POJ-2484 A Funny Game （对称博弈）

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本文介绍了一种基于对称博弈理论的游戏策略问题。通过分析游戏规则，得出结论：当硬币数量大于3时，后手玩家可通过特定策略确保获胜。文章提供了一个简单的C语言程序实现，用于判断不同游戏场景下哪位玩家最终会赢得比赛。

A Funny Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3843		Accepted: 2298

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

Sample Output

Alice
Alice
Bob

思路：

n > 3 时，先手必败。题目很简单，而关键在于思想；

对称博弈的典型题目：

当n > 3 时，无论A怎么选择，B可以选一个特殊的位置拿走1个连续的2个，将A取剩的环路剪成两条相同的链路，然后，无论A从那条链路中那个位置取几个，B都可以在另一条链路中采用相同的取法。此时，B慢A一步，即最后一个一定是B取到的，B必胜！

代码：

#include <stdio.h>

int main()
{
	int n;
	while(scanf("%d", &n), n){
		if(n <= 2)
			printf("Alice\n");
		else
			printf("Bob\n");
	}

	return 0;
}