poj 2484 A Funny Game
A Funny Game
http://poj.org/problem?id=2484
| Time Limit: 1000MS | Memory Limit: 65536K | |
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.
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
结论：n=1或2，先手必胜，否则，先手必败
当n=1或2，先手一次取走
当n=3，先手取一次，后手再取一次
当n>3时，先手从任何地方取1或2，都会将环变成一个链，后手一定可以把这个链变成 两个长度相等的链
然后后手一定赢
#include<cstdio> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { if(!n) return 0; if(n==1||n==2) printf("Alice\n"); else printf("Bob\n"); } }
