POJ 2484-A Funny Game硬币排成环(博弈-找规律)

男娘i 2022-08-22 03:56 173阅读 0赞

# A Funny Game #

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 4802</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 2980</td> 
  </tr> 
 </tbody> 
</table>

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.)  
![2484_1.jpg][]  
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

1
    2
    3
    0

Sample Output

Alice
    Alice
    Bob

Source

[POJ Contest][],Author:Mathematica@ZSU

## 题目意思： ##

n枚硬币排成一个圈，Alice和Bob轮流从中取1或2枚硬币，不过取两枚时必须是连续的。硬币取走之后留下空位，相隔空位的硬币视为不连续的。  
Alice先取，取走最后一枚硬币的一方获胜。当双方都采取最优策略，谁会获胜？

## 解题思路： ##

Alice在第一步取走1或2枚硬币后，原本的环变成长度是n-1或n-2的链。  
这样只要Bob在中间位置，根据链长的奇偶性，取走1或2枚硬币，就可以正好把所有硬币分成两个长度相同的链。  
而两个长度相同的链这是必败态，因为不论自己采取什么策略，对手只要在另一组采取相同的策略，就又回到了分成两个如同长度链的状态。这样循环下去，总会轮到某次自己没有硬币可以取，即对手在上一次拿走了最后一枚硬币，这时候自己就输了。  
  
所以只有n=0/1/2时，是先手必胜态；其它均为先手必败态。

#include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int n;
    void solve()
    {
        if(n<=2)puts("Alice");
        else puts("Bob");
    }
    
    int main()
    {
        while(cin>>n,n)
        solve();
        return 0;
    }

[2484_1.jpg]: /images/20220724/be9299f16be740a78a9f677d06280de1.png
[POJ Contest]: http://poj.org/searchproblem?field=source&key=POJ+Contest