The child and binary tree- open root inversion of polynomials

We set up F(x) The total weight is x Number of tree existence schemes

Then we enumerate the root nodes of the tree , The size of the left and right subtrees , You can get F(x)

Into a formula  

namely   F(x)=G(x) F(x)^2

However, it should be noted that ,x=0, Constant term of time 1

F(x)= G(x)F(x)^2+1

G(x)F(x)^2-F(x)+1=0

/  2G(x)

If we recursively find G(x) Inverse element , We will find that when we recurse to g(0) When ,g(0)=0, Therefore, mathematical transformation is needed

g(x) It can be preprocessed , Then find out 1-4G(x), Re root , Then find the inverse element , Double it again , Lots of details , Code length

# include<iostream>
# define GG 3
# define MAXN 262145
# define mod 998244353
# define inv 499122177
using namespace std;


typedef long long int ll;

ll temp[MAXN],B[MAXN],rev[MAXN],tp[MAXN],tq[MAXN],ta[MAXN],tb[MAXN];
ll A[MAXN];
ll G[MAXN];

void init(int k)
{
    int len=(1<<k);

    for(int i=0;i<len;i++)
    {
        rev[i]=0;

    }
    for(int i=0;i<len;i++)
    {
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(k-1));

    }
    return ;

}
ll qp(ll base, ll pow)
{
    ll ans=1;

    while(pow)
    {

        if(pow&1)
        {
            ans=ans*base%mod;

        }

        pow>>=1;

        base=base*base%mod;

    }

    return ans;

}

void NTT(ll *a,int n,int flag)
{
    for(int i=0;i<n;i++)
    {
        if(i<rev[i])
        {
            swap(a[i],a[rev[i]]);

        }
    }

    for(int h=1;h<n;h<<=1)
    {
        ll gn=qp(GG,(mod-1)/(h<<1));

        if(flag==-1)
        {
            gn=qp(gn,mod-2);

        }

        for(int j=0;j<n;j+=2*h)
        {
            ll g=1;

            for(int k=j;k<j+h;k++)
            {
                ll x=a[k];

                ll y=a[k+h]*g%mod;

                a[k]=(x+y)%mod;

                a[k+h]=((x-y)%mod+mod)%mod;

                g=g*gn%mod;

            }
        }
    }

   if(flag==-1)
   {
       ll invn=qp(n,mod-2);

       for(int i=0;i<n;i++)
       {
           a[i]=a[i]*invn%mod;

       }
   }

   return ;
}

void mul(ll *a,ll*b,ll*c,int n,int m)
{
    int k=1,s=2;

    while((1<<k)<n+m-1)
    {
        k++;

        s<<=1;

    }

    init(k);

    for(int i=0;i<n;i++)
    {
        tp[i]=a[i];

    }
    for(int i=n;i<s;i++)
    {
        tp[i]=0;

    }
    for(int i=0;i<n;i++)
    {
        tq[i]=b[i];

    }

    for(int i=n;i<s;i++)
    {
        tq[i]=0;

    }

    NTT(tp,s,1);

    NTT(tq,s,1);

    for(int i=0;i<s;i++)
    {
        c[i]=tp[i]*tq[i]%mod;

    }

    NTT(c,s,-1);

}

void getinv(ll *a,ll*b,int n)
{
    if(n==1)
    {
        b[0]=qp(a[0],mod-2);

        return ;

    }

    getinv(a,b,(n+1)>>1);

    int k=1,s=2;

    while((1<<k)<n+n-1)
    {
        k++;

        s<<=1;

    }

    init(k);

    for(int i=0;i<n;i++)
    {
        ta[i]=a[i];
    }

    for(int i=n;i<s;i++)
    {
        ta[i]=0;

    }

    NTT(ta,s,1);

    NTT(b,s,1);

    for(int i=0;i<s;i++)
    {

        b[i]=((2ll-ta[i]*b[i])%mod+mod)%mod*b[i]%mod;

    }

    NTT(b,s,-1);

    for(int i=n;i<s;i++)
    {
        b[i]=0;

    }


    return ;

}

void sqrt(ll *a,ll*b,int n)
{
    if(n==1)
    {
        b[0]=1;

        return ;

    }

    sqrt(a,b,(n+1)>>1);

    int k=1,s=2;

    while((1<<k)<n+n-1)
    {
        k++;
        s<<=1;

    }

    init(k);

    for(int i=0;i<s;i++)
    {
        tb[i]=0;
    }

    getinv(b,tb,n);

    mul(a,tb,tb,n,n);

    for(int i=0;i<s;i++)
    {
        b[i]=(b[i]+tb[i])*inv%mod;

    }



}
int main ()
{

int n,m;

cin>>n>>m;

for(int i=0;i<n;i++)
{
    int x;

    cin>>x;

    G[x]=1;
}

temp[0]=1;

for(int i=1;i<=m;i++)
{
    temp[i]=mod-4*G[i]%mod;

}


sqrt(temp,B,m+1);


B[0]++;

getinv(B,A,m+1);


for(int i=1;i<=m;i++)
{
A[i]=A[i]*2%mod;

cout<< A[i]<<'\n';

}
    return 0;
}