[BZOJ3625][CF438E]小朋友和二叉樹 （多項式開根，求逆）

題面

題解

設多項式的第a項爲權值和爲a的二叉樹個數，多項式的第a項表示是否爲真，即

則，所以F是三個多項式的卷積，其中包括自己：

，1是F的常數項，即。

我們發現這是一個一元二次方程，可以求出，因爲g的常數項爲零，所以1-4g的常數項爲1，的常數項也爲1，的常數項就爲零，就跑不了逆元，所以舍掉。

最終，跑一個多項式開根和一個多項式求逆就行。

CODE

大常數TLE的代碼， 自己優化吧（逃 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define LL long long
#define MAXN 800005
//#define int LL
#define LENTH (LERTH*sizeof(int))
#pragma GCC optimize(2)
#pragma G++ optimize(3)
inline int read() {
	int f = 1,x = 0;char s = getchar();
	while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x = x * 10 + s - '0';s = getchar();}
	return x * f;
}
int fan[MAXN],mod = 998244353,n,m,q,i,j,s,o,k,t,a[MAXN],b[MAXN],f[MAXN],g[MAXN],LERTH;
inline int readm() {
	int f = 1,x = 0;char s = getchar();
	while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x = (x * 10ll % mod + s - '0') % mod;s = getchar();}
	return x * f;
}
inline int qkpow(int a,int b) {
	int res = 1;
	while(b>0) {
		if(b&1) res = 1ll*res*a%mod;
		a = 1ll*a*a%mod;
		b>>=1;
	}
	return res;
}
int INV[MAXN];
int mod2 = 998244353;
inline void NTT(int len,int *a,int temp) {
	int ugly = (int)round(log2(len));
	for(int i = 1;i < len;i ++) {
		fan[i] = (fan[i>>1]>>1)|((i&1)<<(ugly-1));
		if(i<fan[i]) 
			swap(a[i],a[fan[i]]);
	}
	for(int k = 2;k <= len;k <<= 1) {
		int t = k / 2;
		int vw = (temp == 1)? qkpow(3,(mod -1) / k):qkpow(3,(mod-1) - (mod-1)/k);
		for(int i = 0;i < len;i += k) {
			int w2 = 1;
			for(int j = 0;j < t;j ++,w2 = 1ll*w2*vw % mod) {
				int w3 = a[i + j],w4 = a[i + j + t];
				a[i + j] = (0ll+w3 + w4*1ll*w2) % mod;
				a[i + j + t] = ((0ll+w3 - w2*1ll*w4)%mod + mod) % mod;
			}
		}
	}
	if(temp == 1) return ;
	int invmod = qkpow(len,mod - 2);
	for(int i = 0;i < len;i ++) {
		a[i] = 1ll*a[i] * invmod % mod;
	}
	return ;
}
inline void DXSC(int *A,int *B,int *C,int la,int lb,int &lc) {
	int le = 1,nm = la + lb;
	while(le <= nm) le <<= 1;
	NTT(le,A,1);
	NTT(le,B,1);
	for(int i = 0;i <= le;i ++) C[i] = 1ll*A[i]*B[i] % mod;
	NTT(le,C,-1);
	lc = nm;
	return ;
}
inline void work(int n) {
	int le = 1;while(le <= 2*n) le <<= 1;
	for(int i = 0;i < le;i ++) f[i] = g[i] = 0;
	for(int i = 0;i < n;i ++) f[i] = a[i];
	for(int i = 0;i < n;i ++) g[i] = b[i];
	NTT(le,f,1);NTT(le,g,1);
	for(int i = 0;i < le;i ++) f[i] = g[i] * 2ll - 1ll*g[i]*g[i] % mod * f[i] % mod,f[i]%= mod;
	NTT(le,f,-1);
	for(int i = 0;i < n;i ++) b[i] = (f[i]%mod + mod) % mod;
}
