题意：

在这里插入图片描述

思路：

写一下式子，发现每个 $f_i$ 只有左边的 $f$ 对他有影响，所以考虑分治 $F F T$ 来解决这个问题。
先递归左边，让后计算对右边贡献，再递归右边。
式子化简一下就是 $fi=∑j=lmidfj∗gi−jf_i=\sum_{j=l}^{mid}f_j*g_{i-j}$ ，直接写就好，注意卡常。。如果偷懒直接长度为 $n$ 的乘起来会 $T$ 很多点，所以需要偏移一下。

// Problem: P3803 【模板】多项式乘法（FFT）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3803
// Memory Limit: 500 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=4000010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
LL g[N],f[N];
LL a[N], b[N];
struct NTT {int P = 998244353, G = 3, Gi = 332748118;int n, m, limit = 1, L, r[N];LL qmi[2][N];inline LL fastpow(LL a, LL k) {LL base = 1;while(k) {if(k & 1) base = (base * a ) % P;a = (a * a) % P;k >>= 1;}return base % P;}inline void ntt(LL *A, int type) {for(int i = 0; i < limit; i++) if(i < r[i]) swap(A[i], A[r[i]]);for(int mid = 1; mid < limit; mid <<= 1) {	LL Wn=qmi[type==1? 1:0][mid];// LL t = fastpow( type == 1 ? G : Gi , (P - 1) / (mid << 1));for(int j = 0; j < limit; j += (mid << 1)) {LL w = 1;for(int k = 0; k < mid; k++, w = (w * Wn) % P) {int x = A[j + k], y = w * A[j + k + mid] % P;A[j + k] = (x + y) % P,A[j + k + mid] = (x - y + P) % P;}}}}void init_qmi() {for(int mid=1;mid<=4e5;mid<<=1) {qmi[1][mid]=fastpow(G,(P - 1) / (mid << 1));qmi[0][mid]=fastpow(Gi,(P - 1) / (mid << 1));}}void init(int n,int m) {limit=1; L=0;while(limit <= m + n) limit <<= 1, L++;for(int i = 0; i < limit; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));}int mul(LL *ans,LL *x,int ln,int rn,LL *y,int rm) {for(int i=ln;i<=rn;i++) a[i-ln]=x[i]%P;for(int i=1;i<=rm;i++) b[i]=y[i]%P; int n=rn-ln+1,m=rm;init(rn,rm);ntt(a, 1); ntt(b, 1);	ntt(a, -1);	LL inv = fastpow(limit, P - 2);for(int i = 0; i <= ln + rm; i++) ans[i]=a[i]*inv%P;for(int i=0;i<limit;i++) a[i]=0,b[i]=0;return rn+rm;}}NT;void solve(int l,int r) {if(l==r) {if(l==0) f[l]=1;return;} int mid=(l+r)>>1;solve(l,mid);int len=r-l+1;for(int i=l;i<=mid;i++) a[i-l]=f[i];for(int i=0;i<=len;i++) b[i]=g[i];NT.init(len+1,len); NT.ntt(a, 1); NT.ntt(b, 1);	for(int i = 0; i < NT.limit; i++) a[i] = (a[i] * b[i]) % NT.P;NT.ntt(a, -1);	LL inv = NT.fastpow(NT.limit, NT.P - 2);for(int i = mid+1; i <= r; i++) f[i]+=a[i-l]*inv%NT.P,f[i]%=mod;for(int i=0;i<NT.limit;i++) a[i]=0,b[i]=0;solve(mid+1,r);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d",&n);for(int i=1;i<=n-1;i++) scanf("%lld",&g[i]);NT.init_qmi();solve(0,n-1);for(int i=0;i<n;i++) printf("%lld ",f[i]);return 0;
}
/**/