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  • hdu 5119 Happy Matt Friends dp

    Happy Matt Friends

    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.
     
    Source
    题意:N个数,选异或和大于等于M的方案数;
    思路:类似求背包方案总数;
    #include<bits/stdc++.h>
    using namespace std;
    #define ll __int64
    #define esp 1e-13
    const int N=2e3+10,M=1e6+50000,inf=1e9+10,mod=1000000007;
    ll dp[42][M];
    int a[N];
    void init()
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
    }
    int main()
    {
        int x,y,i,z,t;
        int T,cas=1;
        scanf("%d",&T);
        while(T--)
        {
            init();
            scanf("%d%d",&x,&y);
            for(i=1;i<=x;i++)
            scanf("%d",&a[i]);
            for(i=1;i<=x;i++)
            {
                for(t=0;t<M;t++)
                dp[i][t^a[i]]+=dp[i-1][t],dp[i][t]+=dp[i-1][t];
            }
            ll ans=0;
            for(i=y;i<M;i++)
            ans+=dp[x][i];
            printf("Case #%d: %I64d
    ",cas++,ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5760064.html
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