CF438E The Child and Binary Tree

$f (x)$ 表示权值和为 $x$ 的答案，$g(x)$ 表示值 $x$ 是否在集合中。考虑当前点，左子树，右子树的权值，有 $f(x) = \sum _ {i + j + k = x} g(i) f(j) f(k)$ 。特别地，单独考虑 $x = 0$ ，因为一定有 $G(0) = 0$ ，但是在答案意义上需要考虑。考虑 $f$ 的生成函数 $F$ ，那么有 $F = 1 + GF ^ 2$ ，解得 $F = \frac {1 \pm \sqrt {1 - 4G}} {2G}$ ，检验得 $F = \frac {1 - \sqrt {1 - 4G}} {2G}$ 。发现 $G(0) = 0$ ，无法求逆。那么有 $F = \frac {1 - \sqrt {1 - 4G}} {2G} = \frac 2 {1 + \sqrt {1 - 4G}} $ 。

查看代码

#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, mod = 998244353, inv2 = mod + 1 >> 1;
int rev[N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void ntt (int *x, int bit, int op)
{
    int tot = 1 << bit;
    for (int i = 1; i < tot; ++i)
        if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
            swap(x[rev[i]], x[i]);
    for (int mid = 1; mid < tot; mid <<= 1)
    {
        int w1 = binpow(3, (mod - 1) / (mid << 1));
        if (!~op) w1 = binpow(w1);
        for (int i = 0; i < tot; i += mid << 1)
            for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
            {
                int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
                x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
            }
    }
    if (~op) return;
    int itot = binpow(tot);
    for (int i = 0; i < tot; ++i)
        x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
    int bit = 0;
    while (1 << bit < n + m - 1) ++bit;
    int tot = 1 << bit;
    for (int i = n; i < tot; ++i) f[i] = 0;
    for (int i = m; i < tot; ++i) g[i] = 0;
    ntt(f, bit, 1), ntt(g, bit, 1);
    for (int i = 0; i < tot; ++i)
        res[i] = (LL)f[i] * g[i] % mod;
    ntt(res, bit, -1);
    for (int i = nm; i < tot; ++i) res[i] = 0;
}
void PolyInv(int n, int *x, int *g)
{
    if (n == 1) return void(g[0] = binpow(x[0]));
    int m = n + 1 >> 1;
    int bit = 0;
    while (1 << bit < n + m + m - 2) ++bit;
    int tot = 1 << bit;
    PolyInv(m, x, g);
    for (int i = m; i < tot; ++i) g[i] = 0;
    static int A[N];
    for (int i = 0; i < n; ++i) A[i] = x[i];
    for (int i = n; i < tot; ++i) A[i] = 0;
    ntt(g, bit, 1), ntt(A, bit, 1);
    for (int i = 0; i < tot; ++i)
        g[i] = (2 - (LL)g[i] * A[i]) % mod * g[i] % mod;
    ntt(g, bit, -1);
    for (int i = n; i < tot; ++i) g[i] = 0;
}
void PolySqrt(int n, int *x, int *g)
{
    if (n == 1) return void(g[0] = 1);
    int m = n + 1 >> 1;
    int bit = 0;
    while (1 << bit < n + n - 1) ++bit;
    int tot = 1 << bit;
    PolySqrt(m, x, g);
    for (int i = m; i < n; ++i) g[i] = 0;
    static int A[N], B[N];
    PolyInv(n, g, A);
    for (int i = n; i < tot; ++i) A[i] = 0;
    for (int i = 0; i < tot; ++i) B[i] = x[i];
    for (int i = n; i < tot; ++i) B[i] = 0;
    ntt(A, bit, 1), ntt(B, bit, 1), ntt(g, bit, 1);
    for (int i = 0; i < tot; ++i)
        g[i] = (g[i] + (LL)B[i] * A[i]) % mod * inv2 % mod;
    ntt(g, bit, -1);
    for (int i = n; i < tot; ++i) g[i] = 0;
}
int main ()
{
    static int n, m, A[N], B[N];
    read(n, m);
    for (int a; n; --n)
        read(a), A[a] = -4;
    A[0] = 1;
    n = 1e5 + 1;
    PolySqrt(n, A, B);
    ++B[0] %= mod;
    PolyInv(n, B, A);
    for (int i = 1; i <= m; ++i)
        write((A[i] * 2 % mod + mod) % mod), puts("");
    return 0;
}