POJ 2484 A Funny Game (博弈)

题意：Alice 和 Bob 在玩一个无聊的游戏，n个棋子围成一圈，两人轮流从中取走一或两个棋子，不过取两个时必须是连续的棋子。

棋子取走之后留下空位，相隔空位的棋子不连续。Alice先取，取走最后一个棋子的人赢。如果都采取最优策略，谁会赢？

析：如果是小于两个，很明显是Alice胜，如果是3个是Bob胜，如果大于 3 个呢，Alice 先拿一两个，然后Bob从中间分开，使得两部分一样，

那么就一定是Bob胜，因为Alice拿了之后Bob从另一组中拿一样的即可，最后肯定Bob胜。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
  while(cin >> n && n)  cout << (n <= 2 ? "Alice" : "Bob") << endl;
  return 0;
}