洛谷P5245 【模板】多项式快速幂

悠悠 2021-12-23 05:41 316阅读 0赞

## 题面 ##

[传送门][Link 1]

## 题解 ##

话说现在还用数组写多项式的似乎没几个了……

\\\[B(x)=A^k(x)\\\]

\\\[\\ln B(x)=k\\ln A(x)\\\]

求个\\(\\ln\\)，乘个\\(k\\)，\\(\\exp\\)回去就行了

//minamoto #include<bits/stdc++.h> #define R register #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 C=-1,Z=0; inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;} void print(R int x){ if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x; while(z[++Z]=x%10+48,x/=10); while(sr[++C]=z[Z],--Z);sr[++C]=' '; } const int N=(1<<18)+5,P=998244353; 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))(y&1)?res=mul(res,x):0; return res; } void read(R int &x){ R char ch; while((ch=getc())>'9'||ch<'0'); for(x=ch-'0';(ch=getc())>='0'&&ch<='9';x=(x*10ll+ch-'0')%P); } int inv[N],r[21][N],rt[2][N<<1],lg[N],lim,d; void Pre(){ fp(d,1,18){ fp(i,1,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1)); lg[1<<d]=d; } inv[0]=inv[1]=1; fp(i,2,262144)inv[i]=mul(P-P/i,inv[P%i]); for(R int t=(P-1)>>1,i=1,x,y;i<=262144;i<<=1,t>>=1){ x=ksm(3,t),y=ksm(332748118,t),rt[0][i]=rt[1][i]=1; fp(k,1,i-1) rt[1][i+k]=mul(rt[1][i+k-1],x), rt[0][i+k]=mul(rt[0][i+k-1],y); } } void NTT(int *A,int ty){ fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]); for(R int mid=1;mid<lim;mid<<=1) for(R int j=0,t;j<lim;j+=(mid<<1)) fp(k,0,mid-1) A[j+k+mid]=dec(A[j+k],t=mul(rt[ty][mid+k],A[j+k+mid])), A[j+k]=add(A[j+k],t); if(!ty)fp(i,0,lim-1)A[i]=mul(A[i],inv[lim]); } void Inv(int *a,int *b,int len){ if(len==1)return b[0]=ksm(a[0],P-2),void(); Inv(a,b,len>>1); static int A[N],B[N];lim=(len<<1),d=lg[lim]; fp(i,0,len-1)A[i]=a[i],B[i]=b[i]; fp(i,len,lim-1)A[i]=B[i]=0; NTT(A,1),NTT(B,1); fp(i,0,lim-1)A[i]=mul(A[i],mul(B[i],B[i])); NTT(A,0); fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),A[i]); fp(i,len,lim-1)b[i]=0; } void Ln(int *a,int *b,int len){ static int A[N],B[N]; fp(i,1,len-1)A[i-1]=mul(a[i],i);A[len-1]=0; Inv(a,B,len);lim=(len<<1),d=lg[lim]; fp(i,len,lim-1)A[i]=B[i]=0; NTT(A,1),NTT(B,1); fp(i,0,lim-1)A[i]=mul(A[i],B[i]); NTT(A,0); fp(i,1,len-1)b[i]=mul(A[i-1],inv[i]);b[0]=0; fp(i,len,lim-1)b[i]=0; } void Exp(int *a,int *b,int len){ if(len==1)return b[0]=1,void(); Exp(a,b,len>>1); static int A[N];Ln(b,A,len); lim=(len<<1),d=lg[lim]; A[0]=dec(a[0]+1,A[0]); fp(i,1,len-1)A[i]=dec(a[i],A[i]); fp(i,len,lim-1)A[i]=b[i]=0; NTT(A,1),NTT(b,1); fp(i,0,lim-1)b[i]=mul(A[i],b[i]); NTT(b,0); fp(i,len,lim-1)b[i]=0; } void ksm(int *a,int *b,int len,int k){ static int A[N]; Ln(a,A,len); fp(i,0,len-1)A[i]=mul(A[i],k); Exp(A,b,len); } int A[N],B[N],n,k; int main(){ // freopen("testdata.in","r",stdin); n=read(),read(k),Pre(); int len=1;while(len<=n)len<<=1; fp(i,0,n-1)A[i]=read(); ksm(A,B,len,k); fp(i,0,n-1)print(B[i]); return Ot(),0; }

转载于:https://www.cnblogs.com/bztMinamoto/p/10578760.html

[Link 1]: https://www.luogu.org/problemnew/show/P5245