POJ 2484:A Funny Game__博弈論

A Funny Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5099		Accepted: 3180

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

1230

Sample Output

AliceAliceBob

題意：有n枚硬幣圍成一個圈，每個人只能取走連續的一個或者兩個硬幣，取走的地方為空白，Alice為先手，問最終誰會獲勝。

當n==1 || n==2時，明顯先手必勝。

當n==3時，明顯先手必敗。

由於每次只可取1或2個，而取2個時，2個必須相鄰，

推斷有：當n>3時，若n為偶數，先手無論如何取，後手可在先手對稱的位置上取同等數量，於是先手必敗。

若n為奇數，先手取1個時，後手可在先手對稱的位置上取2個，之後無論先手如何取，後手都可在先手對稱的位置上取同等數量，先手必敗。

如果先手一開始取2個時，後手可在先手對稱的位置上取1個，之後還剩下偶數個，可如上推出先手必敗。故當 n>3時，先手必敗。

AC代碼：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;int main(){    long long int n;    while(cin>>n&&n)    {        if(n<=2)cout<<"Alice"<<endl;        else cout<<"Bob"<<endl;    }    return 0;}