HOJ 5199 Happy Matt Friends （暴力dp）

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本文介绍了一种基于异或运算的游戏算法，Matt和他的朋友们通过选择特定的魔法数值组合来赢得游戏。文章详细解释了如何计算可能获胜的方式数量，并提供了一个示例案例及代码实现。

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Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 4863 Accepted Submission(s): 1841

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

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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#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 3000010;
using ll = long long;
ll dp[2][maxn];
int n, m;
int a[50];
int main() {
    //freopen("in.txt", "r", stdin);
    int t, cur;
    cin >> t;
    for (int ca = 1; ca <= t; ++ca) {
        memset(dp, 0, sizeof(dp));
        cin >> n >> m;
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        dp[1][0] = 1;
        cur = 1;
        for (int i = 0; i < n; ++i) {
            cur = 1 - cur;
            for (int j = 0; j <= 300000; ++j) {
                dp[cur][j] = dp[1 - cur][j] + dp[1 - cur][j ^ a[i]];
            }
        }
        ll res = 0;
        for (int i = m; i < maxn; ++i) {
            res += dp[cur][i];
        }
        printf("Case #%d: %lld\n", ca, res);
    }
}