数学这一块还是要补补的,做一做生成函数题。

设 $f_i$ 表示权值为 $i$ 的点权二叉树的数量,且 $f_0 = 1$;设 $g_i$ 表示权值 $i$ 是否在集合 $\{c_1, c_2, \ldots, c_n\}$ 中,显然有 $g_0 = 0$。则 $f_k$ 可以建立以下关系

设 $F(x)$ 和 $G(x)$ 分别是 $f_i$ 和 $g_i$ 的生成函数,则上式可以转化为

所以可以解得

若符号为正,则有

取极限得式子发散,故舍去。若符号为负,则有

因此取负号,再由于 $g_0 = 0$,故 $G(x)$ 无法直接求逆,将分子有理化后得到

只需要多项式求逆和开根就能在 $O(n + k \log k)$ 的时间下求出答案。进一步的,开根的多项式的零次项为 $1$,因此可以使用简单版的开根即可。


完整代码如下:

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#include <bits/stdc++.h>
using namespace std;
constexpr int MAX = 1e5 + 5;
constexpr int MOD = 998244353;
constexpr int INF = 0x3f3f3f3f;
using i64 = long long;
using u64 = unsigned long long;
using pii = pair <int, int>;
inline int read()
{
int s = 0; int f = 1;
char ch = getchar();
while ((ch < '0' || ch > '9') && ch != EOF)
{
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
s = s * 10 + ch - '0';
ch = getchar();
}
return s * f;
}
template <int MOD = 998244353>
class NTT
{
vector <int> rev; int n = 1;
const int P = MOD;//998244353 469762049 50000000001507329LL 4179340454199820289
const int G = 3;
//1945555039024054273 for G = 5
int qpow(int x, int y)
{
int res = 1;
while (y)
{
if (y & 1) res = 1ll * res * x % P;
x = 1ll * x * x % P;
y >>= 1;
}
return res;
}
void init_rev(int tot)
{
int L = 0; n = 1;
while (n < tot) n <<= 1, ++L;
rev.resize(n);
for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (L - 1));
}
void ntt(vector <int> &a, int op)
{
for (int i = 0; i < n; ++i)
if (i < rev[i]) swap (a[i] ,a[rev[i]]);
for (int len = 1; len < n; len <<= 1)
{
int wn = qpow(G, (P - 1) / (len << 1));
if (op == -1) wn = qpow (wn, P - 2);
for (int i = 0; i < n; i += len << 1)
{
int w = 1;
for(int j = 0; j < len; ++j)
{
int u = a[i + j] % P, v = 1ll * a[i + j + len] * w % P;
a[i + j] = (0ll + u + v) % P;
a[i + j + len] = (0ll + u - v + P) % P;
w = 1ll * w * wn % P;
}
}
}
if (op == -1)
{
int inv_n = qpow(n, P - 2);
for (auto &x : a) x = 1ll * x * inv_n % P;
}
}
public:
vector <int> Conv(vector <int> a, vector <int> b)
{
int tot = (int)a.size() + (int)b.size() - 1;
init_rev(tot);
a.resize(n); b.resize(n);
ntt(a, 1); ntt(b, 1);
for (int i = 0; i < n; ++i) a[i] = 1ll * a[i] * b[i] % P;
ntt(a, -1); a.resize(tot);
return a;
}
vector <int> Inv(vector <int> a, int m)
{
int inv = qpow(a[0], P - 2);
vector <int> b {inv};
for (int len = 2; (len >> 1) < m; len <<= 1)
{
int sz = min((int)a.size(), len);
vector <int> f(a.begin(), a.begin() + sz);
init_rev(len << 1);
f.resize(n);
vector <int> h = b; h.resize(n);
ntt(f, 1); ntt (h, 1);
for (int i = 0; i < n; ++i)
h[i] = 1ll * h[i] * (2ll - 1ll * f[i] * h[i] % P + P) % P;
ntt(h, -1);
b.assign (h.begin(), h.begin() + min(len, m));
}
b.resize (m);
return b;
}
pair <vector <int>, vector <int>> Divmod(vector <int> a, vector <int> b)
{
int n_sz = a.size(), m_sz = b.size();
if (n_sz < m_sz) return {{0},a};
int dq = n_sz - m_sz + 1;
vector <int> ra = a, rb = b;
reverse(ra.begin(), ra.end()); reverse(rb.begin(), rb.end());
ra.resize(dq); rb.resize(dq);
vector <int> inv_rb = Inv(rb, dq);
vector <int> q = Conv(ra, inv_rb);
q.resize(dq);
reverse(q.begin(), q.end());
vector <int> qb = Conv(q, b);
vector <int> r(m_sz - 1);
for (int i = 0; i < m_sz - 1; ++i)
r[i] = (0ll + a[i] - (i < (int)qb.size() ? qb[i] : 0) + P) % P;
return {q,r};
}
vector <int> Sqrt (vector <int> a, int m) // b[0] = 1
{
vector <int> b {1};
int inv2 = qpow(2, P - 2);
for (int len = 2; (len >> 1) < m; len <<= 1)
{
int sz = min((int)a.size(), len);
vector <int> f(a.begin(), a.begin() + sz);
vector <int> inv_b = Inv(b, len);
init_rev(len << 1);
f.resize(n); inv_b.resize(n);
ntt(f, 1); ntt(inv_b, 1);
for (int i = 0; i < n; ++i)
f[i] = 1ll * f[i] * inv_b[i] % P;
ntt(f, -1);
b.resize(len);
for (int i = 0; i < len; ++i)
b[i] = 1ll * (b[i] + f[i]) % P * inv2 % P;
}
b.resize(m);
return b;
}
};
constexpr int V = 1e5;
NTT <MOD> calc;
void solve()
{
int n = read(), m = read();
vector <int> G(m + 1);
for (int i = 1; i <= n; ++i)
{
int x = read();
if (x <= m) G[x] = MOD - 4;
}
G[0] = 1;
auto A = calc.Sqrt(G, m + 1);
A[0] = (A[0] + 1) % MOD;
auto B = calc.Inv(A, m + 1);
for (int i = 1; i <= m; ++i) printf("%d\n", 2 * B[i] % MOD);
}
int main()
{
int t = 1;
while (t--) solve();
return 0;
}