G. The jar of divisors
time limit per test:2 seconds
memory limit per test:64 megabytes
input:standard input
output:standard output
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.
Examples
Input
2
5
1
Output
Alice
Bob
题目链接：codeforces.com/gym/100952/…
题意：有一个容器里装着1-n n个数，A和B每次任意说一个数m，那么他要拿走容器里m的所有因子，如果谁拿空了容器，那么他输了，求先手赢还是后手赢
思路：只有1是后手赢，因为1只能拿一次，大于1的情况，假设a和b都不是聪明的，假设先手出x，后手出y，结果是后手赢，那么现在a和b都是聪明的，先手可以直接出x*y（即先手可以复制后手的操作，先手的操作可以包括后手的操作），则可以赢后手，所以大于1的情况一定是先手赢！
下面给出AC代码：
1 #include <bits/stdc++.h>
2 using namespace std;
3 inline int read()
4 {
5 int x=0,f=1;
6 char ch=getchar();
7 while(ch<'0'||ch>'9')
8 {
9 if(ch=='-')
10 f=-1;
11 ch=getchar();
12 }
13 while(ch>='0'&&ch<='9')
14 {
15 x=x*10+ch-'0';
16 ch=getchar();
17 }
18 return x*f;
19 }
20 inline void write(int x)
21 {
22 if(x<0)
23 {
24 putchar('-');
25 x=-x;
26 }
27 if(x>9)
28 write(x/10);
29 putchar(x%10+'0');
30 }
31 int main()
32 {
33 int T;
34 T=read();
35 while(T--)
36 {
37 int n;
38 n=read();
39 if(n>1)
40 printf("Alice\n");
41 else printf("Bob\n");
42 }
43 return 0;
44 }