inline void inv(int *A,int *B,int n) {
	memset(a,0,LENTH);
	memset(b,0,LENTH);
	memset(B,0,LENTH);
	for(int i = 0;i < n;i ++) a[i] = A[i];
	b[0] = qkpow(a[0],mod - 2);
	int m = 1;
	while(m < n) {
		m <<= 1;
		work(m);
	}
	for(int i = 0;i < n;i ++) B[i] = b[i];
	return ;
}
inline void deriv(int *A,int *B,int len) {
	for(int i = 0;i < len;i ++) B[i] = (1ll + i) * A[i + 1] % mod;B[len] = 0;
}
inline void integ(int *A,int *B,int len) {
	for(int i = len + 1;i > 0;i --) B[i] = 1ll * INV[i] * A[i - 1] % mod;B[0] = 0;
}
inline void putd(int *a,int len) {
	for(int i = 0;i <= len;i ++) printf("%d ",a[i]);putchar('\n');
}
int ai[MAXN],bi[MAXN],c[MAXN],d[MAXN],h[MAXN];
inline void polyln(int *A,int *B,int n) {
	int m = n + 1;
	memset(ai,0,LENTH);
	memset(bi,0,LENTH);
	memset(c,0,LENTH);
	memset(d,0,LENTH);
	memset(h,0,LENTH);
	for(int i = 0;i <= n;i ++) ai[i] = A[i];
	deriv(ai,c,n);
	inv(ai,d,m);
	DXSC(c,d,h,n-1,n,k);
	integ(h,bi,n-1);
	memset(B,0,LENTH);
	for(int i = 0;i <= n;i ++) {
		B[i] = bi[i];
	}
}
int cc[MAXN],ee[MAXN],dd[MAXN],inv2 = qkpow(2ll,mod - 2) % mod;
inline void polyexp(int *A,int *B,int n) {
	int m = 1;
	memset(B,0,LENTH);
	B[0] = 1;
	while(m < n+1) {
		m<<= 1;
		memcpy(cc,B,LENTH);
		memcpy(ee,B,LENTH);
		polyln(cc,cc,m-1);
		for(int i = 0;i < m;i ++) cc[i] = (0ll+A[i] + mod - cc[i]) % mod;cc[0] = (cc[0] + 1ll) % mod;
		DXSC(ee,cc,ee,m-1,m-1,k);
		for(int i = 0;i < m;i ++) B[i] = (ee[i]%mod + mod) % mod;
	}
	for(int i = n+1;i <= m;i ++) B[i] = 0;
	return ;
}
void polysqrt(int *A,int *B,int n) {
	int m = 1;
	memset(B,0,LENTH);
	B[0] = 1;
	while(m < n+1) {
		m <<= 1;
		memcpy(ee,B,LENTH);
		memset(dd,0,LENTH);
		for(int i = 0;i < m;i ++) dd[i] = A[i];
		inv(B,cc,m);
		int le = 1;while(le <= m*2) le<<=1;
		NTT(le,cc,1);NTT(le,dd,1);NTT(le,ee,1);
		for(int i = 0;i <= le;i ++) ee[i] = 1ll*inv2*(0ll + ee[i] + 1ll*dd[i] * cc[i] % mod) % mod;
		NTT(le,ee,-1);
		for(int i = 0;i < m;i ++) B[i] = (ee[i]%mod + mod) % mod;
	}
	for(int i = n+1;i <= m;i ++) B[i] = 0;
	return ;
}
int FF[MAXN],deltag[MAXN],sqrtg[MAXN],doubleg[MAXN];
int main() {
	n = read();m = read();
	for(LERTH=1;LERTH <= m;LERTH<<=1);
	for(int i = 1;i <= n;i ++) {
		s = read();
		if(s <= m) deltag[s] = mod - 4;
	}deltag[0] = 1;
	polysqrt(deltag,sqrtg,m);
	sqrtg[0] = (sqrtg[0] + mod + 1ll) % mod;
	inv(sqrtg,FF,m + 1);
	for(int i = 1;i <= m;i ++) {
		printf("%d\n",(FF[i]*2ll) % mod);
	}
	return 0;
}

我本來A了只是T了