Description
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
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.
Output
Sample Input
Sample Output
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.
分析:首先看题发现并没有什么明显的规律。然后考虑dp
dp[i][j]表示 前i个数里面异或值为j的方法数
找到递推关系就可以;
#include <iostream> #include <stdio.h> #include <string> #include <string.h> #include <algorithm> using namespace std; typedef long long ll; const int maxn=1e6+100; int n,m,test; int a[maxn]; ll dp[44][maxn]; ll solve() { memset(dp,0,sizeof(dp)); dp[1][0]=dp[1][a[1]]=1; for(int i=2;i<=n;i++) { for(int j=0;j<maxn;j++)dp[i][j]+=dp[i-1][j]; for(int j=0;j<maxn;j++)dp[i][a[i]^j]+=dp[i-1][j]; } ll ans=0; for(int i=m;i<maxn;i++) ans+=dp[n][i]; return ans; } int main() { int T,test=1; scanf("%d",&T); while( T-- ) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&a[i]); ll ans=solve(); printf("Case #%d: %lld\n",test++,ans); } return 0; }