luoguP4492 [HAOI2018]苹果树 组合计数 + dp

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首先，每个二叉树对应着唯一的中序遍历，并且每个二叉树的概率是相同的

这十分的有用

考虑(dp)求解

令(f_i)表示(i)个节点的子树，根的深度为(1)时，所有点的期望深度之和（乘(i!)）的值

令(g_i)表示(i)个节点的子树，期望两两路径之和（乘(i!)）的值

那么(f_i = i * i! + sum limits_{L = 0}^{i - 1} inom{i - 1}{L} (f_L * R! + f_R * L!))，(L, R)分别表示左右子树的值

(g_i = sum limits_{L = 0}^{i - 1} inom{i - 1}{L} (g_L * R! + g_R * L! + f_L * R! * (R + 1) + f_R * L! * (L + 1)))

复杂度(O(n^2))

由于没有逆元，因此组合数要预处理

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好像可以用多项式优化到(O(n log^2 n))QAQ

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 2e3 + 5;

int n, mod;
int C[sid][sid], fac[sid], f[sid], g[sid];

inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline void dec(int &a, int b) { a -= b; if(a < 0) a += mod; }
inline int Inc(int a, int b) { return (a + b >= mod) ? a + b - mod : a + b; }
inline int Dec(int a, int b) { return (a - b < 0) ? a - b + mod : a - b; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }

inline void init() {
    fac[0] = C[0][0] = 1;
    rep(i, 1, n) {
        C[i][0] = C[i][i] = 1;
		fac[i] = mul(fac[i - 1], i);
        rep(j, 1, i - 1) C[i][j] = Inc(C[i - 1][j], C[i - 1][j - 1]);
    }
}

inline void dp() {
    f[1] = 1;
    rep(i, 2, n) {
        rep(L, 0, i - 1) {
            int R = i - 1 - L, F = 0, G = 0;
			F = (1ll * f[L] * fac[R] + 1ll * f[R] * fac[L]) % mod;
            G = (1ll * f[L] * fac[R] % mod * (R + 1) % mod);
			inc(G, 1ll * f[R] * fac[L] % mod * (L + 1) % mod);
			inc(G, 1ll * g[L] * fac[R] % mod);
			inc(G, 1ll * g[R] * fac[L] % mod);
            inc(f[i], mul(F, C[i - 1][L])); inc(g[i], mul(G, C[i - 1][L]));
        }
		inc(f[i], 1ll * i * fac[i] % mod);
    }
    printf("%d
", g[n]);
}

int main() {
    cin >> n >> mod;
    init(); dp();
    return 0;
}