poj2484A Funny Game+对称博弈

Description
Alice and Bob decide to play a funny game. At the beginning of the game they pick n(1 <= n <= 106) coins in a circle, as Figure 1 shows. A move consists in removing one or two adjacent coins, leaving all other coins untouched. At least one coin must be removed. Players alternate moves with Alice starting. The player that removes the last coin wins. (The last player to move wins. If you can’t move, you lose.)

Figure 1

Note: For n > 3, we use c1, c2, …, cn to denote the coins clockwise and if Alice remove c2, then c1 and c3 are NOT adjacent! (Because there is an empty place between c1 and c3.)

Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.

Input
There are several test cases. Each test case has only one line, which contains a positive integer n (1 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input.

Output
For each test case, if Alice win the game,output “Alice”, otherwise output “Bob”.

Sample Input

1
2
3
0

Sample Output

Alice
Alice
Bob
/*
0000   0000
000       000
00           00
0                0
*/

由上式可得如果存在一种状态能是两个对称的链，先手怎么取
后手都能让这个状态再次回到对称的状态
环状取了之后便是链装，只要取走之后变成对称链，便能赢。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int n;    while(scanf("%d",&n),n!=0){        if(n<3) printf("Alice\n");        else printf("Bob\n");    }    return 0;}