Problem 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.
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 ≤ 106).
In the second line, there are N integers ki (0 ≤ ki ≤ 106), indicating the i-th friend’s magic number.
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.
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.
Sample Input
2
3 2
1 2 3
3 3
1 2 3
Sample Output
Case #1: 4
Case #2: 2
题意：给出n个数，求n个数中几个数通过异或运算得到m的个数。
思路：设dp[i][j]为前i个数中异或为j的个数，dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]];这dp[i-1][j]很容易懂，就不解释了，对于dp[i-1][j^a[i]]，因为j^a[i]^a[i]=j;所以j^a[i]就是前i-1个数中异或能的到j的个数。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 long long dp[3][2000006]; 6 int main() 7 { 8 long long t,n,m,i,j,k=1; 9 long long ans; 10 int a[44]; 11 scanf("%I64d",&t); 12 while (t--) 13 { 14 scanf("%I64d%I64d",&n,&m); 15 for (i=1;i<=n;i++) scanf("%I64d",&a[i]); 16 memset(dp,0,sizeof(dp)); 17 dp[0][0]=1; 18 for (i=1;i<=n;i++) 19 { 20 for (j=0;j<=2000004;j++) 21 dp[i%2][j]=dp[(i-1)%2][j]+dp[(i-1)%2][j^a[i]]; 22 } 23 ans=0; 24 n=n%2; 25 for (i=m;i<=1400000;i++) ans+=dp[n][i]; 26 printf("Case #%I64d: %I64d\n",k,ans); 27 k++; 28 } 29 }
