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.
 

 

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.
 

 

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 }

 

posted on 2015-09-22 15:15  pb2016  阅读(187)  评论(0编辑  收藏  举报