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<iostream>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
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<n;i++)
25 cin>>a[i];
26
27 for(int i=0;i<n;i++)
28 {
29 roll=roll^1;
30 for(int j=0;j<=(C/2);j++)
31 dp[roll][j]=dp[roll^1][j]+dp[roll^1][j^a[i]];
32 }
33 long long ans=0;
34 for(int i=m;i<=(C/2);i++)
35 ans=ans+dp[roll][i];
36 cout<<"Case #"<<tt-t<<": "<<ans<<endl;
37
38 }
39 return 0;
40 }
41