answer
Immortal title.
Finally, consideration must be put a \ (1 \) , and then the other positions are deleted.
Taking into account the sequence for each position corresponds to a single operation.
We can put for each \ (1 \) operation, it deleted the corresponding operating position as its son nodes.
This is a tree structure, it should be the last remaining only a \ (1 \) , so this is a rooted tree.
We then transformed to the problem rooted tree count problems.
We first set up a \ (f [i] \) represents (I \) \ Number Scheme rooted tree of nodes, \ (G [I] \) represents \ (I \) Number Scheme forests nodes ( It can be only one tree).
\ (G \) transfer is relatively simple:
\ [G_i = F_i + \ sum_ {J =. 1} ^ {I-2} \ Binom {I-. 1} {J-. 1} F_j * G_ {ij of} \]
Transfer \ (F \) , then need to consider multiple trees spliced together.
\ [F_i = \ sum_ {j
\ subset A} \ binom {i-1} {j} G_ {ij-1} + [i-1 \ subset B] \] boundary: \ (F [. 1] = G [ 1] = 0 \) .
This thing can divide and conquer \ (FFT \) demand, pay attention to discuss the partition of the interval contains the left point.
Code
#include<bits/stdc++.h>
#define N 270009
using namespace std;
typedef long long ll;
const int mod=998244353;
const int _G=3;
const int Gi=332748118;
int rev[N],n,A,B;
ll f[N],g[N],F[N],G[N],a[N],b[N],ni[N],jie[N],fn[N];
inline void MOD(ll &x){x=x>=mod?x-mod:x;}
inline ll rd(){
ll x=0;char c=getchar();bool f=0;
while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
return f?-x:x;
}
inline ll power(ll x,ll y){
ll ans=1;
while(y){
if(y&1)ans=ans*x%mod;
x=x*x%mod;
y>>=1;
}
return ans;
}
inline void NTT(ll *a,int l,int tag){
for(int i=1;i<l;++i)if(i>rev[i])swap(a[i],a[rev[i]]);
for(int i=1;i<l;i<<=1){
ll wn=power(tag?_G:Gi,(mod-1)/(i<<1));
for(int j=0;j<l;j+=(i<<1)){
ll w=1;
for(int k=0;k<i;++k,w=w*wn%mod){
ll x=a[j+k],y=a[i+j+k]*w%mod;
MOD(a[j+k]=x+y);MOD(a[i+j+k]=x-y+mod);
}
}
}
if(!tag){
ll ny=power(l,mod-2);
for(int i=0;i<l;++i)a[i]=a[i]*ny%mod;
}
}
void CDQ(int l,int r){
if(l==r){
if(l==1){
f[l]=g[l]=0;
return;
}
g[l]=g[l]*jie[l-1]%mod;
f[l]=f[l]*jie[l-1]%mod;
if(b[l-1])MOD(f[l]+=1);
MOD(g[l]+=f[l]);
g[l]=g[l]*ni[l]%mod;
fn[l]=f[l]*ni[l-1]%mod;
return;
}
int mid=(l+r)>>1;
CDQ(l,mid);
int len=1,L=0;
while(len<=(r-l+1+mid-l+1))len<<=1,L++;
for(int i=1;i<len;++i)rev[i]=rev[i>>1]>>1|((i&1)<<(L-1));
for(int i=0;i<len;++i)F[i]=a[i];
for(int i=l;i<=mid;++i)G[i-l+1]=g[i];
NTT(F,len,1);NTT(G,len,1);
for(int i=0;i<len;++i)F[i]=F[i]*G[i]%mod;
NTT(F,len,0);
for(int i=mid+1;i<=r;++i)MOD(f[i]+=F[i-l]);
for(int i=0;i<len;++i)F[i]=G[i]=0;
if(l==1){
len=1;L=0;int x=(mid-l+1)*2;
while(len<=x)len<<=1,L++;
for(int i=1;i<len;++i)rev[i]=rev[i>>1]>>1|((i&1)<<(L-1));
for(int i=l;i<=mid;++i)G[i-l]=g[i];
for(int i=0;i<r-l&&i<=mid;++i)F[i]=fn[i];
NTT(F,len,1);NTT(G,len,1);
for(int i=0;i<len;++i)F[i]=F[i]*G[i]%mod;
NTT(F,len,0);
for(int i=mid+1;i<=r;++i)MOD(g[i]+=F[i-l]);
for(int i=0;i<len;++i)F[i]=G[i]=0;
}
else{
len=1;L=0;int x=mid-l+1+r-l+1;
while(len<=x)len<<=1,L++;
for(int i=1;i<len;++i)rev[i]=rev[i>>1]>>1|((i&1)<<(L-1));
for(int i=l;i<=mid;++i)F[i-l]=fn[i];
for(int i=0;i<=r-l&&i<=mid;++i)G[i]=g[i];
NTT(F,len,1);NTT(G,len,1);
for(int i=0;i<len;++i)F[i]=F[i]*G[i]%mod;
NTT(F,len,0);
for(int i=mid+1;i<=r;++i)MOD(g[i]+=F[i-l]);
for(int i=0;i<len;++i)F[i]=G[i]=0;
for(int i=l;i<=mid;++i)G[i-l]=g[i];
for(int i=0;i<=r-l&&i<=mid;++i)F[i]=fn[i];
NTT(F,len,1);NTT(G,len,1);
for(int i=0;i<len;++i)F[i]=F[i]*G[i]%mod;
NTT(F,len,0);
for(int i=mid+1;i<=r;++i)MOD(g[i]+=F[i-l]);
for(int i=0;i<len;++i)F[i]=G[i]=0;
}
CDQ(mid+1,r);
}
int main(){
n=rd();A=rd();B=rd();
jie[0]=1;
for(int i=1;i<=n;++i)jie[i]=jie[i-1]*i%mod;
ni[n]=power(jie[n],mod-2);
for(int i=n-1;i>=0;--i)ni[i]=ni[i+1]*(i+1)%mod;
for(int i=1;i<=A;++i){int x=rd();a[x]=1;}
for(int i=0;i<=n;++i)a[i]=a[i]*ni[i];
for(int i=1;i<=B;++i){int x=rd();b[x]=1;}
if(n==1){
puts("1");
return 0;
}
CDQ(1,n);
cout<<f[n];
return 0;
}