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:http://codeforces.com/gym/100952/problem/G

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:

 #include <bits/stdc++.h>
using namespace std;
inline int read()
{
int x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')
f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<)
{
putchar('-');
x=-x;
}
if(x>)
write(x/);
putchar(x%+'');
}
int main()
{
int T;
T=read();
while(T--)
{
int n;
n=read();
if(n>)
printf("Alice\n");
else printf("Bob\n");
}
return ;
}

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