Editorial: Arrays and Queries

Notations & Definitions

Let's denote:

• $n$ = number of elements in array $a$
• $b$ = the final constructed array of size $2^n - 1$
• $S(l, r)$ = sum of elements $b_l + b_{l+1} + \ldots + b_r$
• For any position $p$ (1-indexed), we can represent it in binary
• $\text{MOD} = 10^9 + 7$

Key Observation: The value at position $p$ (1-indexed) in array $b$ is determined by the position of the lowest set bit (trailing zeros) in $p$. Specifically, if position $p$ has $k$ trailing zeros in its binary representation, then $b_p = a_{k+1}$.

Subtask 1: All elements equal ($a_i = c$ for all $i$)

Editorial: When all elements are equal to some constant $c$, every position in array $b$ has value $c$. Therefore:

$S(l, r) = c \cdot (r - l + 1) \bmod \text{MOD}$

Time Complexity: $O(1)$ per query
Space Complexity: $O(1)$

Subtask 2: Queries where $l = 1, r = 2^k - 1$

Editorial: These queries ask for the sum of the entire array built after $k$ rounds. We can use the recurrence:

• After round 1: sum = $a_1$
• After round $i$: sum = $2 \cdot \text{sum}_{i-1} + a_i$

This is because $b := b + [a_i] + b$, so the new sum is twice the old sum plus $a_i$.

Time Complexity: $O(n)$ per query
Space Complexity: $O(1)$

Subtask 3: $n \leq 20$

Editorial: For small $n$, we can actually construct the array $b$ explicitly since $|b| = 2^n - 1 \leq 2^{20} - 1 \approx 10^6$. Then we can compute prefix sums and answer each query in .

Time Complexity: $O(2^n + q)$
Space Complexity: $O(2^n)$

Subtask 4: Point queries ($l = r$)

Editorial: For a point query at position $p$, we need to find which element of $a$ is at position $p$. The key insight is that $b_p = a_{k+1}$ where is the number of trailing zeros in the binary representation of (equivalently, the position of the lowest set bit, 0-indexed from right).

Algorithm:

Example:

• Position 1 (binary: 001): 0 trailing zeros → $a_1$
• Position 2 (binary: 010): 1 trailing zero → $a_2$
• Position 3 (binary: 011): 0 trailing zeros → $a_1$
• Position 4 (binary: 100): 2 trailing zeros → $$

Time Complexity: $O(1)$ per query
Space Complexity: $O(n)$

Subtask 5: Full Solution

Editorial: The key insight is to compute prefix sums efficiently without building the array. For a position $n$ (1-indexed), we can compute $\text{PrefixSum}(n)$ by analyzing the binary representation of $n$.

Mathematical Insight: The contribution of each $a_i$ to the prefix sum can be computed by examining the bits of $n$:

$\text{PrefixSum}(n) = \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^i} \right\rfloor \cdot a_i - \left\lfloor \frac{n}{2^i} \right\rfloor \cdot a_{i-1}$

where $a_{-1} = 0$.

The algorithm works as follows:

1. Initialize $\text{ans} = 0$, $\text{prev} = 0$
2. For each bit position $i$ from 0 to $n-1$ (while $n > 0$):

To answer a query $(l, r)$: compute $\text{PrefixSum}(r) - \text{PrefixSum}(l-1)$.

Why this works: Each element $a_i$ appears at positions where the number of trailing zeros equals $i$. The number of such positions in the first $n$ positions follows a pattern based on the binary representation of $n$. The term $\lfloor n/2^i \rfloor$ counts occurrences, and subtracting the previous element's contribution accounts for the transition between levels.

Time Complexity: $O(n)$ per query
Space Complexity: $O(n)$