B$. This simplifies directly to $k \ge B$, so the answer is simply the number of bits in the classical memory.

The only subtlety is converting megabytes into bits. One megabyte is defined as $10^6$ bytes, and one byte is 8 bits, so:

$$B = M \cdot 10^6 \cdot 8$$

The constraints allow $M$ up to $10^{10}$, which makes the total bit count up to $8 \cdot 10^{16}$. This fits comfortably in 64-bit integers, so arithmetic is safe without big integer libraries.

A naive mistake would be trying to simulate powers of two directly and incrementing qubits until capacity is reached. For large $M$, that would require iterating up to around $10^{16}$, which is infeasible.

Another possible mistake is misunderstanding the base definition of megabytes. If someone incorrectly assumes $2^{20}$ bytes per MB instead of $10^6$, the result would differ significantly. For example, $M = 1$ would give either $8 \cdot 10^6$ bits or $8 \cdot 2^{20}$ bits depending on interpretation, which diverges quickly.

Approaches

The brute-force way is to start with $k = 0$ qubits and repeatedly increase it until $2^k$ reaches the number of classical configurations. Each step doubles the representable space. This is correct but immediately becomes infeasible because the answer itself can be on the order of $10^{16}$. Even checking exponent growth step by step is unnecessary because we are not searching for a logarithm through simulation.

The key observation is that qubit capacity grows in powers of two, but we are not asked for an exponential threshold comparison in terms of $2^k$ against a large number; instead, both systems are exponentials of the same base structure. The classical state space is $2^B$, and the quantum state space is $2^k$. Matching them reduces to equating exponents, which collapses the entire problem into a linear conversion.

So instead of exponential search, we directly compute the number of classical bits and output it.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (increment qubits)	$O(B)$	$O(1)$	Too slow
Optimal (direct formula)	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Read the integer $M$, which represents memory in megabytes. The goal is to convert this into an equivalent number of bits, because qubit capacity matches bit-level state space.
2. Convert megabytes into bytes by multiplying by $10^6$. This follows the problem’s explicit definition of a megabyte, not the binary convention.
3. Convert bytes into bits by multiplying by 8, since each byte stores 8 bits. This gives the total number of classical bits $B$.
4. Return $B$ directly as the answer, because $k$ qubits can represent $2^k$ states, and $B$ bits represent $2^B$ states, so equality of exponents determines the minimum $k$.

Why it works

The core property is that both systems express capacity as powers of two. A classical memory of $B$ bits has $2^B$ possible configurations. A quantum register of $k$ qubits spans a state space of size $2^k$. The requirement is that the quantum system covers all classical configurations, so we need $2^k \ge 2^B$. Taking logarithms base 2 collapses this to $k \ge B$, making the minimal valid value exactly $B$. No approximation or rounding issues appear because both sides are exact integers.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    M = int(input().strip())
    bits = M * 10**6 * 8
    print(bits)

if __name__ == "__main__":
    solve()

The implementation is intentionally minimal because the reasoning already resolves the structure completely. The only computation is the conversion chain from megabytes to bits.

The multiplication order is safe because Python handles large integers natively, and even in stricter languages this fits in 64-bit signed integer range given the constraints. There are no loops or exponentiation operations, so the solution is constant time.

Worked Examples

Since the statement snippet does not clearly show full samples, we construct representative traces.

Example 1

Input:

1

We track the conversion:

Step	Value
M	1
bytes	$1 \cdot 10^6 = 1{,}000{,}000$
bits	$1{,}000{,}000 \cdot 8 = 8{,}000{,}000$

Output is:

8000000

This confirms that even a single megabyte maps directly into a large fixed qubit requirement.

Example 2

Input:

3

Step	Value
M	3
bytes	$3 \cdot 10^6 = 3{,}000{,}000$
bits	$3{,}000{,}000 \cdot 8 = 24{,}000{,}000$

Output:

24000000

This trace shows linear scaling in memory size, consistent with the fact that the exponential behavior cancels when comparing both systems.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only a fixed number of arithmetic operations are performed regardless of input size
Space	$O(1)$	No auxiliary structures are used

The constraints allow very large values of $M$, but the solution avoids iteration or recursion entirely. This makes it easily safe under a 0.5 second limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline
    M = int(input().strip())
    print(M * 10**6 * 8)
    return sys.stdout.getvalue().strip()

# minimal case
assert run("1\n") == "8000000", "single MB"

# small case
assert run("2\n") == "16000000", "linear scaling"

# boundary-like larger case
assert run("10\n") == "80000000", "ten MB scaling"

# large value stress
assert run("10000000000\n") == str(10000000000 * 10**6 * 8), "max input scaling"

Test input	Expected output	What it validates
1	8000000	base unit conversion correctness
2	16000000	linear scaling
10	80000000	medium scaling sanity
10000000000	large value	overflow safety and correctness

Edge Cases

One edge case is the smallest input $M = 1$. The algorithm directly computes $8 \cdot 10^6$, which correctly represents the minimal non-zero memory unit without any special handling.

Another edge case is the maximum input $M = 10^{10}$. The computation becomes $8 \cdot 10^{16}$, which still fits within Python integers comfortably. The algorithm performs a single multiplication chain, so there is no risk of performance degradation or intermediate overflow in Python.

A conceptual edge case is misunderstanding the unit definition of megabyte. If a competing interpretation uses $2^{20}$ bytes, the computed result would differ significantly. However, the algorithm strictly follows the problem’s definition, ensuring consistency with the intended output model.