CF438E The Child and Binary Tree

题意:给定一个大小为\(n\)的点权集合,所有的点的点权必须是集合里的点权,问这样形成的二叉树的点权和为\([1-m]\)的有多少个

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题解:

首先我们可以设\(G(i)\)表示\(i\)这个数是否在集合中,\(f(i)\)表示点权和为\(i\)的方案数
那么我们可以考虑枚举一个点的点权和ta的两个子树的点权和
\(f(0)=1\\ f(k)=\sum_{i=1}^{k}g(i)\sum_{j=0}^{i+j\le k}{f(j)f(k-i-j)}\)
所以可以发现\(f=g*f^2\)
直接解方程就可以得到\(f=\frac{1 \pm \sqrt{1-4g}}{2g}\)
这个\(\pm\)该怎么处理?
先试试分母有理化
\(f=\frac{2}{1 \pm \sqrt{1-4g}}\)
可以发现一个问题
对于\(f(0):g(0)=0 \; \; => \; \; 1-4g = 1\)
要保证分母必须不为\(0\)
所以应该是\(f=\frac{2}{1 + \sqrt{1-4g}}\)
多项式开根+求逆即可

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
# define LL long long
using namespace std ;
const int M = 400050 ;
const int mod = 998244353 ;
const int G = 3 ;
const int Gi = mod / G + 1 ;

inline int read() {
	char c = getchar() ; int x = 0 , w = 1 ;
	while(c>'9'||c<'0') { if(c=='-') w = -1 ; c = getchar() ; }
	while(c>='0'&&c<='9') { x = x*10+c-'0' ; c = getchar() ; }
	return x * w ; 
}

int n , m ;
int len , lim = 1 , r[M] ;
LL g[M] , a[M] , b[M] , c[M] , ans[M] , inv2 ;

inline LL Fpw(LL Base , LL k) {
	LL temp = 1 ;
	while(k) {
		if(k & 1) temp = temp * Base % mod ;
		Base = Base * Base % mod ; k >>= 1 ;
	}
	return temp ;
}
inline void NTT(LL *A , int unit) {
	for(int i = 0 ; i < lim ; i ++) if(r[i] > i) swap(A[i] , A[r[i]]) ;
	for(int mid = 1 ; mid < lim ; (mid <<= 1)) {
		LL R = (mid << 1) , W = Fpw(unit > 0 ? G : Gi , (mod - 1) / R) ;
		for(int j = 0 ; j < lim ; j += R) {
			LL w = 1 ;
			for(int k = 0 ; k < mid ; k ++ , w = (w * W) % mod) {
				LL x = A[j + k] , y = w * A[j + k + mid] % mod ;
				A[j + k] = (x + y) % mod ; A[j + k + mid] = (x - y) % mod ;
			}
		}
	}
}

void Inv(int d , LL *A , LL *b) {
	if(d == 1) { b[0] = Fpw(A[0] , mod - 2) ; return ; }
	Inv((d + 1) >> 1 , A , b) ;
	lim = 1 ; len = 0 ;
	while(lim < (d << 1)) lim <<= 1 , ++ len ;
	for(int i = 1 ; i < lim ; i ++) r[i] = ((r[i >> 1] >> 1) | ((i & 1) << (len - 1))) ;
	for(int i = 0 ; i < d ; i ++) a[i] = A[i] ;
	for(int i = d ; i < lim ; i ++) a[i] = 0 ;
	NTT(b , 1) ; NTT(a , 1) ;
	for(int i = 0 ; i < lim ; i ++) b[i] = ((2 - a[i] * b[i]) % mod * b[i] % mod + mod) % mod ;
	NTT(b , -1) ; LL inv = Fpw(lim , mod - 2) ;
	for(int i = 0 ; i < d ; i ++) b[i] = (b[i] * inv % mod + mod) % mod ;
	for(int i = d ; i < lim ; i ++) b[i] = 0 ;
}
void Sqrt(int d , LL *A , LL *b) {
	if(d == 1) { b[0] = 1 ; return ; }
	Sqrt((d + 1) >> 1 , A , b) ;
	for(int i = 0 ; i < (d << 1) ; i ++) c[i] = 0 ;
	Inv(d , b , c) ;
	lim = 1 ; len = 0 ;
	while(lim < (d << 1)) lim <<= 1 , ++ len ;
	for(int i = 1 ; i < lim ; i ++) r[i] = ((r[i >> 1] >> 1) | ((i & 1) << (len - 1))) ;
	for(int i = 0 ; i < d ; i ++) a[i] = A[i] ;
	for(int i = d ; i < lim ; i ++) a[i] = 0 ;
	NTT(a , 1) ; NTT(c , 1) ;
	for(int i = 0 ; i < lim ; i ++) a[i] = (a[i] * c[i] % mod * inv2 % mod + mod) % mod ;
	NTT(a , -1) ; LL inv = Fpw(lim , mod - 2) ;
	for(int i = 0 ; i < d ; i ++) a[i] = (a[i] * inv % mod + mod) % mod ;
	for(int i = d ; i < lim ; i ++) a[i] = 0 ;
	for(int i = 0 ; i < d ; i ++) b[i] = ((b[i] * inv2 % mod + a[i]) % mod + mod) % mod ;
	for(int i = d ; i < lim ; i ++) b[i] = 0 ;
}
int main() {
	n = read() ; m = read() ; inv2 = Fpw(2 , mod - 2) ;
	for(int i = 1 , x ; i <= n ; i ++) {
		x = read() ;
		g[x] = (g[x] - 4 + mod) % mod ;
	}
	g[0] ++ ; Sqrt(m + 1 , g , b) ;
	b[0] ++ ; Inv(m + 1 , b , ans) ;
	for(int i = 1 ; i <= m ; i ++)
		printf("%lld\n",(2LL * ans[i] % mod + mod) % mod) ;
	return 0 ;
}