LightOJ 1253 Misere NIM(反NIM博弈)

Alice and Bob are playing game of Misère Nim. Misère Nim is a game playing on k piles of stones, each pile containing one or more stones. The players alternate turns and in each turn a player can select one of the piles and can remove as many stones from that pile unless the pile is empty. In each turn a player must remove at least one stone from any pile. Alice starts first. The player who removes the last stone loses the game.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer k (1 ≤ k ≤ 100). The next line contains k space separated integers denoting the number of stones in each pile. The number of stones in a pile lies in the range [1, 10⁹].

Output

For each case, print the case number and 'Alice' if Alice wins otherwise print 'Bob'.

Sample Input

3

4

2 3 4 5

5

1 1 2 4 10

1

1

Sample Output

Case 1: Bob

Case 2: Alice

Case 3: Bob

题解：反NIM博弈板题；（anti-NIM博弈见前面博弈总结）知道：先手必胜的状态有两种： 1：全是一个，且ans=0;

2 : 不全是一个，且ans!=0;(ans为所有的异或和);

其他的情况均为先手必败；

参考代码：

 #include<bits/stdc++.h>
 using namespace std;
 int T,n,x,ans,cnt;
 int main()
 {
     scanf("%d",&T);
     for(int cas=;cas<=T;cas++)
     {
         scanf("%d",&n);
         cnt=ans=;
         for(int i=;i<=n;++i)
         {
             scanf("%d",&x);
             if(x==) cnt++;
             ans^=x;
         }
         if(cnt==n)
         {
             if(ans==) printf("Case %d: Alice\n",cas);
             else  printf("Case %d: Bob\n",cas);
         }
         else
         {
             if(ans) printf("Case %d: Alice\n",cas);
             else printf("Case %d: Bob\n",cas);
         }
     } 

     return ;
 }