poj 2484 A Funny Game （博弈）

最新推荐文章于 2021-09-12 22:29:31 发布

clover_hxy

最新推荐文章于 2021-09-12 22:29:31 发布

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本文介绍了一个基于移除硬币的简单游戏，并通过分析确定了获胜策略。在该游戏中，玩家轮流从圆圈中移除一到两枚相邻的硬币，目标是最先移除最后一枚硬币的玩家获胜。文章提供了具体的实现代码，展示了如何根据硬币数量判断哪位玩家最终获胜。

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A Funny Game

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 <= 10 ⁶) 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 <= 10 ⁶). 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

Source

POJ Contest,Author:Mathematica@ZSU

[Submit] [Go Back] [Status] [Discuss]

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");
	}
}