Codeforces 438E The Child and Binary Tree - 生成函数 - 多项式

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$c\in {c_{1}, c_{2}, \cdots, c_{n}}$，问对于每个$1\leqslant s\leqslant m$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$f\otimes f$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$A\cdot F^2(x) - F(x) + 1$。

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

　　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 = ;
 const int bzmax = ;
 const int M = ;
 const int g = ;
 const int inv2 = (M + ) >> ;

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

 class NTT {
     public:
         int gn[bzmax], _gn[bzmax];

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

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

             for (int b = , t = ; b <= len; b <<= , t++) {
                 int wn = ((sgn > ) ? (gn[t]) : (_gn[t])), w = , hb = b >> ;
                 for (int i = ; i < len; i += b, w = )
                     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 < ) {
                 int invlen = qpow(len, -);
                 for (int i = ; i < len; i++)
                     f[i] = f[i] * 1ll * invlen % M;
             }
         }

         int correctLen(int n) {
             int m = ;
             while (m < n)    m <<= ;
             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 = ; i < n; i++)
         cerr << f[i] << " ";
     cerr << endl;
 }

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

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

 void pol_sqrt(int *f, int *g, int n) {
     static int C[N], D[N];
     if (n == )
         g[] = ;
     else {
         pol_sqrt(f, g, (n + ) >> );

         int t = NTT.correctLen(n << );
         pcopy(C, C + n, f);
         pfill(C + n, C + t, );
         pfill(D, D + t, );
         pol_inverse(g, D, n);
         NTT(C, t, );
         NTT(D, t, );
         NTT(g, t, );
         for (int i = ; i < t; i++)
             g[i] = add(C[i] * 1ll * D[i] % M, g[i]) * 1ll * inv2 % M;
         NTT(g, t, -);
         pfill(g + n, g + t, );
     }
 }

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

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

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

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