Luogu5245 【模板】多项式快速幂（多项式exp）

　　A(x)^k=e^klnA(x)。泰勒展开之后容易发现k并非在指数上，所以对p取模。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 600010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(10ll*x+c-48)%P,c=getchar();
	return x*f;
}
int n,m,a[N],r[N],b[N],c[N],d[N],A[N],B[N],t;
int ksm(int a,int k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
int inv(int a){return ksm(a,P-2);}
void DFT(int *a,int n,int g)
{
	for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1);
	for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
	for (int i=2;i<=n;i<<=1)
	{
		int wn=ksm(g,(P-1)/i);
		for (int j=0;j<n;j+=i)
		{
			int w=1;
			for (int k=j;k<j+(i>>1);k++,w=1ll*w*wn%P)
			{
				int x=a[k],y=1ll*w*a[k+(i>>1)]%P;
				a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P;
			}
		}
	}
}
void IDFT(int *a,int n)
{
	DFT(a,n,inv(3));
	int u=inv(n);
	for (int i=0;i<n;i++) a[i]=1ll*a[i]*u%P;
}
void mul(int *a,int *b,int n)
{
	DFT(a,n,3),DFT(b,n,3);
	for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
	IDFT(a,n);
}
void Inv(int *a,int *b,int n)
{
	if (n==1) {for (int i=0;i<t;i++) b[i]=0;b[0]=inv(a[0]);return;}
	Inv(a,b,n>>1);
	for (int i=0;i<n;i++) A[i]=a[i];
	for (int i=n;i<(n<<1);i++) A[i]=0;
	n<<=1;
	DFT(A,n,3),DFT(b,n,3);
	for (int i=0;i<n;i++) b[i]=1ll*b[i]*(P+2-1ll*A[i]*b[i]%P)%P;
	IDFT(b,n);
	n>>=1;
	for (int i=n;i<(n<<1);i++) b[i]=0;
}
//G(B(x))=A(x)B(x)-1
//B(x)=B0(x)-G(B0(x))/G'(B0(x))
//B(x)=B0(x)-(A(x)B0(x)-1)/A(x)
//B(x)=B0(x)(2-A(x)B0(x))
void trans(int *a,int *b,int n){for (int i=n-1;i>=0;i--) b[i]=1ll*a[i+1]*(i+1)%P;}
void dx(int *a,int *b,int n){b[0]=0;for (int i=1;i<n;i++) b[i]=1ll*a[i-1]*inv(i)%P;}
void Ln(int *a,int n)
{
	for (int i=0;i<n;i++) b[i]=c[i]=0;
	trans(a,b,n>>1);
	Inv(a,c,n>>1);
	mul(b,c,n);
	dx(b,a,n);
}
//ln(F(x))=G(x)
//F'(x)/F(x)=G'(x)
//dx F'(x)/F(x)=G(x)
void Exp(int *a,int *b,int n)
{
	if (n==1){b[0]=1;return;}
	Exp(a,b,n>>1);
	for (int i=0;i<(n>>1);i++) B[i]=b[i];
	for (int i=(n>>1);i<n;i++) B[i]=0;
	Ln(B,n);
	for (int i=0;i<n;i++) B[i]=(a[i]-B[i]+P)%P;
	B[0]=(B[0]+1)%P;
	n<<=1;
	for (int i=(n>>1);i<n;i++) B[i]=0;
	mul(b,B,n);
	n>>=1;
	for (int i=n;i<(n<<1);i++) b[i]=0;
}
//exp(A(x))=B(x)
//A(x)=ln(B(x))
//G(B(x))=ln(B(x))-A(x)
//B(x)=B0(x)(A(x)+1-ln(B0(x)))
int main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d
";
#else
	const char LL[]="%lld
";
#endif
	n=read(),m=read();
	for (int i=0;i<n;i++) a[i]=read();
	t=1;while (t<=(n<<1)) t<<=1;
	Ln(a,t);
	for (int i=0;i<t;i++) a[i]=1ll*a[i]*m%P;
	Exp(a,d,t);
	for (int i=0;i<n;i++) printf("%d ",d[i]);
	return 0;
}