POJ2484

A Funny Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6178		Accepted: 3861

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

1
2
3
0

Sample Output

Alice
Alice
Bob

Source

POJ Contest,Author:Mathematica@ZSU

 

题意：每人每次可以取一个或两个连续的硬币，硬币之间有空位视作不连续。取完最后一个硬币的人获胜。问先手必胜还是后手必胜。

思路：n<=2时，先手必胜。n>2时，后手以中心对称模仿先手，必胜。

 //2017-10-26
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 int main()
 {
     int n;
     while(cin>>n && n){
         if(n <= )cout<<"Alice"<<endl;
         else cout<<"Bob"<<endl;
     }

     return ;
 }

 import java.util.*;

 public class Main{
     public static void main(String args[]){
         int n;
         Scanner cin = new Scanner(System.in);
         while(cin.hasNext()){
             n = cin.nextInt();
             if(n == 0)break;
             if(n <= 2)System.out.println("Alice");
             else System.out.println("Bob");
         }
     }
 }