傳送門

咱用的是拆系數(shù)\(FFT\)因為咱真的不會三模數(shù)\(NTT\)……

簡單來說就是把每一次多項式乘法都改成拆系數(shù)\(FFT\)就行了

如果您還不會多項式求逆的左轉(zhuǎn)->這里

順帶一提，因為求逆的時候要乘兩次，兩次分開乘，否則會像咱一樣炸精度

//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; }

轉(zhuǎn)載于:https://www.cnblogs.com/bztMinamoto/p/10284693.html

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