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。
1 #include<stdio.h> 2 3 int main() 4 { 5 int t,tt=1; 6 scanf("%d",&t); 7 while(t--) 8 { 9 int n,xx; 10 scanf("%d",&n); 11 int x=0,flag=0; 12 for(int i=0; i<n; i++) 13 { 14 scanf("%d",&xx); 15 if(xx!=1) 16 { 17 flag=1; 18 } 19 x^=xx; 20 } 21 if(flag==0) 22 { 23 if(n%2==0) 24 printf("Case %d: Alice\n",tt++); 25 else 26 printf("Case %d: Bob\n",tt++); 27 } 28 else 29 { 30 if(x!=0) 31 printf("Case %d: Alice\n",tt++); 32 else 33 printf("Case %d: Bob\n",tt++); 34 } 35 } 36 return 0; 37 }