[jzoj3303]【集训队互测2013】城市规划 分治FFT

版权声明：本文为博主原创文章，转载时请标明出处。 https://blog.csdn.net/FarmerJohnOfZS/article/details/80915747

分治比想象当中容易=w=
这题也可以多项式求逆 推式子和逆多项式做法都在我的这篇博客

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#define oo 2139062143
#define fo(i,x,y) for (int i=(x);i<=(y);++i)
#define fd(i,x,y) for (int i=(x);i>=(y);--i)
using namespace std;
const int N=262144+100,mo=1004535809;
int n;
int jc[N],ny[N],w[N],wz[N];
int qsm(int x,int y)
{
    int rt=1;
    fo(i,0,30)
    {
        if(y&(1<<i)) rt=1ll*rt*x%mo;
        x=1ll*x*x%mo;   
    }
    return rt;
}
int c[N];
struct ffst{
    int a[N],cs;
    void clear()
    {
        fo(i,0,cs-1) a[i]=0;
        cs=0;
    }
    void dft(int flag)
    {
        wz[0]=0;fo(i,1,cs-1)wz[i]=(wz[i>>1]>>1)+((cs>>1)*(i&1)); 
        w[0]=1,w[1]=qsm(3,(mo-1)/cs);
        fo(i,2,cs) w[i]=1ll*w[i-1]*w[1]%mo;int tp;
        if(flag==-1) fo(i,1,cs>>1) swap(w[i],w[cs-i]); 
        fo(i,0,cs-1) c[wz[i]]=a[i];
        for (int wv=2,hf=1;wv<=cs;hf=wv,wv<<=1)
        fo(i,0,hf-1) for(int j=i;j<cs;j+=wv)
        tp=1ll*w[cs/wv*i]*c[j+hf]%mo,c[j+hf]=(c[j]-tp+mo)%mo,c[j]=(c[j]+tp)%mo;
        int nv=qsm(cs,mo-2);
        fo(i,0,cs-1) 
        {
            a[i]=c[i];
            if(flag==-1)a[i]=1ll*a[i]*nv%mo;
        }
    }
}a,b;
int f[N],g[N];
void fft(ffst &a,ffst &b)//a*=b
{
    int ct=max(a.cs,b.cs)*2;
    a.cs=b.cs=ct;
    a.dft(1),b.dft(1);
    fo(i,0,ct-1) a.a[i]=1ll*a.a[i]*b.a[i]%mo; 
    a.dft(-1);
}
void solve(int l,int r)
{
    if(l==r)
    {
        f[l]=(0ll+g[l]+mo-(1ll*jc[l-1]*f[l]%mo))%mo;
        return;
    }
    int mid=(l+r)>>1;
    solve(l,mid);
    a.clear(),b.clear();
    a.cs=(mid-l+1)*2;
    a.a[0]=0;fo(i,l,mid) a.a[i-l+1]=1ll*f[i]*ny[i-1]%mo;
    b.cs=r-l+1;
    b.a[0]=0;fo(i,1,r-l) b.a[i]=1ll*g[i]*ny[i]%mo;
    fft(a,b);
    fo(i,mid+1,r) f[i]=(f[i]+a.a[i-l+1])%mo;
    solve(mid+1,r);
}
int main()
{
    scanf("%d",&n);
    g[0]=1;fo(i,1,N-100) g[i]=qsm(2,(1ll*(i-1)*i/2)%(mo-1));
    jc[0]=1;fo(i,1,N-100) jc[i]=1ll*jc[i-1]*i%mo;
    ny[N-100]=qsm(jc[N-100],mo-2);fd(i,N-100-1,0) ny[i]=1ll*ny[i+1]*(i+1)%mo;
    int css=1;while(css<n)css<<=1;
    solve(1,css);
    printf("%d\n",f[n]);
    return 0;
}