LightOJ-1253-Misere Nim-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

题意:

有n堆石子，每堆石子里面至少有一个石子，有A、B两人。A先取，取完所有石子的一方获胜，问当双方都采取最优策略时，谁能获胜。

思路：

nim博弈模板，谁面临平衡态势谁就会输。

特判一种情况：当每一堆石子的个数全部都为1的时候，这个时候只能每次拿一个，根据1的奇偶性进行判断。

 #include<stdio.h>

 int main()
 {
     int t,tt=;
     scanf("%d",&t);
     while(t--)
     {
         int n,xx;
         scanf("%d",&n);
         int x=,flag=;
         for(int i=; i<n; i++)
         {
             scanf("%d",&xx);
             if(xx!=)
             {
                 flag=;
             }
             x^=xx;
         }
         if(flag==)
         {
             if(n%==)
                 printf("Case %d: Alice\n",tt++);
             else
                 printf("Case %d: Bob\n",tt++);
         }
         else
         {
             if(x!=)
                 printf("Case %d: Alice\n",tt++);
             else
                 printf("Case %d: Bob\n",tt++);
         }
     }
     return ;
 }