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Gym - 100952G (博弈)

2024-11-22 13:06:02   203人阅读


Statements

Alice and Bob play the following game. They choose a number N to play with. The rules are as follows:

- They write each number from 1 to N on a paper and put all these papers in a jar.

- Alice plays first, and the two players alternate.

- In his/her turn, a player can select any available number M and remove its divisors including M.

- The person who cannot make a move in his/her turn wins the game.

Assuming both players play optimally, you are asked the following question: who wins the game?

Input

The first line contains the number of test cases T (1 ?≤? T ?≤? 20). Each of the next T lines contains an integer (1 ?≤? N ?≤? 1,000,000,000).

Output

 

Output T lines, one for each test case, containing Alice if Alice wins the game, or Bob otherwise.

 

Example

 
Input
2
5
1
 
Output
Alice
Bob

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 
 5 int main(){
 6     int T; scanf("%d", &T);
 7     while(T--){
 8         int n;
 9         scanf("%d", &n);
10         if(n>1) printf("Alice\n");
11         else printf("Bob\n");
12     }
13     return 0;
14 }
15 /*
16 题意：有一个容器里装着 n个数，
17 A和B每次任意说一个数m，
18 那么他要拿走容器里m的所有因子，
19 如果谁拿空了容器，那么他输了，
20 求先手赢还是后手赢
21 
22 思路：只有1是后手赢，
23 因为1只能拿一次，
24 大于1的情况，
25 假设a和b都不是聪明的，
26 假设先手出x，后手出y
27 结果是后手赢，
28 那么现在a和b都是聪明的，
29 先手可以直接出x*y（
30 即先手可以复制后手的操作，
31 先手的操作可以包括后手的操作），
32 则可以赢后手，
33 所以大于1的情况一定是先手赢
34 */

 

Gym - 100952G (博弈)


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