description
Matt hzs 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.
Entry
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 ≤ 106).
In the second line, there are N integers ki (0 ≤ ki ≤ 106), indicating the i-th friend’s magic number.
Export
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.
Sample input
2
3 2
1 2 3
3 3
1 2 3
Sample Output
Case #1: 4
Case #2: 2
prompt
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.
The meaning of problems
Selecting the number N and any number of divergent or greater than the number M of the program.
answer
Counting DP.
DP [i] [j] represents the i-th and j is XOR article number scheme.
Obviously dp [i] [j] = dp [i-1] [j] + dp [i-1] [j ^ a [i]] (i installed or not installed i).
Since j up to 1e6, so the need to scroll through a dimension i.
Code
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 #define LL long long 5 LL dp[2][1<<20]; 6 int main() 7 { 8 ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); 9 int t,n,m,ca=1,a[45]; 10 cin>>t; 11 while(t--) 12 { 13 cin>>n>>m; 14 int mx=0; 15 for(int i=1;i<=n;i++)cin>>a[i],mx=max(mx,a[i]); 16 int sta=1; 17 while(sta<=mx)sta<<=1; 18 for(int i=0;i<sta;i++)dp[0][i]=dp[1][i]=0; 19 dp[0][0]=1; 20 int cur=1; 21 for(int i=1;i<=n;i++,cur^=1) 22 { 23 for(int j=0;j<sta;j++) 24 dp[cur][j]=dp[cur^1][j]+dp[cur^1][j^a[i]]; 25 for(int j=0;j<sta;j++) 26 dp[cur^1][j]=0; 27 } 28 LL sum=0; 29 for(int i=m;i<sta;i++)sum+=dp[cur^1][i]; 30 cout<<"Case #"<<ca++<<": "<<sum<<endl; 31 } 32 return 0; 33 }