[Codeforces Round #250]小朋友和二叉树

题目描述：

bzoj

luogu

题解：

生成函数ntt。

显然这种二叉树应该暴力薅掉树根然后分裂成两棵子树。

所以$f(x)= \sum_{i \in c} \sum _{j=0}^{x-c} f(i)*f(x-i-j)$

这是个不好看的卷积。

所以我们再引入$c$的生成函数$G$，那么有$$F(x)=1+G(x)*F(x)*F(x)$$

注意常数项，因为正常$f(0)=1$但是$G(0)=0$，所以要人为配常数。

然后$F= \frac{1 \pm \sqrt{1-4G}} {2G} = \frac{2}{1 \pm \sqrt{1-4G}}$

还是因为系数，正负号只能取正。

bz严重卡常：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = ;
const int MOD = ;
const int inv_2 = ((MOD+)>>);
template<typename T>
inline void read(T&x)
{
    T f = ,c = ;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){c=c*+ch-'';ch=getchar();}
    x = f*c;
}
inline void Mod(unsigned int&x){if(x>=MOD)x-=MOD;}
inline unsigned int fastpow(unsigned int x,int y)
{
    int ret = ;
    while(y)
    {
        if(y&)ret=1ll*ret*x%MOD;
        x=1ll*x*x%MOD;y>>=;
    }
    return ret;
}
int inv(const unsigned int x){return fastpow(x,MOD-);}
int to[N],lim,L,LL[N],n,m;
inline void init(const int len)
{
    lim=LL[]=;
    while(lim<=len)lim<<=,LL[lim<<]=LL[lim]+;
}
unsigned int W[N],INV;
inline void get_lim(const int len)
{
    lim = len,L = LL[len];
    for(register int i=;i<lim;++i)to[i]=((to[i>>]>>)|((i&)<<(L-)));
    INV = inv(len);
    for(register int i=;i<lim;i<<=)W[i]=fastpow(,(MOD-)/(i<<));
}
inline void ntt(unsigned int*a,const int len,const int k)
{
    for(register int i=;i<len;++i)
        if(i<to[i])swap(a[i],a[to[i]]);
    for(register int i=;i<len;i<<=)
    {
        unsigned int w0 = W[i];
        for(register int j=;j<len;j+=(i<<))
        {
            unsigned int w = ;
            for(register int o=;o<i;++o,w=1ll*w*w0%MOD)
            {
                unsigned int w1 = a[j+o],w2 = 1ll*a[j+o+i]*w%MOD;
                Mod(a[j+o]=w1+w2);
                Mod(a[j+o+i]=w1+MOD-w2);
            }
        }
    }
    if(k==-)
    {
        for(register int i=;i<len>>;++i)swap(a[i],a[len-i]);
        for(register int i=;i<len;++i)a[i]=1ll*a[i]*INV%MOD;
    }
}
unsigned int a[N],b[N],c[N];
void mul(unsigned int*F,unsigned int*G,unsigned int*H,const int len)
{
    get_lim(len<<);
    for(register int i=;i<lim;++i)a[i]=b[i]=;
    for(register int i=;i<len;++i)a[i]=F[i],b[i]=G[i];
    ntt(a,lim,),ntt(b,lim,);
    for(register int i=;i<lim;++i)c[i]=1ll*a[i]*b[i]%MOD;
    ntt(c,lim,-);
    for(register int i=;i<lim;++i)H[i]=c[i];
}
unsigned int T[N];
void get_inv(unsigned int*F,unsigned int*G,const int len)
{
    if(len==){G[]=inv(F[]),G[]=;return ;}
    get_inv(F,G,len>>);
    get_lim(len<<);for(register int i=;i<lim;++i)a[i]=b[i]=;
    for(register int i=;i<len;++i)a[i]=F[i];for(register int i=;i<(len>>);++i)b[i]=G[i];
    ntt(a,lim,),ntt(b,lim,);for(register int i=;i<lim;++i)c[i]=1ll*a[i]*b[i]%MOD*b[i]%MOD;ntt(c,lim,-);
    for(register int i=;i<len;++i)G[i]=(2ll*G[i]+MOD-c[i])%MOD,G[i+len]=;
}
unsigned int H[N];
void get_sqrt(unsigned int*F,unsigned int*G,const int len)
{
    if(len==){G[]=;return ;}
    get_sqrt(F,G,len>>);get_inv(G,H,len);
    mul(G,G,T,len);for(register int i=;i<len;++i)T[i]=1ll*(T[i]+F[i])*inv_2%MOD;
    mul(H,T,G,len);for(register int i=len;i<(len<<);++i)G[i]=;
}
unsigned int F[N],G[N],S[N];
int main()
{
//    freopen("tt.in","r",stdin);
    read(n),read(m);
    init(m);int c,mx = lim;
    for(register int i=;i<=n;++i)
        read(c),G[c]=MOD-;
    G[]++;
    get_sqrt(G,S,mx);
    S[]++;
    for(register int i=;i<mx;++i)S[i]=1ll*S[i]*inv_2%MOD;
    get_inv(S,F,mx);
    for(register int i=;i<=m;++i)
        printf("%u\n",F[i]);
    return ;
}