hdu5119（dp）

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Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)

Total Submission(s): 667 Accepted Submission(s): 253

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: 2Hint
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：给你n个数，让你求这n个数中任意k个数抑或和大于等于m的方案数。
思路：dp[i][j]表示i个数中任意k个数抑或和大于等于m的方案数。注意方案数可能会爆LL。转移方程：dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ a[i]]
代码：
#include<bits/stdc++.h>using namespace std;const int maxn=1e6+5;typedef long long LL;LL dp[50][maxn];int a[50];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%d",&n,&m);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        dp[0][0]=1;        for(int i=1;i<=n;i++)        {            for(int j=0;j<maxn;j++)            {                dp[i][j^a[i]]+=dp[i-1][j];                dp[i][j]+=dp[i-1][j];            }        }        LL ans=0;        for(int i=m;i<maxn;i++)        {            ans+=dp[n][i];        }        static int t=1;        printf("Case #%d: %lld\n",t++,ans);    }    return 0;}