Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 5248 Accepted Submission(s): 2014
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.
For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10
6).
In the second line, there are N integers ki (0 ≤ k
i ≤ 10
6), indicating the i-th friend’s magic number.
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.
#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 45;const int maxm = (1<<20);int a[maxn];LL dp[2][maxm];int main(){ int t; scanf("%d", &t); for(int k = 1; k <= t; k++) { int n, m; scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } memset(dp, 0, sizeof(dp)); dp[0][0] = 1; int pre = 0, now = 1; for(int i = 1; i <= n; i++) { for(int j = 0; j < maxm; j++) { int x = a[i] ^ j; dp[now][x] = dp[pre][x] + dp[pre][j]; } swap(now, pre); } LL ans = 0; for(int i = m; i < maxm; i++) { ans += dp[pre][i]; } printf("Case #%d: %lld\n", k, ans); } return 0;}
