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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?
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 T lines, one for each test case, containing Alice if Alice wins the game, or Bob otherwise.
2 5 1
Alice Bob
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 #include2 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 }
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