题面

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
    
     #define R register#define fp(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<
      
       >1]>>1)|((i&1)<<(d-1));        lg[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
        
         >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;}