hdoj 5119 Happy Matt Friends 背包DP-优快云博客

本文介绍了一个关于选择玩家以达到特定异或值的游戏问题。通过动态规划方法，详细解析了如何计算赢得游戏的方法数，并提供了具体的实现代码。

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Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 700 Accepted Submission(s): 270

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.

题意

给你N个人，然后让你选一些人，然后问你，选的这些人，异或值大于m的方法数有多少个

题解

大概就是类似背包的思想，每个人有选择和不选择两种选择，然后我们就可以根据这个写出转移方程，dp[i][j]表示选择前i个人中，得到答案为j的方法数有多少，由于j^a[i]^a[i]=j，所以
dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]]

代码

#define RD(n) scanf("%d",&n)    
#define REP(i, n) for (int i=0;i<n;++i)
#define REP_1(i, n) for (int i=1;i<=n;++i)
int dp[50][maxn+1];
int a[50];
int main()
{
    int t;
    RD(t);
    REP_1(ti,t)
    {
        int n,m;
        RD(n),RD(m);
        REP_1(i,n)
        {
            RD(a[i]);
        }
        dp[0][0]=1;
        REP_1(i,n)
        {
            REP(j,maxn)
            {
                dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]];
            }
        }
        LL ans=0;
        FOR_1(j,m,maxn)
            ans+=dp[n][j];
        printf("Case #%d: %lld\n",ti,ans);

    }
}

转载于:https://www.cnblogs.com/qscqesze/p/4325829.html