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，每次可以从一堆中取任意个石子，至少一个，拿最后一个石子的人输（经典Nim博弈为拿走最后一颗石子胜）
当N堆石子都为1时，奇数个石子堆，先手必败；其他情况符合Nim博弈性质。Nim博弈的必胜态在这个游戏中任然是必胜态，因为处于必胜态的人一定可以把最后一颗石子留给对方
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn=120;
LL a[120];
int main(){
int T;
int ca=1;
scanf("%d",&T);
while(T--){
int n,m;
scanf("%d",&n);
LL ans=0;
bool flag=true;
for(int i=0;i<n;i++){
scanf("%lld",&a[i]);
if(a[i]!=1) flag=false;
ans^=a[i];
}
printf("Case %d: ",ca++);
if(flag){
if(n%2==1)
printf("Bob\n");
else
printf("Alice\n");
continue;
}
if(ans==0)
printf("Bob\n");
else
printf("Alice\n");
}
return 0;
}