poj 2484 A Funny Game （博弈）

A Funny Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5460 Accepted: 3422

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

1230

Sample Output

AliceAliceBob

Source

POJ Contest,Author:Mathematica@ZSU

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题解：博弈

这道题其实不用sg函数，可以直接分析出答案。

我们发现，只要个数不是1或者2，那么后手一定有办法把剩下的全部取走或者分成两个相同的子状态。

所以只1，2是先手胜。

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<cmath>using namespace std;int main(){int n;while (true){scanf("%d",&n);if (!n) break;if (n==1||n==2) printf("Alice\n");else printf("Bob\n");}}