Codeforces Round 1007 (Div. 2) |
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Finished |
This is the easy version of the problem. The difference between the versions is that in this version, $$$l=r$$$. You can hack only if you solved all versions of this problem.
You are given a positive integer $$$n$$$ and the first $$$n$$$ terms of an infinite binary sequence $$$a$$$, which is defined as follows:
Your task is to compute the sum of elements in a given range $$$[l, r]$$$: $$$a_l + a_{l + 1} + \ldots + a_r$$$.
$$$^{\text{∗}}$$$$$$\oplus$$$ denotes the bitwise XOR operation.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The first line of each test case contains three integers $$$n$$$, $$$l$$$, and $$$r$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le l=r\le 10^{18}$$$).
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$\color{red}{a_i \in \{0, 1\}}$$$) — the first $$$n$$$ terms of the sequence $$$a$$$.
It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, output a single integer — the sum of elements in the given range.
91 1 112 3 31 03 5 51 1 11 234 23405 1111 11111 0 1 0 11 1000000000000000000 1000000000000000000110 87 870 1 1 1 1 1 1 1 0 012 69 691 0 0 0 0 1 0 1 0 1 1 013 46 460 1 0 1 1 1 1 1 1 0 1 1 1
1 1 0 0 1 0 1 0 0
In the first test case, the sequence $$$a$$$ is equal to $$$$$$[\underline{\color{red}{1}}, 1, 1, 0, 0, 1, 1, 1, 1, 1, \ldots]$$$$$$ where $$$l = 1$$$, and $$$r = 1$$$. The sum of elements in the range $$$[1, 1]$$$ is equal to $$$$$$a_1 = 1.$$$$$$
In the second test case, the sequence $$$a$$$ is equal to $$$$$$[\color{red}{1}, \color{red}{0}, \underline{1}, 1, 1, 0, 0, 1, 1, 0, \ldots]$$$$$$ where $$$l = 3$$$, and $$$r = 3$$$. The sum of elements in the range $$$[3, 3]$$$ is equal to $$$$$$a_3 = 1.$$$$$$
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