版权声明：转载请声明出处，谢谢配合。 https://blog.csdn.net/zxyoi_dreamer/article/details/88647635
传送门
解析：
直接做的话还是可以 后做快速幂，那样的话就有两个
实际上由于 ，可以考虑两边取对数：
取对数，一乘，再exp回去就做完了。
代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
#define gc get_char
#define cs const
namespace IO{
inline char get_char(){
static cs int Rlen=1<<20|1;
static char buf[Rlen],*p1,*p2;
return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,Rlen,stdin),p1==p2)?EOF:*p1++;
}
inline int getint(){
re char c;
while(!isdigit(c=gc()));re int num=c^48;
while(isdigit(c=gc()))num=(num+(num<<2)<<1)+(c^48);
return num;
}
}
using namespace IO;
cs int mod=998244353;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(int a,int b){return a<b?a-b+mod:a-b;}
inline int mul(int a,int b){return (ll)a*b%mod;}
inline int quickpow(int a,int b,int res=1){
while(b){
if(b&1)res=mul(res,a);
a=mul(a,a);
b>>=1;
}
return res;
}
cs int N=410000;
int inv[N],r[N];
inline void NTT(int *A,int n,int typ){
for(int re i=0;i<n;++i)if(i<r[i])swap(A[i],A[r[i]]);
for(int re i=1;i<n;i<<=1){
int wn=quickpow(typ==-1?(mod+1)/3:3,(mod-1)/i/2);
for(int re j=0;j<n;j+=i<<1)
for(int re k=0,x,y,w=1;k<i;++k,w=mul(w,wn)){
x=A[j+k],y=mul(w,A[j+k+i]);
A[j+k]=add(x,y);
A[j+k+i]=dec(x,y);
}
}
if(typ==-1)for(int re i=0;i<n;++i)A[i]=mul(A[i],inv[n]);
}
inline void deriv(int *A,int *B,int len){
for(int re i=1;i<len;++i)B[i-1]=mul(i,A[i]);B[len-1]=0;
}
inline void integ(int *A,int *B,int len){
for(int re i=1;i<len;++i)B[i]=mul(inv[i],A[i-1]);B[0]=0;
}
inline void Inv(int deg ,int *a,int *b){
static int c[N];
if(deg==1)return (void)(b[0]=quickpow(a[0],mod-2));
Inv((deg+1)>>1,a,b);
memset(c,0,sizeof c);
int re len=1;
while(len<(deg<<1))len<<=1;
for(int re i=0;i<len;++i)r[i]=(r[i>>1]>>1)|((i&1)*(len>>1));
memcpy(c,a,sizeof(int)*deg);
memset(c+deg,0,sizeof(int)*(len-deg));
NTT(c,len,1),NTT(b,len,1);
for(int re i=0;i<=len;++i)b[i]=mul(b[i],dec(2,mul(b[i],c[i])));
NTT(b,len,-1);
memset(b+deg,0,sizeof(int)*(len-deg));
}
inline void Ln(int deg,int *f,int *g){
static int a[N],b[N];
memset(a,0,sizeof(a));memset(b,0,sizeof(b));
deriv(f,a,deg);
Inv(deg,f,b);
int re len=1;
while(len<(deg<<1))len<<=1;
for(int re i=0;i<len;++i)r[i]=(r[i>>1]>>1)|((i&1)*(len>>1));
NTT(a,len,1),NTT(b,len,1);
for(int re i=0;i<len;++i)a[i]=mul(a[i],b[i]);
NTT(a,len,-1);
integ(a,g,len);
memset(g+deg,0,sizeof(int)*(len-deg));
}
inline void Exp(int deg,int *a,int *b){
if(deg==1){b[0]=1;return ;}
static int F[N];
Exp((deg+1)>>1,a,b);
memset(F,0,sizeof F);
Ln(deg,b,F);
F[0]=dec(a[0]+1,F[0]);
for(int re i=1;i<deg;++i)F[i]=dec(a[i],F[i]);
int re len=1;
while(len<=deg)len<<=1;
for(int re i=0;i<len;++i)r[i]=(r[i>>1]>>1)|((i&1)*(len>>1));
NTT(F,len,1),NTT(b,len,1);
for(int re i=0;i<len;++i)b[i]=mul(b[i],F[i]);
NTT(b,len,-1);
memset(b+deg,0,sizeof(int)*(len-deg));
}
inline void init_inv(){
inv[0]=inv[1]=1;
for(int re i=2;i<N;++i)inv[i]=mul(inv[mod%i],mod-mod/i);
}
int n,k;
inline int getmod(){
re char c;
while(!isdigit(c=gc()));re int num=c^48;
while(isdigit(c=gc()))num=add(mul(num,10),c^48);
return num;
}
int a[N],b[N],c[N];
signed main(){
init_inv();
n=getint();
k=getmod();
for(int re i=0;i<n;++i)a[i]=getint();
int re len=1;
while(len<=n)len<<=1;
Ln(n,a,c);
for(int re i=0;i<n;++i)a[i]=mul(c[i],k);
memset(a+n,0,sizeof(int)*(len-n));
Exp(len,a,b);
for(int re i=0;i<n;++i)cout<<b[i]<<" ";
return 0;
}