HDU 5119 Happy Matt Friends

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 5248    Accepted Submission(s): 2014

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 ≤ 10 ⁶).

In the second line, there are N integers ki (0 ≤ k _i ≤ 10 ⁶), 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.

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

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题意：有N个人，每个人有一个权值，挑选一些人并将他们的权值异或，求最后得到的值大于M的取法有多少种；

题解：对于每个值都有取和不取两种状态，用dp[i]表示亦或结果为i的方案数；int x = a[i] ^ j; dp[now][x] = dp[pre][x] + dp[pre][j];

参考代码为：

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int maxn = 45;
const int maxm = (1<<20);
int a[maxn];
LL dp[2][maxm];

int main()
{
    int t;
    scanf("%d", &t);
    for(int k = 1; k <= t; k++)
    {
        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;
        int pre = 0, now = 1;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 0; j < maxm; j++)
            {
                int x = a[i] ^ j;
                dp[now][x] = dp[pre][x] + dp[pre][j];
            }
            swap(now, pre);
        }
        LL ans = 0;
        for(int i = m; i < maxm; i++)
        {
            ans += dp[pre][i];
        }
        printf("Case #%d: %lld\n", k, ans);
    }
    return 0;
}