Codeforces914G Sum the Fibonacci（FWT）

　　FWT大杂烩。跟着模拟做很多次FWT即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N (1<<17)
#define P 1000000007
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,ab[N],c[N],de[N],f[N];
void OR(int *a,int n,int op)
{
    for (int i=;i<=n;i<<=)
        for (int j=;j<n;j+=i)
            for (int k=j;k<j+(i>>);k++)
            {
                int x=a[k],y=a[k+(i>>)];
                a[k]=x;if (op==) a[k+(i>>)]=(y+x)%P;else a[k+(i>>)]=(y-x+P)%P;
            }
}
void AND(int *a,int n,int op)
{
    for (int i=;i<=n;i<<=)
        for (int j=;j<n;j+=i)
            for (int k=j;k<j+(i>>);k++)
            {
                int x=a[k],y=a[k+(i>>)];
                if (op==) a[k]=(x+y)%P;else a[k]=(x-y+P)%P;a[k+(i>>)]=y;
            }
}
void XOR(int *a,int n,int op)
{
    for (int i=;i<=n;i<<=)
        for (int j=;j<n;j+=i)
            for (int k=j;k<j+(i>>);k++)
            {
                int x=a[k],y=a[k+(i>>)];
                a[k]=(x+y)%P,a[k+(i>>)]=(x-y+P)%P;
                if (op) a[k]=1ll*a[k]*%P,a[k+(i>>)]=1ll*a[k+(i>>)]*%P;
            }
}
void FWT(int *a,int *b,int n,int op)
{
    if (op==) OR(a,n,),OR(b,n,);
    else if (op==) AND(a,n,),AND(b,n,);
    else XOR(a,n,),XOR(b,n,);
    for (int i=;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
    if (op==) OR(a,n,),OR(b,n,);
    else if (op==) AND(a,n,),AND(b,n,);
    else XOR(a,n,),XOR(b,n,);
}
int main()
{
    n=read();
    while (n--) c[read()]++;
    memcpy(de,c,sizeof(de));
    FWT(de,c,N,);
    for (int i=;i<N;i++)
    {
        for (int j=i;j;j=j-&i)
        ab[i]=(ab[i]+1ll*c[j]*c[i^j]%P)%P;
        ab[i]=(ab[i]+1ll*c[i]*c[])%P;
    }
    f[]=,f[]=;for (int i=;i<N;i++) f[i]=(f[i-]+f[i-])%P;
    for (int i=;i<N;i++) ab[i]=1ll*ab[i]*f[i]%P,c[i]=1ll*c[i]*f[i]%P,de[i]=1ll*de[i]*f[i]%P;
    FWT(ab,de,N,);FWT(ab,c,N,);
    int ans=;for (int i=;i<;i++) ans=(ans+ab[<<i])%P;
    cout<<ans;
    return ;
}