洛谷P5245 【模板】多项式快速幂

题面

传送门

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)=kln 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;
}