Codeforces 438E The Child and Binary Tree

题目传送门

　　传送点I

　　传送点II

　　传送点III

题目大意

　　每个点的权值$cin {c_{1}, c_{2}, cdots, c_{n}}$，问对于每个$1leqslant sleqslant m$有多少种不同的这样的有根二叉树满足所有点的点权和等于$s$。

　　先考虑一下怎么用dp来做。

　　设$f_{n}$表示点权和为$n$的满足条件的二叉树的个数，那么有：

$f_{n} = sum_{c in C}sum_{i = 0}^{n - c}f_{i}f_{n - c - i}$

　　初值满足$f_{0} = 1$。

　　注意到右边的式子有点像卷积，考虑$f_{n}$的普通生成函数$F(x) = sum_{n = 0}f_{n}x^{n}$。

　　将$F(x)$平方就能得到$fotimes f$的生成函数。我们用它替换右半边，得到了：

$F(x) = sum_{c in C}x^{c}F^2(x) + 1$

　　设某个常数$A = sum_{c in C}x^{c}$，则有$Acdot F^2(x) - F(x) + 1$。

　　解得$F_{1}(x) = frac{1 + sqrt{1 - 4A}}{2A}, F_{2}(x) = frac{1 - sqrt{1 - 4A}}{2A} = frac{1 - (1 - 4A)}{2Asqrt{1 + 4A}} = frac{2}{sqrt{1 + 4A}}$。

　　我们考虑$F_{1}(x)$的分母的常数项，它是2，分子的常数项为0，然后求逆的时候就会出事情。

　　但是$F_{2}(x)$可以通过恒等变换使得分母的常数项为非0。

　　然后做一次多项式开根，一次多项式求逆就做完了。

Code

/**
 * Codeforces
 * Problem#438E
 * Accepted
 * Time: 1091ms
 * Memory: 5100k
 */
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

const int N = 262144;
const int bzmax = 20;
const int M = 998244353;
const int g = 3;
const int inv2 = (M + 1) >> 1;

int qpow(int a, int p) {
    if (p < 0)
        p += M - 1;
    int rt = 1, pa = a;
    for ( ; p; p >>= 1, pa = pa * 1ll * pa % M)
        if (p & 1)
            rt = rt * 1ll * pa % M;
    return rt;
}

int add(int a, int b) {
    return ((a += b) >= M) ? (a - M) : (a);
}

int sub(int a, int b) {
    return ((a -= b) < 0) ? (a + M) : (a);
}

class NTT {
    public:
        int gn[bzmax], _gn[bzmax];
        
        NTT() {
            for (int i = 0, L = 2; i < bzmax; i++, L <<= 1) {
                gn[i] = qpow(g, (M - 1) / L);
                _gn[i] = qpow(g, -(M - 1) / L);
            }
        }

        void operator () (int* f, int len, int sgn) {
            for (int i = 1, j = len >> 1, k; i < len - 1; i++, j += k) {
                if (i < j)
                    swap(f[i], f[j]);
                for (k = len >> 1; j >= k; j -= k, k >>= 1);
            }

            for (int b = 2, t = 0; b <= len; b <<= 1, t++) {
                int wn = ((sgn > 0) ? (gn[t]) : (_gn[t])), w = 1, hb = b >> 1;
                for (int i = 0; i < len; i += b, w = 1)
                    for (int j = i; j < i + hb; j++, w = w * 1ll * wn % M) {
                        int a = f[j], b = f[j + hb] * 1ll * w % M;
                        f[j] = add(a, b);
                        f[j + hb] = sub(a, b);
                    }
            }

            if (sgn < 0) {
                int invlen = qpow(len, -1);
                for (int i = 0; i < len; i++)
                    f[i] = f[i] * 1ll * invlen % M;
            }
        }

        int correctLen(int n) {
            int m = 1;
            while (m < n)    m <<= 1;
            return m;
        }
}NTT;

template<typename T>
void pcopy(T* ns, T* ne, const T* os) {
    for ( ; ns != ne; *ns = *os, ns++, os++);
}

template<typename T>
void pfill(T* ps, T* pt, T val) {
    for ( ; ps != pt; *ps = val, ps++);
}

void debug(const int* f, int n) {
    for (int i = 0; i < n; i++)
        cerr << f[i] << " ";
    cerr << endl;
}

void pol_inverse(int *f, int *g, int n) {
    static int h[N];
    if (n == 1)
        g[0] = qpow(f[0], -1);
    else {
        pol_inverse(f, g, (n + 1) >> 1);

        int t = NTT.correctLen(n << 1 | 1);
        pcopy(h, h + n, f);
        pfill(h + n, h + t, 0);
        NTT(h, t, 1);
        NTT(g, t, 1);
        for (int i = 0; i < t; i++)
            g[i] = g[i] * 1ll * sub(2, g[i] * 1ll * h[i] % M) % M;
        NTT(g, t, -1);
        pfill(g + n, g + t, 0);
    }
}

void pol_sqrt(int *f, int *g, int n) {
    static int C[N], D[N];
    if (n == 1)
        g[0] = 1;
    else {
        pol_sqrt(f, g, (n + 1) >> 1);
        
        int t = NTT.correctLen(n << 1);
        pcopy(C, C + n, f);
        pfill(C + n, C + t, 0);
        pfill(D, D + t, 0);
        pol_inverse(g, D, n);
        NTT(C, t, 1);
        NTT(D, t, 1);
        NTT(g, t, 1);
        for (int i = 0; i < t; i++)
            g[i] = add(C[i] * 1ll * D[i] % M, g[i]) * 1ll * inv2 % M;
        NTT(g, t, -1);
        pfill(g + n, g + t, 0);
    }
}

int n, m;
int C[N], qC[N];

inline void init() {
    scanf("%d%d", &n, &m);
    for (int i = 1, x; i <= n; i++) {
        scanf("%d", &x);
        if (x <= m)
            C[x] = -4;
    }
}

inline void solve() {
    C[0]++, m++;
    pol_sqrt(C, qC, m);
    qC[0]++;
    pfill(C, C + m, 0);
    pol_inverse(qC, C, m);
    for (int i = 1; i < m; i++)
        printf("%d
", add(C[i], C[i]));
}

int main() {
    init();
    solve();
    return 0;
}