AtCoder Regular Contest 163
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Problem Statement
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Alice and Bob are playing a game using a length-$N$ sequence of non-negative integers $A=(A_1,A_2,\dots,A_N)$.
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Starting with Alice, they take turns performing the following operation. The player who cannot make a move first loses.
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- Choose a non-negative integer
$X$such that there is an integer$i$satisfying$A_i > A_i \oplus X$. - For each
$1 \le i \le N$, replace$A_i$with$\min(A_i,A_i \oplus X)$.
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Determine who wins when both players play optimally.
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Here, $\oplus$ represents the bitwise XOR.
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You are given $T$ test cases. Find the answer for each of them.
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Constraints
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$1 \le T \le 100$$1 \le N \le 100$$0 \le A_i \le 10^9$
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Input
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The input is given from Standard Input in the following format:
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T
\mathrm{case}_1
\mathrm{case}_2
\vdots
\mathrm{case}_T翻译中…
Here, $\mathrm{case}_i$ is the $i$-th test case. Each test case is given in the following format:
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N
A_1 A_2 \dots A_N翻译中…
Output
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Print $T$ lines.
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The $i$-th line $(1 \le i \le T)$ should contain Alice if Alice wins, and Bob if Bob wins in the $i$-th test case.
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样例
Input #1
5
2
3 1
5
1 1 1 1 1
4
0 0 0 0
4
8 1 6 4
5
3 8 7 12 15
Output #1
Bob
Alice
Bob
Bob
Alice
In the first test case, one possible game progression could be as follows.
- Alice chooses $X=3$. This choice is valid because $3 > 3 \oplus 3(=0)$ for $i=1$.
- $A=(3,1)$ becomes $A=(0,1)$.
- Bob chooses $X=1$. This choice is valid because $1 > 1 \oplus 1(=0)$ for $i=2$.
- $A=(0,1)$ becomes $A=(0,0)$.
- Since Alice cannot choose any $X$, the game ends.
In this case, Bob wins.
In the second test case, one possible game progression could be as follows.
- Alice chooses $X=1$. This choice is valid because $1 > 1 \oplus 1(=0)$ for $i=1$.
- $A=(1,1,1,1,1)$ becomes $A=(0,0,0,0,0)$.
- Since Bob cannot choose any $X$, the game ends.
In this case, Alice wins.
题目信息
题面与样例版权归 AtCoder 及出题人所有；中文翻译由 AI 生成（CI 管线），如有出入请以原文为准。