CF438E - The Child and Binary Tree

给某些权值的结点，可以无限使用。$f(x)$为总权值为$x$的二叉树的种类数，询问$f(1)…f(m)$

只有权值$1$的时候，即卡特兰数，考虑生成函数推导。

$g(i)$表示是否有$i$这个权值。

$$f(x)=\sum_{i=1}^x g(i)\sum _{t+j=x-i}f(t)f(j)$$

观察为三个卷积形式。注意边界$g(0)=0,f(0)=1$

$$f=g\times f \times f+1\\ f=\frac{1\pm \sqrt{1-4g}}{2g}$$

带入$0$值可知取负号=$\frac{0}{0}$

$$f=\frac{2}{1+\sqrt{1+4g}}\rightarrow f(0)=1$$

带入板子即可。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;

const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}
void NTTX(int *a, int n, int *b, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = 1ll * a[i] * b[i] % P;
    NTT(a, ML, -1);
}
int Inv2;
int C[N], D[N];

void Finv(int *a, int *b, int n)
{
    b[0] = qpow(a[0], P - 2, P);
    int len, ML;
    int bit = 0;

    for (len = 1; len < (n << 1); len <<= 1)
    {
        ML = len << 1;
        bit++;
        for (int i = 0; i < len; i++)
            C[i] = a[i], D[i] = b[i];
        for (int i = 0; i < ML; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        NTT(C, ML, 1), NTT(D, ML, 1);
        for (int i = 0; i < ML; i++)
            b[i] = ((2ll - 1ll * C[i] * D[i] % P) * D[i] % P + P) % P;
        NTT(b, ML, -1);
        for (int i = len; i < ML; i++)
            b[i] = 0;
    }
    for (int i = 0; i < len; i++)
        C[i] = D[i] = 0;
    for (int i = n; i < len; i++)
        b[i] = 0;
}
int E[N], F[N];
void Sqrt(int *a, int *b, int n)
{
    Inv2 = qpow(2, P - 2, P);
    b[0] = 1;
    int bit = 0;
    int len;

    for (len = 1; len < (n << 1); len <<= 1)
    {
        int ML = len << 1;
        bit++;
        for (int i = 0; i < len; i++)
            E[i] = a[i];
        Finv(b, F, len);
        for (int i = 0; i < ML; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        NTT(E, ML, 1), NTT(F, ML, 1);
        for (int i = 0; i < ML; i++)
            E[i] = 1ll * E[i] * F[i] % P;
        NTT(E, ML, -1);
        for (int i = 0; i < len; i++)
            b[i] = 1LL * (b[i] + E[i]) % P * Inv2 % P;
        for (int i = len; i < ML; i++)
            b[i] = 0;
    }
    for (int i = 0; i < len; i++)
        E[i] = F[i] = 0;
    for (int i = n; i < len; i++)
        b[i] = 0;
}

// int aa[N], ia[N]; // aa 表示a的导数，ia表示a的逆
// void Ln(int *a, int *res, int n)
// {
//     for (int i = 0; i < n << 1; i++)
//         aa[i] = ia[i] = 0;
//     for (int i = 1; i < n; i++)
//         aa[i - 1] = 1ll * i * a[i] % P;
//     Finv(a, ia, n);
//     NTTX(aa, n, ia, n);
//     res[1] = 0;
//     for (int i = 1; i < n; i++)
//         res[i] = 1ll * aa[i - 1] * qpow(i, P - 2, P) % P;
// }
// int G[N], H[N], M[N];
// void EXP(int *a, int *b, int n)
// {
//     Inv2 = qpow(2, P - 2, P);
//     b[0] = 1;
//     int bit = 0;
//     int len;

//     for (len = 1; len < (n << 1); len <<= 1)
//     {
//         int ML = len << 1;
//         bit++;
//         for (int i = 0; i < len << 1; i++)
//             H[i] = G[i] = M[i] = 0;
//         for (int i = 0; i < len; i++)
//             H[i] = a[i];
//         for (int i = 0; i < len; i++)
//             G[i] = b[i];
//         Ln(G, M, len); // M(x)=lin(G)
//         for (int i = 0; i < ML; i++)
//             rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
//         NTT(G, ML, 1), NTT(H, ML, 1), NTT(M, ML, 1);
//         for (int i = 0; i < ML; i++)
//             b[i] = 1LL * G[i] * (1ll - M[i] + H[i] + P) % P;
//         NTT(b, ML, -1);
//         for (int i = len; i < ML; i++)
//             b[i] = 0;
//     }
//     for (int i = 0; i < len; i++)
//         G[i] = H[i] = M[i] = 0;
//     for (int i = n; i < len; i++)
//         b[i] = 0;
// }
int f[N], g[N], sg[N];
int main()
{
    int n = read(), m = read();
    for (int i = 1; i <= n; i++)
    {
        int x = read();
        g[x] = 1;
    }
    for (int i = 1; i <= m; i++)
        g[i] = (-4ll * g[i] + P) % P;
    g[0] = 1;
    Sqrt(g, sg, m + 1);
    sg[0] = inc(sg[0], 1, P);
    Finv(sg, f, m + 1);
    for (int i = 1; i <= m; i++)
        f[i] = 2ll * f[i] % P;
    for (int i = 1; i <= m; i++)
        printf("%d\n", f[i]);
}