POJ 2484(博弈-对称博弈)

A Funny Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3345   Accepted: 1960

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.) 

 

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

1
2
3
0

Sample Output

Alice
Alice
Bob

Source

POJ Contest,Author:Mathematica@ZSU

博弈第一招——对称博弈。

思路:无论对方怎么取,我都取与它对称(或者相反/一样的)

#include<cstdio>

include<cstdlib>

include<algorithm>

include<functional>

include<cstring>

include<iostream>

using namespace std; int main() { int n; while(scanf("%d",&n)&&n) { if (n>=3) cout<<"Bob"<<endl; else cout<<"Alice"<<endl; } return 0; }