hdu5119（dp）

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 4177    Accepted Submission(s): 1587

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 ≤ k_i ≤ 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

23 21 2 33 31 2 3

 

Sample Output

Case #1: 4Case #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.
 

题意：在n个数中随意找几个数进行异或运算，求结果大于等于m的情况数。

思路：dp，本以为需要三重循环，其实不需要异或两次就变回了原来的数。但是不知道为什么判断一下j^num[i]是不是非负数就答案错误了。。

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;#define ll long longll dp[45][1148576];int num[45];int n,m;void solve(){memset(dp,0,sizeof(dp));dp[0][0]=1;for(int i=1;i<=n;i++){for(int j=0;j<1148576;j++){//if(j^num[i]>=0)dp[i][j]=dp[i-1][j]+dp[i-1][j^num[i]];//else//dp[i][j]=dp[i-1][j];}}}int main(){int t;scanf("%d",&t);for(int k=1;k<=t;k++){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&num[i]);solve();ll all=pow(2,n);ll ans=0;for(int i=0;i<m;i++)ans+=dp[n][i];printf("Case #%d: %I64d\n",k,all-ans);}}