多项式对数函数

已知$f(x)$,求$g(x)=\ln f(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$
保证常数项为$1$

$g(x{)}^{\prime }=\frac{f(x{)}^{\prime }}{f(x)}$
$g(x)=\int \frac{f(x{)}^{\prime }}{f(x)}$
结束了。

$\mathrm{A}\mathrm{C}\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}$

#include<bits/stdc++.h>
#define maxn 3000005
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define mod 998244353
using namespace std;

int Wl,W[maxn],lg[maxn],r[maxn],inv[maxn];
int Pow(int b,int k){ int r=1;for(;k;k>>=1,b=1ll*b*b%mod) if(k&1) r=1ll*r*b%mod;return r; }
void init(int n){
	for(W[0]=inv[0]=inv[1]=Wl=1;n>=2*Wl;Wl<<=1);int pw=Pow(3,(mod-1)/Wl/2);
	rep(i,1,Wl<<1) W[i]=1ll*W[i-1]*pw%mod,(i>1)&&(lg[i]=lg[i>>1]+1,inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod);
}
void FFT(int *A,int n,int tp){
	rep(i,1,n-1) (i<(r[i]=(r[i>>1]>>1)|((i&1)<<(lg[n]-1))))&&(swap(A[i],A[r[i]]),0);
	for(int L=1,B=Wl;L<n;L<<=1,B>>=1) for(int s=0;s<n;s+=L<<1) for(int k=s,x=0,t;k<s+L;k++,x+=B) 
		t=1ll*(tp==1?W[x]:W[(Wl<<1)-x])*A[k+L]%mod,A[k+L]=(A[k]-t)%mod,A[k]=(A[k]+t)%mod;
	if(tp^1) rep(i,0,n-1) A[i]=1ll*A[i]*inv[n]%mod;
}
void INV(int *A,int *B,int n){
	B[B[1]=0]=Pow(A[0],mod-2);static int t[maxn];
	for(int k=2,L=4;k<(n<<1);k<<=1,L<<=1){
		rep(i,0,L-1) t[i]=(i<k?A[i]:B[i]=0);FFT(t,L,1),FFT(B,L,1);
		rep(i,0,L-1) B[i]=B[i]*(2-1ll*t[i]*B[i]%mod)%mod;FFT(B,L,-1);
		rep(i,min(n,k),L-1) B[i]=0;
	}
}
void LN(int *A,int *B,int n){
	INV(A,B,n);static int t[maxn];int L=1<<lg[2*n-2]+1;
	rep(i,0,L-1) t[i]=(i<n-1?A[i+1]*(i+1ll)%mod:0),(i>=n)&&(B[i]=0);FFT(B,L,1),FFT(t,L,1);
	rep(i,0,L-1) t[i]=1ll*t[i]*B[i]%mod;FFT(t,L,-1);
	rep(i,0,L-1) B[i]=(i&&i<n?1ll*t[i-1]*inv[i]%mod:0);
}
int n,m;int A[maxn],B[maxn];
int main(){
	scanf("%d",&n);init(2*n);
	rep(i,0,n-1) scanf("%d",&A[i]);
	LN(A,B,n);
	rep(i,0,n-1) printf("%d%c",(B[i]+mod)%mod," \n"[i==n-1]);
}