Description

Matt has N friends. They are playing a game together. 

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins. 

Matt wants to know the number of ways to win.

 

Input

The first line contains only one integer T , which indicates the number of test cases. 

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10 6). 

In the second line, there are N integers ki (0 ≤ k i ≤ 10 6), indicating the i-th friend’s magic number.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.

 

  正解貌似是高斯消元....不会的说....PYY说不就是解方程么....然后我去看了下.....挺多概念没学线代的话还是挺眼生的...（orzpyy大神）记得高三的时候做过一个用矩阵加速求线性递推式的东西....当时10^18的数据秒过....还是挺让我惊讶了一下...

扯远了... 这题dp也能做，有点类似01背包（dp真是够渣，寒假看看能不能抽时间弄一下，依然是最大的短板）

只不过状态转移方程是 dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]]

然后正常些貌似会MLE。。。。。用到了滚动数组

其实滚动数组,完全...没有理解的难度啊 

竟然是我第一次使用，大概是因为基本没怎么做过DP的题吧，而滚动数组这个优化貌似主要在Dp的题里需要...

然后在搜滚动数组的时候，看到一篇博客里用异或来表示两个状态我觉得这一点也很赞.....

我自己想的话的大概就要又加，又mod，然后还得再来一个变量了吧.....差评满满

还有位运算....是挺神的东西....目前还处于一知半解的阶段....ORZ  M67大神

 

1   2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 const long long C=1<<21; 7 long long dp[2][C]; 8 using namespace std; 9  10 int main()11 {12     int t,a[49];13     cin>>t;14     int tt;15     tt=t;16     while (t--)17  18     {19         int n,m,roll;20         roll=0;21          memset(dp,0,sizeof(dp));22         dp[0][0]=1;23         cin>>n>>m;24         for(int i=0;i
          
           >a[i];26  27         for(int i=0;i