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=1;cas<=T;cas++)
    {
        scanf("%d",&n);
        cnt=ans=0;
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&x);
            if(x==1) cnt++;
            ans^=x;    
        }    
        if(cnt==n)
        {
            if(ans==0) 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 0;
}