hdu5119——Happy Matt Friends

標籤：dp

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 150    Accepted Submission(s): 56


Problem DescriptionMatt 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. 
InputThe 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. 
OutputFor 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

23 21 2 33 31 2 3

 
Sample Output

Case #1: 4Case #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. 

 
Source2014ACM/ICPC亞洲區北京站-重現賽（感謝北師和上交） 
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題解是高斯消元（真心不太會）
用了背包過的，再加個滾動數組，一開始糾結在前i個數異或最大，然後怎麼都A不了，後來乾脆全部把上限改成1 << 20，然後過了

#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;__int64 dp[2][1 << 22];int num[44];int main(){int t, n, m, icase = 1;__int64 ans;scanf("%d", &t);while (t--){memset (dp, 0, sizeof(dp));scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i){scanf("%d", &num[i]);}dp[0][0] = 1;for (int i = 1; i <= n; ++i){for (int j = 0; j <= (1 << 20); ++j){dp[i % 2][j] = dp[1 - (i % 2)][j] + dp[1 - (i % 2)][j ^ num[i]];}}ans = 0;for (int i = m; i <= (1 << 20); ++i){ans += dp[n % 2][i];}printf("Case #%d: %I64d\n", icase++, ans);}return 0;}

hdu5119——Happy Matt Friends