【模板】普通多項式轉下降冪多項式

題目
下降冪多項式$\times e^x$就是其點值多項式的$EGF$。
所以多點求值後乘$e^{-x}$即可。
$\mathrm{AC \ Code}$

#include<bits/stdc++.h>
#define maxn 300005
#define mod 998244353
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
using namespace std;

int Wl,Wl2,W[maxn],lg[maxn],r[maxn],inv[maxn],fac[maxn],invf[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(Wl=W[0]=inv[0]=inv[1]=fac[0]=fac[1]=invf[0]=invf[1]=1;n>=Wl<<1;Wl<<=1);int pw=Pow(3,(mod-1)/(Wl2=Wl<<1));
	rep(i,1,Wl2) W[i]=1ll*pw*W[i-1]%mod,i>1&&(lg[i]=lg[i>>1]+1,inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod,
		fac[i]=1ll*fac[i-1]*i%mod,invf[i]=1ll*invf[i-1]*inv[i]%mod);
}
int& upd(int &u){ return u+=u>>31&mod; }
void NTT(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=W[tp^1?Wl2-x:x]*1ll*A[k+L]%mod,upd(A[k+L]=A[k]-t),upd(A[k]=A[k]+t-mod); 
	if(tp^1) rep(i,0,n-1) A[i]=1ll*A[i]*inv[n]%mod; 
}
void MUL(int *A,int *B,int *C,int n,int m){
	static int t[2][maxn],L;
	rep(i,0,(L=1<<lg[n-1+m-1]+1)-1) t[0][i]=i<n?A[i]:0,t[1][i]=i<m?B[i]:0;
	NTT(t[0],L,1),NTT(t[1],L,1);rep(i,0,L-1) C[i]=1ll*t[0][i]*t[1][i]%mod;
	NTT(C,L,0);
}
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;NTT(t,L,1),NTT(B,L,1);
		rep(i,0,L-1) upd(B[i]=B[i]*(2-1ll*t[i]*B[i]%mod)%mod);
		NTT(B,L,0);rep(i,min(n,k),L-1) B[i]=0;
	} 
}
void DIV(int *A,int *B,int *C,int *R,int n,int m){
	#define REV reverse
	REV(A,A+n),REV(B,B+m),INV(B,C,n-m+1),MUL(A,C,C,n-m+1,n-m+1);
	REV(A,A+n),REV(B,B+m),REV(C,C+n-m+1),MUL(B,C,R,m-1,min(m-1,n-m+1));
	rep(i,0,m-1<<1) upd(R[i]=A[i]-R[i]);
	#undef REV
}
#define lc u<<1
#define rc lc|1
#define NAF(a) new int[1<<lg[a]+2]
int *F[maxn],*M[maxn],SZ[maxn];
void BDM(int u,int l,int r,int *X){
	M[u]=NAF(SZ[u]=r-l+1);
	if(l==r) return (void)(upd(M[u][0]=-X[l]),M[u][1]=1);
	int m=l+r>>1;BDM(lc,l,m,X),BDM(rc,m+1,r,X);
	MUL(M[lc],M[rc],M[u],SZ[lc]+1,SZ[rc]+1);
}
void EVL(int u,int l,int r,int *C){
	if(u>1) DIV(F[u>>1],M[u],F[0],F[u]=NAF(SZ[u]),SZ[u>>1],SZ[u]+1);
	if(l==r) return (void)(C[l]=F[u][0]);
	int m=l+r>>1;EVL(lc,l,m,C),EVL(rc,m+1,r,C);
}
void MVL(int *A,int *X,int *C,int n,int m){ n=max(n,m+1),BDM(1,0,m-1,X),DIV(A,M[1],F[0]=NAF(n),F[1]=NAF(m),n,m+1),EVL(1,0,m-1,C); }

int n,A[maxn],X[maxn],B[maxn],C[maxn];

int main(){
	scanf("%d",&n);init(n<<1);
	rep(i,0,n-1) scanf("%d",&A[i]),X[i]=i,B[i]=i&1?mod-invf[i]:invf[i];
	MVL(A,X,C,n,n);
	rep(i,0,n-1) C[i]=1ll*C[i]*invf[i]%mod;
	MUL(C,B,A,n,n);
	rep(i,0,n-1) printf("%d%c",A[i]," \n"[i==n-1]);
}