POJ 2484 —— 博弈
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A Funny Game
Description
Alice and Bob decide to play a funny game. At the beginning of the game they pick n(1 <= n <= 10
6) 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 <= 10
6). 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
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题意是n枚硬币排一个圈，Alice和Bob轮流取，一次可以取一枚或连续的两枚，已取走的硬币左右硬币不算相连，Alice先走，输出赢家。
思路：本题就是一种对称的思路，我们首先可以证明相同的两列是必败态，因为你无论取什么，对手不断模仿你就行。
本题中Alice无论第一次取什么，Bob都可以取走1或2枚使得硬币形成相同的两列；特例是n<= 2 ， Alice一次取完的情况。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 15 + 5;
const int MAXS = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-10
const long long MOD = 1000000000 + 7;
const int mod = 10007;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
typedef vector<int> vec;
typedef vector<vec> mat;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int main()
{
int n;
while(~scanf("%d" , &n) , n)
{
if(n <= 2)puts("Alice");
else puts("Bob");
}
return 0;
}