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A Funny Game

2013-09-02 12:11 190 查看

                                               1087. A Funny Game

Description
 Aliceand 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 leastone 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 1Note: 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. InputThereare 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. OutputForeach test case, if Alice win the game,output "Alice", otherwise output "Bob". Sample Input Copysample input to clipboard
1
2
3
0
SampleOutput
Alice
Alice
Bob
转：
当n==1 || n==2时，明显先手必胜。
当n==3时，明显先手必败。
由于每次只可取1或2个，而取2个时，2个必须相邻，推断有：
当n>3时，
若n为偶数，先手无论如何取，后手可在先手对称的位置上取同等数量，于是先手必败。
若n为奇数，先手取1个时，后手可在先手对称的位置上取2个，之后无论先手如何取，后手都可在先手对称的位置上取同等数量，先手必败。
先手一开始取2个时，可如上推出先手必败。
#include<iostream>using namespace std;int main(){int num;while(cin>>num){if(num==0) return 0;if(num<=2) cout<<"Alice"<<endl;else cout<<"Bob"<<endl;}return 0;}

                                            

                                        

                        
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                                                sicily
                                                

                        

                    


                

            


            

                
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