[BZOJ3625][Codeforces Round #250]小朋友和二叉树 多项式开根+求逆

https://www.lydsy.com/JudgeOnline/problem.php?id=3625

愉快地列式子。设\(F[i]\)表示权值为\(i\) 的子树的方案数,\(A[i]\)\(i\)在不在集合中。

\[F[n]=\sum_{i=0}^n \sum_{j=0}^{n-i}F[i]\cdot F[j]\cdot A[n-i-j] \]

初始状态\(F[0]=1\)

我们把\(F,A\)看成多项式。

\[F(x)-1=F^2(x)\cdot A(x)\\ A(x)\cdot F^2(x)-F(x)+1=0\\ F(x)=\frac{1\pm\sqrt{1-4A(x)}}{2A(x)} \]

因为\(A[0]=0\)\(F[0]=1\),如果取\(+\)号,末位会不符,舍出。因此只能取\(-\)

这样

\[\begin{align*} F(x)&=\frac{1-\sqrt{1-4A(x)}}{2A(x)}\\ &=\frac{(1-\sqrt{1-4A(x)})(1+\sqrt{1-4A(x)})}{2A(x)(1+\sqrt{1-4A(x)})}\\ &=\frac{4A(x)}{2A(x)(1+\sqrt{1-4A(x)})}\\ &=\frac{2}{1+\sqrt{1-4A(x)}} \end{align*} \]

这样就变成多项式开根+求逆的板子了。


(刚开始转码风,可能有些地方不太自然,也有可能有些地方仍然保留着就码风没有注意)

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
#define isbreak dbg("*")

template<typename A, typename B> inline char SMAX(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char SMIN(A &a, const B &b) {return b < a ? a = b , 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I>
inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 4e5 + 7;
const int P = 998244353;
const int G = 3;
const int Gi = 332748118;
const int Inv2 = 499122177;

int n, m, x, a[N];

inline void SADD(int &x, int y) {x += y;x >= P ? x -= P : 0;}
inline int SMOD(int x) {return x >= P ? x - P : x;}
inline int fpow(int x,int y) {
	int ans = 1;
	for (; y; y >>= 1, x = (ll)x * x % P) if (y & 1) ans = (ll)ans * x % P;
	return ans;
}

namespace DFT {
	int A[N], B[N], C[N];

	inline void NTT(int *a, int n, int f) {
		for (int i = 0, j = 0; i < n; ++i) {
			if (i > j) std::swap(a[i], a[j]);
			for (int l = n >> 1; (j ^= l) < l; l >>= 1);
		}
		for (int i = 1; i < n; i <<= 1) {
			int w = fpow(f > 0 ? G : Gi, (P - 1) / (i << 1));
			for (int j = 0; j < n; j += i << 1)
				for (int k = 0, e = 1; k < i; ++k, e = (ll)e * w % P){
					int x = a[j + k], y = (ll)e * a[i + j + k] % P;
					a[j + k] = SMOD(x + y); a[i + j + k] = SMOD(x + P - y);
				}
		}
		if (f < 0) for (int i = 0, p = fpow(n, P - 2); i < n; ++i) a[i] = (ll)a[i] * p % P;
	}
	namespace Inv {
		int A[N], B[N];
		inline void GetInv(int *a, int n, int *b) {
			memset(B, 0, sizeof(B)); B[0] = fpow(a[0], P - 2);
			for (int deg = 2; deg < (n << 1); deg <<= 1) {
				int L = deg << 1;
				for (int i = 0; i < deg; ++i) A[i] = a[i];
				for (int i = deg; i < L; ++i) A[i] = 0;
				NTT(A, L, 1); NTT(B, L, 1);
				for (int i = 0; i < L; ++i) B[i] = (ll)B[i] * (2 + P - (ll)B[i] * A[i] % P) % P;
				NTT(B, L, -1);
				for (int i = deg; i < L; ++i) B[i] = 0;
			}
			for (int i = 0; i < n; ++i) b[i] = B[i];
		}
	} using Inv::GetInv;

	inline void GetSqrt(int *a, int n, int *b) {
		B[0] = 1;
		for (int deg = 2; deg < (n << 1); deg <<= 1) {
			int L = deg << 1;
			for (int i = 0; i < deg; ++i) A[i] = a[i];
			for (int i = deg; i < L; ++i) A[i] = 0;
			GetInv(B, deg, C);
			NTT(A, L, 1); NTT(C, L, 1);
			for (int i = 0; i < L; ++i) C[i] = (ll)A[i] * C[i] % P;
			NTT(C, L, -1);
			for (int i = 0; i < L; ++i) B[i] = (ll)(B[i] + C[i]) * Inv2 % P;
			for (int i = deg; i < L; ++i) B[i] = 0;
		}
		for (int i = 0; i < n; ++i) b[i] = B[i];
	}
}
using DFT::GetInv;
using DFT::GetSqrt;

int main() {
	#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
	#endif
	read(n), read(m);
	for (int i = 1; i <= n; ++i) read(x), a[x] = 1;
	for (int i = 1; i <= m; ++i) if (a[i]) a[i] = P - SMOD(SMOD(a[i] << 1) << 1);
	a[0] = 1;
	GetSqrt(a, m + 1, a);  SADD(a[0], 1);
	GetInv(a, m + 1, a);
	for (int i = 1; i <= m; ++i) printf("%d\n", SMOD(a[i] << 1));
}
posted @ 2019-02-08 22:14  hankeke  阅读(...)  评论(...编辑  收藏