Happy Matt Friends
Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 3215 Accepted Submission(s): 1261
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 2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）
解析：动态规划。对每个状态，可以扩展出2种状态：取a[i]和不取a[i]。用dp[i][j]表示前i个数里面异或值为j的方法数，则 dp[i][j] += dp[i-1][j]; dp[i][j^a[i]] += dp[i-1][j];
``` #include
const int MAXN = 1e6+5;
int dp[45][2*MAXN]; //dp[i][j]表示前i个数里面异或值为j的方法数
int a[45], n, m;
void solve()
{
memset(dp, 0, sizeof dp);
dp[0][0] = 1;
for(int i = 1; i <= n; ++i){
for(int j = 0; j <= 1e6; ++j){
dp[i][j] += dp[i-1][j];
dp[i][j^a[i]] += dp[i-1][j];
}
}
long long res = 0;
for(int i = m; i <= 1e6; ++i)
res += dp[n][i];
printf("%I64d
", res);
}
int main()
{
int t, cn = 0;
scanf("%d", &t);
while(t--){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
printf("Case #%d: ", ++cn);
solve();
}
return 0;
}