POJ2484 A Funny Game【博弈】

A Funny Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6905		Accepted: 4327

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

Source

POJ Contest,Author:Mathematica@ZSU

问题链接：POJ2484 A Funny Game

问题描述：

n枚硬币排成一圈，Alice和Bob两个人轮流从中取一枚或两枚硬币，不过取两枚硬币时必须连续的。Alice先取，取走最后的硬币则获胜。计算获胜方。

问题分析：

n=1或n=2时，先手Alice全部取走就可以取胜。

n=3时，不论先手Alice怎么取（取一枚或两枚），后手Bob去走剩下的全部获胜。

n>3时，可以分为n为奇数或偶数。采用对称取硬币的方式，后手方总可以获胜。

程序说明：（略）

参考链接：（略）

题记：（略）

AC的C++语言程序如下：

/* POJ2484 A Funny Game */

#include <iostream>
#include <stdio.h>

using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin >> n && n)
        printf("%s\n", n <= 2 ? "Alice" : "Bob");

    return 0;
}

POJ2484 A Funny Game【博弈】

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