BJ模拟 and【容斥】

题目描述：

解题思路：

#include<bits/stdc++.h>
using namespace std;
int getint()
{
    int i=0,f=1;char c;
    for(c=getchar();c!='-'&&(c<'0'||c>'9');c=getchar());
    if(c=='-')f=-1,c=getchar();
    for(;c>='0'&&c<='9';c=getchar())i=(i<<3)+(i<<1)+c-'0';
    return i*f;
}
const int N=51,mx=(1<<13)-1,mod=1e9+7;
int n,a[N],f[2][1<<13][1<<13];
inline void add(int &x,int y){x=x+y>=mod?x+y-mod:x+y;}
inline void dec(int &x,int y){x=x-y>=0?x-y:x-y+mod;}
int main()
{
    //freopen("lx.in","r",stdin);
    //freopen("lx.out","w",stdout);
    n=getint();
    for(int i=1;i<=n;i++)a[i]=getint()^mx;
    int now=0;f[now][0][0]=1;
    for(int i=0;i<n;i++,now^=1)
        for(int s1=0;s1<=mx;s1++)
            for(int s2=s1;;s2=(s2-1)&s1)
            {
                if(!f[now][s1][s2])
                {
                    if(!s2)break;
                    continue;
                }
                add(f[now^1][s1][s2],f[now][s1][s2]);
                add(f[now^1][s1|a[i+1]][s2^(s2&a[i+1])],f[now][s1][s2]);
                dec(f[now^1][s1|a[i+1]][(s2^(s2&a[i+1]))|(a[i+1]^(a[i+1]&s1))],f[now][s1][s2]);
                f[now][s1][s2]=0;
                if(!s2)break;
            }

    cout<<f[now][mx][0]<<'\n';
    return 0;
}