CF 103306K - K-Binary Repetitive Numbers

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Solve time: 48s
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Solution

Problem Understanding

We are given a positive integer $K$. We look at all binary representations of numbers $N$, but not in their usual minimal form. Instead, each number is written using exactly $K$ bits, including leading zeros. So every $N$ corresponds to a fixed-length binary string of length $K$.

A number $N$ is called $K$-binary repetitive if this $K$-bit string can be expressed as a repetition of some shorter binary pattern. In other words, there exists a non-empty binary string $S$, shorter than length $K$, such that repeating $S$ some number of times gives exactly the $K$-bit representation of $N$.

The task is, for each query $K$, to count how many integers $N$ in the range $[0, 2^K - 1]$ produce a $K$-bit representation that is periodic in this sense, and output the count modulo $10^9 + 7$.

The key subtlety is that we are not asking for periodicity in the integer sense, but periodicity of the fixed-length binary string including leading zeros. This means strings like 0011 are valid candidates for repetition patterns even though the integer itself would normally be written differently.

The constraints matter heavily. We have up to $10^5$ test cases and $K$ up to $10^6$, so any solution that processes each $K$ independently in linear time is impossible. Even $O(K)$ per query would lead to $10^{11}$ operations in the worst case, which is far beyond limits. This forces a precomputation or a closed-form combinational observation that reduces each query to $O(1)$.

A few edge cases clarify the definition.

When $K = 1$, there are only two strings: 0 and 1. Neither can be formed by repeating a shorter non-empty string, so the answer is $0$.

When $K = 2$, the strings are 00, 01, 10, 11. Here 00 is repetition of 0, and 11 is repetition of 1, so there are 2 valid numbers.

A common mistake is to treat only “clean” integer representations, ignoring leading zeros. That would incorrectly reject cases like 00, which are crucial for the periodic structure.

Another mistake is assuming any string with repeated characters is valid. For example 0101 is valid, but 0110 is not, even though both contain repeated bits, because only the first has a consistent period dividing the full length.

Approaches

A brute-force approach is straightforward: for each $K$-bit string, try every possible period length $d$ that divides $K$. For each divisor $d < K$, check if the prefix of length $d$ repeated $K/d$ times reconstructs the full string. If any divisor works, the string is counted.

This works because periodic strings are exactly those with a proper period dividing the full length. However, this is catastrophically expensive. For each of the $2^K$ strings, we may check up to $O(K)$ divisors and perform $O(K)$ comparisons, giving $O(K \cdot 2^K)$, which is infeasible even for very small $K$, let alone up to $10^6$.

The key observation is that we are not working with a fixed string, but counting how many strings of length $K$ are periodic. This transforms the problem from per-string checking into a combinatorial counting problem.

A binary string of length $K$ is fully determined by its smallest period $p$, and once we fix $p$, the string is determined by choosing the first $p$ bits, then repeating them. However, overcounting occurs because a string with period $p$ is also counted for all multiples of $p$. This is exactly a divisor lattice structure over $K$.

So instead of counting periodic strings directly, we count all strings and subtract those that are “primitive”, meaning they have no smaller period. The number of all strings is $2^K$. The number of primitive strings can be computed using Möbius inversion over divisors of $K$. Once we know primitive counts for all divisors, we can derive periodic counts efficiently.

This leads to a standard number-theoretic DP over divisors: we precompute $2^K$ and then subtract contributions of smaller divisors using inclusion-exclusion.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(2^K \cdot K)$	$O(1)$	Too slow
Divisor + Möbius / DP	$O(K \log K)$ precompute, $O(1)$ per query	$O(K)$	Accepted

Algorithm Walkthrough

Precompute powers of two up to the maximum $K$ across all queries, since $2^K$ represents all binary strings of length $K$. This gives a base count for every case.
For each value $K$, initialize the idea that all strings are valid candidates before filtering out those that are primitive. The structure of repetition depends only on divisors of $K$, so we focus on divisor relationships.
Iterate over all possible divisors $d$ of each $K$ in increasing order, treating $d$ as a potential period length. A string with period $d$ is formed by repeating a block of size $d$, so there are $2^d$ such blocks before considering overcounting.
Subtract contributions from smaller divisors that already represent more fundamental patterns. This step ensures that each periodic structure is counted exactly once at its minimal period. The subtraction follows inclusion-exclusion over the divisor lattice.
After computing the number of primitive strings for all divisors, derive the number of periodic strings as total minus primitive strings for the full length. This isolates exactly those strings that have some proper repetition structure.
Answer each query in $O(1)$ using the precomputed arrays.

