HDU

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.

InputThe 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.OutputFor 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.Sample Input

2
3 2
1 2 3
3 3
1 2 3

Sample Output

Case #1: 4
Case #2: 2


        
 

Hint

In the first sample, Matt can win by selecting:
friend with number 1 and friend with number 2. The xor sum is 3.
friend with number 1 and friend with number 3. The xor sum is 2.
friend with number 2. The xor sum is 2.
friend with number 3. The xor sum is 3. Hence, the answer is 4.

        
 这题确实没想到是冒泡。
另外注意数组开的时候小心点，在如果for循环1<<20 到1<<21的话，会越界。（因为是异或
还有就是最后答案会爆int（这个应该都能注意到
果然dp这方面自己还是只会模板，改改就蒙了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1<<20;
int nu[maxn];
int dp[41][maxn+10];
int main()
{
    ios::sync_with_stdio(false);
    int t,n,m;
  //  cout<<setiosflags(ios::fixed)<<setprecision(6);
    cin>>t;
    for(int cas=1;cas<=t;++cas)
    {
        cin>>n>>m;
        for(int i=1;i<=n;++i)
            cin>>nu[i];
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1;i<=n;++i)
        {
            for(int j=0;j<maxn;++j)
            {
                dp[i][j]=dp[i-1][j]+dp[i-1][j^nu[i]];
            }
        }
        ll ans=0;
        
        for(int i=m;i<maxn;++i)
            ans+=dp[n][i];
        cout<<"Case #"<<cas<<": "<<ans<<endl;
    }
    return 0;
}