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=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),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=2;i<=n;i<<=1)
        for (int j=0;j<n;j+=i)
            for (int k=j;k<j+(i>>1);k++)
            {
                int x=a[k],y=a[k+(i>>1)];
                a[k]=x;if (op==0) a[k+(i>>1)]=(y+x)%P;else a[k+(i>>1)]=(y-x+P)%P;
            }
}
void AND(int *a,int n,int op)
{
    for (int i=2;i<=n;i<<=1)
        for (int j=0;j<n;j+=i)
            for (int k=j;k<j+(i>>1);k++)
            {
                int x=a[k],y=a[k+(i>>1)];
                if (op==0) a[k]=(x+y)%P;else a[k]=(x-y+P)%P;a[k+(i>>1)]=y;
            }
}
void XOR(int *a,int n,int op)
{
    for (int i=2;i<=n;i<<=1)
        for (int j=0;j<n;j+=i)
            for (int k=j;k<j+(i>>1);k++)
            {
                int x=a[k],y=a[k+(i>>1)];
                a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P;
                if (op) a[k]=1ll*a[k]*500000004%P,a[k+(i>>1)]=1ll*a[k+(i>>1)]*500000004%P;
            }
}
void FWT(int *a,int *b,int n,int op)
{
    if (op==0) OR(a,n,0),OR(b,n,0);
    else if (op==1) AND(a,n,0),AND(b,n,0);
    else XOR(a,n,0),XOR(b,n,0);
    for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
    if (op==0) OR(a,n,1),OR(b,n,1);
    else if (op==1) AND(a,n,1),AND(b,n,1);
    else XOR(a,n,1),XOR(b,n,1);
}
int main()
{
    n=read();
    while (n--) c[read()]++;
    memcpy(de,c,sizeof(de));
    FWT(de,c,N,2);
    for (int i=0;i<N;i++)
    {
        for (int j=i;j;j=j-1&i)
        ab[i]=(ab[i]+1ll*c[j]*c[i^j]%P)%P;
        ab[i]=(ab[i]+1ll*c[i]*c[0])%P;
    }
    f[0]=0,f[1]=1;for (int i=2;i<N;i++) f[i]=(f[i-1]+f[i-2])%P;
    for (int i=0;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,1);FWT(ab,c,N,1);
    int ans=0;for (int i=0;i<17;i++) ans=(ans+ab[1<<i])%P;
    cout<<ans;
    return 0;
}

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转载自www.cnblogs.com/Gloid/p/10218828.html
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