洛谷P4239 【模板】多项式求逆（加强版）（多项式求逆）

传送门

咱用的是拆系数\(FFT\)因为咱真的不会三模数\(NTT\)……

简单来说就是把每一次多项式乘法都改成拆系数\(FFT\)就行了

如果您还不会多项式求逆的左转->这里

顺带一提，因为求逆的时候要乘两次，两次分开乘，否则会像咱一样炸精度

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int K=-1,Z=0;
inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}
void print(R int x){
    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++K]=z[Z],--Z);sr[++K]=' ';
}
const int N=5e5+5,P=1e9+7;const double Pi=acos(-1.0);
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
    R int res=1;
    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);
    return res;
}
struct cp{
    double x,y;
    cp(double xx=0,double yy=0){x=xx,y=yy;}
    inline cp operator +(const cp &b)const{return cp(x+b.x,y+b.y);}
    inline cp operator -(const cp &b)const{return cp(x-b.x,y-b.y);}
    inline cp operator *(const cp &b)const{return cp(x*b.x-y*b.y,x*b.y+y*b.x);}
    inline cp operator *(const double &b)const{return cp(x*b,y*b);}
}A[N],B[N],C[N],D[N],E[N],G[N],F[N],H[N],w[N],a[N],b[N],c[N],d[N];
int r[N];
int n,len;
void FFT(cp *A,int ty,int lim){
    fp(i,0,lim-1)if(i<r[i])swap(A[i],A[r[i]]);
    for(R int mid=1;mid<lim;mid<<=1)
        for(R int j=0;j<lim;j+=(mid<<1))
            for(R int k=0;k<mid;++k){
                cp x=A[j+k],y=w[mid+k]*A[j+k+mid];
                A[j+k]=x+y,A[j+k+mid]=x-y;
            }
    if(ty==-1){
        reverse(A+1,A+lim);
        double k=1.0/lim;fp(i,0,lim-1)A[i]=A[i]*k;
    }
}
void Mul(cp *a,cp *b,int len,cp *d){
    int lim=1,l=0;while(lim<(len<<1))lim<<=1,++l;
    fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    for(R int i=1;i<lim;i<<=1)fp(k,0,i-1)w[i+k]=cp(cos(Pi*k/i),sin(Pi*k/i));
    fp(i,0,len-1){
        A[i].x=(ll)(a[i].x+0.5)>>15,B[i].x=(ll)(a[i].x+0.5)&32767;
        C[i].x=(ll)(b[i].x+0.5)>>15,D[i].x=(ll)(b[i].x+0.5)&32767;
        A[i].y=B[i].y=C[i].y=D[i].y=0;
    }fp(i,len,lim-1)A[i]=B[i]=C[i]=D[i]=0;
    FFT(A,1,lim),FFT(B,1,lim),FFT(C,1,lim),FFT(D,1,lim);
    fp(i,0,lim-1){
        F[i]=A[i]*C[i],G[i]=A[i]*D[i]+C[i]*B[i],H[i]=B[i]*D[i];
    }
    FFT(F,-1,lim),FFT(G,-1,lim),FFT(H,-1,lim);
    fp(i,0,lim-1){
    	d[i].x=(((ll)(F[i].x+0.5)%P<<30)+((ll)(G[i].x+0.5)<<15)+((ll)(H[i].x+0.5)))%P;
    	d[i].y=0;
	}
}
void Inv(cp *a,cp *b,int len){
    if(len==1)return b[0].x=ksm(a[0].x,P-2),void();
    Inv(a,b,len>>1);
    Mul(a,b,len,c);
    Mul(c,b,len,d);
    fp(i,0,len-1)b[i].x=((ll)(b[i].x+b[i].x-d[i].x)%P+P)%P;
}
int main(){
//	freopen("testdata.in","r",stdin);
    n=read();fp(i,0,n-1)a[i].x=read();
    int len=1;while(len<n)len<<=1;
    Inv(a,b,len);
    fp(i,0,n-1)print((ll)(b[i].x+0.5)%P);
    return Ot(),0;
}