4469: C. Even Number Addicts
内存限制:512 MB
时间限制:2 S
标准输入输出
题目类型:传统
评测方式:Special Judge
上传者:
提交:47
通过:15
题目描述
Alice and Bob are playing a game on a sequence a1, a2, ..., an of length n. They move in turns and Alice moves first.
In the turn of each player, he or she should select an integer and remove it from the sequence. The game ends when there is no integer left in the sequence.
Alice wins if the sum of her selected integers is even; otherwise, Bob wins.
Your task is to determine who will win the game, if both players play optimally.
In the turn of each pla
Alice wins if the sum of her selected integers is even; otherwise, Bob wins.
Your task is to determine who will win the game, if both pla
输入格式
Each test contains multiple test cases. The first line contains an integer t (1 <= t <= 100) — the number of test cases. The following lines contain the description of each test case.
The first line of each test case contains an integer n (1 <= n <= 100), indicating the length of the sequence.
The second line of each test case contains n integers a1, a2, ..., an (-109 <= ai <= 109), indicating the elements of the sequence.
The first line of each test case contains an integer n (1 <= n <= 100), indicating the length of the sequence.
The second line of each test case contains n integers a1, a2, ..., an (-109 <= ai <= 109), indicating the elements of the sequence.
输出格式
For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.
输入样例 复制
4
3
1 3 5
4
1 3 5 7
4
1 2 3 4
4
10 20 30 40
输出样例 复制
Alice
Alice
Bob
Alice
数据范围与提示
In the first and second test cases, Alice always selects two odd numbers, so the sum of her selected numbers is always even. Therefore, Alice always wins.
In the third test case, Bob has a winning strategy that he always selects a number with the same parity as Alice selects in her last turn. Therefore, Bob always wins.
In the fourth test case, Alice always selects two even numbers, so the sum of her selected numbers is always even. Therefore, Alice always wins.
In the third test case, Bob has a winning strategy that he always selects a number with the same parity as Alice selects in her last turn. Therefore, Bob always wins.
In the fourth test case, Alice always selects two even numbers, so the sum of her selected numbers is always even. Therefore, Alice always wins.