HDU-5119-Happy Matt Friends

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HDU-5119-Happy Matt Friends

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

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

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.

题目链接：HDU 5119

题目大意：有N个人，每个人有一个权值，你可以挑任意多的人并将他们的权值异或(也可以不选)，求最后得到的值大于M的取法有多少种。

题目思路：dp[i][j]为从1到i的长度中，答案为j的有多少种情况。有两种可能性，第i个位置取或者不取。取的情况就是dp[i][j ^ a[i]] = dp[i - 1][j],不取则是dp[i][j] = dp[i - 1][j]

以下是代码：

#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;int a[45];long long dp[45][1000010];int main(){    int t;    scanf("%d",&t);    int zl = 1;    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        for (int i = 1; i <= n; i++) scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        dp[0][0] = 1;        for (int i = 1; i <= n; i++)        {            for (int j = 0; j < 1000000; j++)            {                dp[i][j] += dp[i - 1][j];                dp[i][j ^ a[i]] += dp[i - 1][j];            }        }        long long ans = 0;        for (int i = m; i < 1000000; i++)        {            ans += dp[n][i];        }        printf("Case #%d: %lld\n",zl++,ans);    }    return 0;}