首页 > 代码库 > hdu-5119-Happy Matt Friends

hdu-5119-Happy Matt Friends

2024-08-05 13:05:32   217人阅读

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 ?rst 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
2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

 

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, 0, sizeof(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\n", Case++, ans);
40     }
41     return 0;
42 }

 

hdu-5119-Happy Matt Friends


联系
我们