思路：用字典树来统计异或和大于M的种数就行了
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1100010;
const int INF=0x3f3f3f3f;
const int maxc=2;
const int BIT=21;
int cas=1,T;
int n,m,k[50],kk[50],half;
int ch[N*20][maxc],si[N*20];
void init()
{
memset(ch[0],0,sizeof(ch[0]));
si[0]=1;
}
void insert(int s)
{
int u=0;
for(int i=BIT-1;i>=0;i--)
{
int c=(s&(1<<i)?1:0);
if(!ch[u][c])
{
memset(ch[si[0]],0,sizeof(ch[0]));
si[si[0]]=0;
ch[u][c]=si[0]++;
}
u=ch[u][c];
si[u]++;
}
}
int find(int s)
{
int u=0,ans=0;
for(int i=BIT-1;i>=0;i--)
{
int c=(s&(1<<i)?1:0);
if(m&(1<<i))
{
if(!ch[u][!c]) return ans;
u=ch[u][!c];
}
else
{
if(ch[u][!c]) ans+=si[ch[u][!c]];
if(!ch[u][c]) return ans;
u=ch[u][c];
}
}
return ans+si[u];
}
int main()
{
//freopen("1.in","w",stdout);
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
scanf("%d",&T);
while(T--)
{
init();
scanf("%d%d",&n,&m);
half=n/2;
int r=n-half;
//printf("%d %d\n",half,r);
for(int i=0;i<half;i++) scanf("%d",&k[i]);
for(int i=0;i<r;i++) scanf("%d",&kk[i]);
LL ans=0;
for(int i=0;i< 1<<half;i++)
{
int sum=0;
for(int j=0;j<half;j++) if(i&(1<<j)) sum^=k[j];
//printf("s:%d\n",sum);
insert(sum);
}
for(int i=0;i< 1<<r;i++)
{
int sum=0;
for(int j=0;j<r;j++) if(i&(1<<j)) sum^=kk[j];
//printf("sum:%d\n",sum);
ans+=find(sum);
}
printf("Case #%d: %lld\n",cas++,ans);
}
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
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 ≤ 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
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.
