hdu-5119-Happy Matt Friends
浏览数：45 /
时间：2015年06月11日
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.
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.
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: 2HintIn 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
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 }
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 }
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