zoukankan      html  css  js  c++  java
  • hdoj 5119 Happy Matt Friends 背包DP

    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
    ",ti,ans);
    
    	}
    }
  • 相关阅读:
    ubuntu 9.04更新源
    想学一下asp.net,跟着书本做了个bbs
    [转]ubuntu系统中遇到的一些问题及解决
    第一篇,打个招呼
    人际交往的书籍推荐
    程序员的五层境界,你在哪一层?
    HTTP报文之"请求报文"和"响应报文"详解
    如何提高你的工作效率?
    面对焦虑我们怎么办 ?
    CEO要看的书籍推荐
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4325829.html
Copyright © 2011-2022 走看看