HDU 5119 Happy Matt Friends 递推_a - happy matt friends hdu 5119 高斯消元-优快云博客

本文链接：https://blog.youkuaiyun.com/canghai_wangzi/article/details/52972663

本文介绍了一种算法问题，即Matt和他的朋友们通过选择特定的玩家来赢得游戏的方法数量。问题中，每个朋友都有一个魔法数字，Matt的目标是选出一些朋友，使他们魔法数字的异或和大于等于指定值M以获胜。文章提供了完整的代码实现，并解释了一个示例案例。

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Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 3294 Accepted Submission(s): 1285

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）

Recommend

liuyiding

这个，先找到最大可能的数，然后的就因为年代久远不记得了...不过代码很短，应该很简单，看一看应该就差不多了

#include <bits/stdc++.h>

using namespace std;

int m,n,T,temp;
long long dp[1050000],dp1[1050000],ans;

int main()
{
    cin>>T;
    for(int k=1;k<=T;k++)
    {
        cin>>m>>n;
        memset(dp,0,sizeof(dp));
        memset(dp1,0,sizeof(dp1));
        dp[0]=dp1[0]=1;
        ans=0;
        for(int i=1;i<=m;i++)
        {
            cin>>temp;
            for(int j=0;j<=1048577;j++)
                if(dp[j])
                    dp1[j^temp]+=dp[j];
            for(int j=0;j<=1048577;j++)
                dp[j]=dp1[j];
        }
        for(int i=n;i<=1048577;i++)
            ans+=dp[i];
        cout<<"Case #"<<k<<": "<<ans<<endl;
    }
    return 0;
}