zoukankan      html  css  js  c++  java
  • hdu-5119-Happy Matt Friends

    http://acm.hdu.edu.cn/showproblem.php?pid=5119

    Happy Matt Friends

    Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
    Total Submission(s): 341    Accepted Submission(s): 130


    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

    解题思路:DP

      1 #include <stdio.h>

     2 #include <stdlib.h>
     3 #include <string.h>
     4 
     5 
     6 const int MAX = (1<<20);
     7 
     8 long long dp[41][MAX];
     9 int n, w, a[41];
    10 
    11 int main(){
    12     int T, i, j;
    13     int Case = 1;
    14     scanf("%d", &T);
    15     while(T--){
    16         scanf("%d %d", &n, &w);
    17         for(i = 0; i < n; i++){
    18             scanf("%d", &a[i]);
    19         }
    20         memset(dp, 0sizeof(dp));
    21         dp[0][0] = 1;
    22         long long ans = 0;
    23         if(w == 0){
    24             ans++;
    25         }
    26         for(i = 0; i < n; i++){
    27             for(j = 0; j < MAX; j++){
    28                 if(dp[i][j] == 0){
    29                     continue;
    30                 }
    31                 dp[i + 1][j] += dp[i][j];
    32                 int nj = j^a[i];
    33                 dp[i + 1][nj] += dp[i][j];
    34                 if(nj >= w){
    35                     ans += dp[i][j];
    36                 }
    37             }
    38         }
    39         printf("Case #%d: %I64d ", Case++, ans);
    40     }
    41     return 0;
    42 }
  • 相关阅读:
    PHP 把字符转换为 HTML 实体
    CSS 不换行 white-space 属性详解
    JQuery 事件器的介绍
    maven 添加自己的包
    MYSQL 安装
    优秀系统
    JSP中的相对路径和绝对路径(转)
    Eclipse RCP扩展
    JSTL与EL(转)
    el表达式跟ognl表达式的区别(转)
  • 原文地址:https://www.cnblogs.com/angle-qqs/p/4135898.html
Copyright © 2011-2022 走看看