Why it works

Every $K$-bit string has a unique minimal period, which is the smallest prefix that generates it by repetition. This induces a partition of all strings by their minimal period. The brute-force counts overlap because a string with minimal period $d$ is also compatible with every multiple of $d$, but it should only be attributed once. Möbius inversion corrects this overcounting by systematically canceling contributions from smaller divisors, ensuring each string is assigned to exactly one minimal period class. This guarantees correctness of the final subtraction-based count.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

MAX_K = 10**6

# precompute powers of 2
pow2 = [1] * (MAX_K + 1)
for i in range(1, MAX_K + 1):
    pow2[i] = (pow2[i - 1] * 2) % MOD

# mobius function via linear sieve
mu = [1] * (MAX_K + 1)
is_prime = [True] * (MAX_K + 1)
primes = []

for i in range(2, MAX_K + 1):
    if is_prime[i]:
        primes.append(i)
        mu[i] = -1
    for p in primes:
        if i * p > MAX_K:
            break
        is_prime[i * p] = False
        if i % p == 0:
            mu[i * p] = 0
            break
        else:
            mu[i * p] = -mu[i]

# precompute periodic counts using divisor sums
periodic = [0] * (MAX_K + 1)

for i in range(1, MAX_K + 1):
    # strings with period dividing i
    total = 0
    j = 1
    while j * j <= i:
        if i % j == 0:
            d1 = j
            d2 = i // j
            total = (total + mu[i // d1] * pow2[d1]) % MOD
            if d1 != d2:
                total = (total + mu[i // d2] * pow2[d2]) % MOD
        j += 1
    periodic[i] = (pow2[i] - total) % MOD

t = int(input())
for _ in range(t):
    k = int(input())
    print(periodic[k] % MOD)

The code separates the problem into two conceptual layers. First, it precomputes powers of two, since every binary string corresponds to a free choice of bits. Second, it uses a Möbius-based inclusion-exclusion over divisors to isolate primitive strings, which are exactly those not formed by repeating a smaller block. The final answer subtracts primitive counts from the total space.

A common implementation pitfall is forgetting that leading zeros are part of the string space, which is why all counts are over $2^K$, not over integers in a reduced representation.

Another subtle point is handling divisor enumeration efficiently. Precomputing Möbius avoids recomputing divisor structures per query, which would otherwise lead to timeouts under $10^5$ queries.

Worked Examples

Example 1

Input:

$K = 1$

Step	Total strings	Primitive count	Periodic count
1	2 (`0`, `1`)	2	0

The only two strings have no shorter repeating structure, so none qualify.

This confirms that the algorithm correctly excludes trivial strings with no proper period.

Example 2

Input:

$K = 2$

Step	Total strings	Primitive count	Periodic count
1	4 (`00`, `01`, `10`, `11`)	2 (`01`, `10`)	2

Only 00 and 11 are periodic since they are repetitions of single-bit patterns. The table shows how the method cleanly separates structured strings from non-structured ones.

This demonstrates that the divisor-based decomposition correctly captures repetition even with leading zeros.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(MAX_K \log MAX_K)$	Sieve for Möbius + divisor processing over all $K$
Space	$O(MAX_K)$	Arrays for powers, Möbius values, and DP

The preprocessing fits comfortably within limits since $MAX_K = 10^6$. Each query is answered in constant time, making $10^5$ queries feasible.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    # call solution logic here (wrap main in function in practice)
    ...

# provided samples
# assert run("2\n1\n2\n") == "0\n2\n"

# custom cases
# K = 3: 000,111,010101 patterns
# K = 4: multiple periodic structures
# K = 5: prime length edge case (only full-length repeats)

Test input	Expected output	What it validates
K=1	0	base case, no repetition possible
K=2	2	smallest non-trivial periodic cases
K=3	2	prime-length behavior constraints
K=4	6	multiple divisor structure
K=5	2	prime length edge behavior

Edge Cases

For $K = 1$, the algorithm correctly returns 0 because there are no proper divisors to form repetition, so inclusion-exclusion leaves no periodic structures.

For prime $K$, only strings that repeat a length-1 pattern or full-length patterns exist, and the subtraction correctly removes all full-length-only structures, leaving only constant strings.

For powers of two, the divisor lattice is dense, and the Möbius cancellation ensures that nested repetitions like period 2 inside period 4 are not double counted, preserving correctness across all hierarchical repetitions.