zoukankan      html  css  js  c++  java
  • 水题:HDU 5119 Happy Matt Friends

    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.
      水DP。
     1 #include <iostream>
     2 #include <cstring>
     3 #include <cstdio>
     4 using namespace std;
     5 const int Max=1050000;
     6 int dp[2][Max],n,M;
     7 int main(){
     8     int T,cas=0;
     9     long long ans,tot,x;
    10     scanf("%d",&T);
    11     while(T--){
    12         scanf("%d%d",&n,&M);
    13         memset(dp,0,sizeof(dp));
    14         dp[0][0]=1;int now,pre;
    15         for(int i=1;i<=n;i++){
    16             scanf("%lld",&x);now=i%2,pre=now^1;
    17             memset(dp[now],0,sizeof(dp[now]));
    18             for(int j=0;j<Max;j++){
    19                 if((j^x)<Max)dp[now][j]+=dp[pre][j^x];
    20                 dp[now][j]+=dp[pre][j];
    21             }
    22         }
    23         ans=0;
    24         for(int i=0;i<M;i++)ans+=dp[now][i];
    25         tot=1;x=2;
    26         while(n){
    27             if(n&1)tot=tot*x;
    28             n>>=1;x=x*x;
    29         }
    30         printf("Case #%d: %lld
    ",++cas,tot-ans);    
    31     }
    32     return 0;
    33 }
  • 相关阅读:
    Oracle Dataguard原理
    [转]TOKUDB® VS. INNODB FLASH MEMORY
    [转]什么是简约设计
    [转]DAS、NAS、SAN存储系统分析
    [转]ocp|ocm考证系列文章!
    [转]数据库范式的设计
    Block Media Recovery, BMR
    [转]开启闪回以及闪回的四种原理
    [转]Oracle DB 执行表空间时间点恢复
    Losing All Members of an Online Redo Log Group
  • 原文地址:https://www.cnblogs.com/TenderRun/p/5943501.html
Copyright © 2011-2022 走看看