题意：多项式除法,A(x)=C(x)*B(x)+D(x),给定A(x),B(x)，求C(x),D(x)
题解：A(x)的度是n，B(x)的度是m
定义(A'(x)=x^n*A(frac{1}{x})),可以发现(A'(x)=A(n-x))
(A(frac{1}{x})=C(frac{1}{x})*B(frac{1}{x})+D(frac{1}{x}))
(x^nA(frac{1}{x})=x^{n-m}*C(frac{1}{x})*x^{m}*B(frac{1}{x})+x^{n-m+1}*x^{m}D(frac{1}{x}))
(A'(x)=C'(x)*B'(x)+x^{n-m+1}*D'(x))
那么我们考虑在(mod(x^{n-m+1}))的等式，即可消去D'(x)
这样我们便可求出(C'(x)=frac{A'(x)}{B'(x)}(mod(x^{n-m+1})))，用多项式求逆，再回退C，带入可求出D
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=300000+10,maxn=400000+10,inf=0x3f3f3f3f;
ll a[N<<3],b[N<<3],c[N<<3],d[N<<3],e[N<<3],tmp[N<<3];
int rev[N<<3];
void getrev(int bit)
{
for(int i=0;i<(1<<bit);i++)
rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void ntt(ll *a,int n,int dft)
{
for(int i=0;i<n;i++)
if(i<rev[i])
swap(a[i],a[rev[i]]);
for(int step=1;step<n;step<<=1)
{
ll wn=qp(3,(mod-1)/(step*2));
if(dft==-1)wn=qp(wn,mod-2);
for(int j=0;j<n;j+=step<<1)
{
ll wnk=1;
for(int k=j;k<j+step;k++)
{
ll x=a[k];
ll y=wnk*a[k+step]%mod;
a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod;
wnk=wnk*wn%mod;
}
}
}
if(dft==-1)
{
ll inv=qp(n,mod-2);
for(int i=0;i<n;i++)a[i]=a[i]*inv%mod;
}
}
void pol_inv(int deg,ll *a,ll *b)
{
if(deg==1){b[0]=qp(a[0],mod-2);return ;}
pol_inv((deg+1)>>1,a,b);
int sz=0;while((1<<sz)<=deg)sz++;sz++;
getrev(sz);int len=1<<sz;
for(int i=0;i<deg;i++)tmp[i]=a[i];
for(int i=deg;i<len;i++)tmp[i]=0;
ntt(tmp,len,1),ntt(b,len,1);
for(int i=0;i<len;i++)
b[i]=(2ll-tmp[i]*b[i]%mod+mod)%mod*b[i]%mod;
ntt(b,len,-1);
for(int i=deg;i<len;i++)b[i]=0;
}
void pol_div(int n,int m,ll *a,ll *b,ll *c,ll *d)
{
for(int i=0;i<=n-m;i++)if(m>=i)c[i]=b[m-i];
int sz=0;while((1<<sz)<=MAX(n-m+1,m))sz++;sz++;
int len=1<<sz;
pol_inv(len,c,d);
for(int i=0;i<=n-m;i++)c[i]=a[n-i];
for(int i=n-m+1;i<len;i++)c[i]=d[i]=0;
getrev(sz);
ntt(c,len,1);ntt(d,len,1);
for(int i=0;i<len;i++)c[i]=c[i]*d[i]%mod;
ntt(c,len,-1);
reverse(c,c+n-m+1);
for(int i=n-m+1;i<len;i++)c[i]=0;
sz=0;while((1<<sz)<=n+1)sz++;sz++;
len=1<<sz;getrev(sz);
ntt(c,len,1);ntt(b,len,1);
for(int i=0;i<len;i++)d[i]=c[i]*b[i]%mod;
ntt(d,len,-1);ntt(c,len,-1);
for(int i=0;i<len;i++)d[i]=(a[i]-d[i]+mod)%mod;
for(int i=m;i<len;i++)d[i]=0;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++)scanf("%lld",&a[i]);
for(int i=0;i<=m;i++)scanf("%lld",&b[i]);
pol_div(n,m,a,b,c,d);
for(int i=0;i<=n-m;i++)printf("%lld ",c[i]);puts("");
for(int i=0;i<m;i++)printf("%lld ",d[i]);puts("");
return 0;
}
/********************
5 1
1 9 2 6 0 8
1 7
********************/