[sicily online1087. A Funny Game
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
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 twoadjacent 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
题目解析：
一直希望能往动态规划上靠，结果靠了半天靠不上。
网上牛人解答：
不管先选的人怎么选，候选的人都在对称位置选择，使得剩下的数量为偶数，之后不管对方怎么选，候选者都在对称位置坐相同操作。总是后者胜。
比如：有偶数个，先选选了一个，候选者就在对称位置选2个。
早知道，多试几组数据，估计也能看出来
http://blog.csdn.net/titikdhu/article/details/5746349
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<iomanip>
#include<list>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#include <sstream>
#include <stack>
#include<queue>
#include<string.h>
using namespace std;
int main()
{
int n;
while(cin>>n&&n!=0)
{
if(n==1||n==2)cout<<"Alice"<<endl;
else cout<<"Bob"<<endl;
}
}