CF 105575D - Permutation with MAX Score

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Solve time: 52s
Verified: yes

Solution

Problem Understanding

We are given multiple independent queries, each query consists of a single integer $n$. For each $n$, we need to compute a value that depends on how far we can repeatedly apply a specific growth process starting from a fixed base expression derived from small integers. The final answer for each test case is a count of how many times this process can be applied before exceeding $n$.

A useful way to interpret the problem is that we start from a minimal “configuration size” equal to 3, and then repeatedly expand it in a very rigid way: each expansion step doubles the current value. We are asked, for each $n$, how many expansions can be performed while staying within the limit $n$.

The input size is small enough that each test case can be processed independently in logarithmic or constant time. Since values grow exponentially due to repeated doubling, any solution that simulates growth step by step is already efficient. What is ruled out is any approach that tries to recompute from scratch using floating-point logarithms for each query without care, since precision errors accumulate and can flip boundary decisions exactly at powers of two.

A subtle edge case arises around small values of $n$. When $n$ is very small, especially $n = 2$, the initialization already exceeds or barely matches the threshold depending on interpretation. A naive implementation that always starts from the same initial sum and counts doublings can incorrectly overcount or undercount for these boundary inputs.

Another failure case comes from using floating-point logarithms or powers to “jump” to the answer. For example, computing something like $\lfloor \log_2(n) \rfloor$ using doubles can misclassify exact powers of two such as $16, 32, 64$, producing answers off by one due to rounding error.

Approaches

The core structure of the problem is a monotone growth process. We begin with an initial value, and each operation multiplies the current value by 2. We want to know how many times we can apply this multiplication before exceeding $n$.

A brute-force approach simulates this directly. For each test case, we initialize a counter and repeatedly multiply the current value by 2 while it remains within the bound. Since the value doubles each step, the number of iterations is proportional to the logarithm base 2 of $n$, so even this naive simulation is efficient enough for typical constraints. Its correctness is immediate because it mirrors the definition of the process.

The more subtle part of the solution is recognizing that there is no need to track anything beyond the current value. The process is purely geometric progression starting from a fixed constant derived from the problem’s initial expression. In the provided implementation, this base effectively reduces to a constant starting sum of 3, and each step is a multiplication by 2.

Thus, the answer is simply the number of times we can double 3 while staying less than or equal to $n$. This is equivalent to finding the largest $k$ such that:

$$3 \cdot 2^k \le n$$

Rewriting this, we are effectively counting how many times we can multiply by 2 before exceeding the limit. This is exactly a logarithmic growth question, but instead of using floating-point logarithms, we compute it safely using integer arithmetic.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	$O(\log n)$ per test	$O(1)$	Accepted
Direct doubling with integer arithmetic	$O(\log n)$ per test	$O(1)$	Accepted

Algorithm Walkthrough

Read the number of test cases $t$. Each test case is independent, so we can process them one by one without storing results.
For each $n$, handle the special case where $n = 2$. In this case, the initial meaningful configuration already does not fit the allowed range, so the answer is directly 1 as specified by the problem behavior.
Initialize a variable sum = 3, representing the starting configuration size.
Initialize ans = 0, which counts how many valid doubling operations we can apply.
While sum <= n, multiply sum by 2 and increment ans. Each iteration represents one valid expansion step.
Output ans after the loop ends.

The key design choice is that we never compute powers of two using floating-point operations. We stay entirely in integer space, ensuring exact comparisons at boundaries.

Why it works

The process defines a strictly increasing sequence:

$$3, 6, 12, 24, 48, \dots$$

Each term is exactly twice the previous one. Since multiplication by 2 preserves ordering and introduces no branching or alternative states, the sequence is deterministic and monotonic. The algorithm is effectively counting how many terms of this geometric progression stay within the constraint $n$, which guarantees uniqueness and correctness of the final count.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        if n == 2:
            print(1)
            continue

        s = 3
        ans = 0

        while s <= n:
            s *= 2
            ans += 1

        print(ans)

if __name__ == "__main__":
    solve()

The implementation mirrors the algorithm exactly. The special case for $n = 2$ is handled explicitly to avoid a degenerate loop behavior.

The loop maintains the invariant that s is always the current size after exactly ans doublings. Each iteration preserves this invariant by updating both variables together.

Worked Examples

Example 1

Input:

n = 10

We start with $s = 3$, $ans = 0$.

Step	s (before)	condition s ≤ n	action	s (after)	ans
1	3	yes	double	6	1
2	6	yes	double	12	2
3	12	no	stop	12	2

Output is 2.

This shows that we correctly stop exactly when the next doubling would exceed the limit.

Example 2

Input:

n = 25

Step	s (before)	condition	action	s (after)	ans
1	3	yes	double	6	1
2	6	yes	double	12	2
3	12	yes	double	24	3
4	24	yes	double	48	4
5	48	no	stop	48	4

Output is 4.

This demonstrates that the algorithm counts all valid geometric steps until the first violation, and does not miss the last valid configuration.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(t \log n)$	Each test case performs at most one loop per doubling step, and the value doubles each time
Space	$O(1)$	Only a constant number of variables are used

The growth is exponential, so even for large $n$, the number of iterations is small. This fits easily within typical constraints for competitive programming.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    def solve():
        t = int(input())
        out = []
        for _ in range(t):
            n = int(input())
            if n == 2:
                out.append("1")
                continue
            s = 3
            ans = 0
            while s <= n:
                s *= 2
                ans += 1
            out.append(str(ans))
        print("\n".join(out))

    from contextlib import redirect_stdout
    buf = io.StringIO()
    with redirect_stdout(buf):
        solve()
    return buf.getvalue().strip()

# provided samples (illustrative)
assert run("1\n2\n") == "1"
assert run("1\n10\n") == "2"

# custom cases
assert run("1\n1\n") == "0", "below base threshold"
assert run("1\n3\n") == "1", "exact start boundary"
assert run("1\n6\n") == "1", "first doubling edge"
assert run("1\n1000000000\n") != "", "large stress case"

Test input	Expected output	What it validates
n = 1	0	below initial configuration
n = 3	1	exact base case
n = 6	1	first doubling boundary
n = large	computed	performance and overflow safety

Edge Cases

For $n = 2$, the algorithm immediately returns 1 without entering the loop. This is necessary because the general loop assumes a starting value of 3, which would already violate the condition for such small inputs.

For example:

Input:

n = 2

Execution:

The condition n == 2 triggers, so we output 1 directly.

This avoids starting the doubling process, which would incorrectly suggest no valid steps even though the problem explicitly defines a special outcome for this boundary.

For very large $n$, such as $10^9$, the loop runs only around 30 times due to exponential growth, and each iteration preserves exact integer arithmetic, ensuring no precision issues appear at power-of-two boundaries.