【BZOJ3625】【CF438E】小朋友和二叉树

题目

传送门

思路&做法

我们可以用\(v_i\)表示\(i\)\(c\)中出现了几次, 用\(f_i\)表示权值为\(i\)的神犇树的总数, 于是

\[f_x = \sum_{i = 0}^{x}v_i \bigg( \sum_{j = 0}^{x-i}f_jf_{x-i-j} \bigg) \]

\[f_0 = 1 \]

然后我们设\(v\)的生成函数为\(V = \sum_{i = 0} ^{\infty}v_ix^i\),
\(f\)的生成函数为\(F = \sum_{i = 0}^{\infty}f_ix^i\)

所以

\[F(x) = C(x) F(x)F(x) + 1 \]

然后移一下项

\[V(x)F^2(x) - F(x) + 1 = 0 \]

接着直接用求根公式

\[F(x) = {1 \pm \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)} \]

于是有两种情况

\((1)\)

\[F(x) = {1 + \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)} \]

要舍去\((\)原因?不知道,我太弱了\()\)

\((2)\)

\[F(x) = {1 - \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)} \]

可以把分子和分母同时乘上\(1 + \sqrt{1 + 4 \times V(x)}\)

所以

\[\begin {aligned} F(x)&={1^2 - \big(\sqrt{1 - 4 \times V(x)}\big)^2 \over {2 \times V(x) \times \big(1 + \sqrt{1 - 4 \times V(x)} \big)}} \\\ &={4 \times V(x) \over {2 \times V(x) \times \big(1 + \sqrt{1 - 4 \times V(x)}} \big)} \\\ &={2 \over {1 + \sqrt{1 - 4 \times V(x)}}} \end {aligned} \]

多项式求逆和多项式开根可以看我的这篇blog : 多项式的一些东西

代码

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

const LL mod = 998244353LL;

const int N = 400010;

char I[8000010], *pos, *End;

inline char GetChar()
{	if(pos == End)
	{	End = (pos = I) + fread(I, 1, 8000000, stdin);
		if(pos == End) return EOF;
	}
	return *pos++;
}

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


int Bit;

inline LL power(LL a, LL n)
{	LL ans = 1;
	while (n)
	{	if (n & 1) ans = (ans * a) % mod;
		a = (a * a) % mod;
		n >>= 1;
	}
	return ans;
}

int rev[N];

void getReverse(int Len)
{	for (int i = 0; i < Len; i++)
		rev[i] = (rev[i>>1] >> 1) | ((i&1) * (Len>>1));
}

LL w_n[N];

void NTT(LL * a, int opt, int Len)
{	getReverse(Len);
	for (int i = 0; i < Len; i++)
		if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (register int i = 2; i <= Len; i <<= 1)
	{	LL w_n = power(3LL, (mod-1LL) / (LL)i);
		if (opt == -1) w_n = power(w_n, mod-2);
		for (register int j = 0; j < Len; j += i)
		{	LL w = 1;
			for (register int k = 0; k < (i>>1); k++)
			{	LL x = a[j + k];
				LL y = (w * a[j + k + (i>>1)]) % mod;
				a[j + k] = (x + y) % mod;
				a[j + k + (i>>1)] = (x - y + mod) % mod;
				w = (w * w_n) % mod;
			}
		}
	}
	if (opt == -1)
	{	LL num = power(Len, mod-2);
		for (register int i = 0; i < Len; i++)
			a[i] = (a[i] * num) % mod;
	}
}

int getBit(int l)
{	Bit = 0;
	for (; (1 << Bit) <= l; Bit++);
}

inline void cpy(LL * a, LL * b, int n, int Len)
{	for (register int i = 0; i < n; i++) a[i] = b[i];
	for (register int i = n; i < Len; i++) a[i] = 0;
}

LL tmp1[N], tmp2[N];

void get_Inv(LL * a, LL * b, int l)
{	b[0] = power(a[0], mod-2);
	for (register int i = 2; i <= l; i <<= 1)
	{	int Len = i << 1;
		cpy(tmp1, a, i, Len);
		cpy(tmp2, b, i>>1, Len);
		NTT(tmp1, 1, Len);
		NTT(tmp2, 1, Len);
		for (register int j = 0; j < Len; j++)
			tmp1[j] = ((2LL*tmp2[j])%mod + mod - (((tmp2[j]*tmp2[j])%mod)*tmp1[j])%mod) % mod;
		NTT(tmp1, -1, Len);
		for (register int j = 0; j < i; j++)
			b[j] = tmp1[j];
	}
}

LL inv2;

LL tmp3[N], tmp4[N], tmp5[N];

void Sqrt(LL * a, LL * b, int l)
{	b[0] = 1;
	for (register int i = 1; i <= l; i <<= 1)
	{	int Len = i << 1;
		for (register int j = 0; j < Len; j++)
			if (j < i) tmp3[j] = (2LL * b[j]) % mod;
			else tmp3[j] = 0;
		get_Inv(tmp3, tmp4, i);
		for (register int j = i; j < Len; j++) tmp4[j] = 0;
		cpy(tmp5, a, i, Len);
		NTT(tmp4, 1, Len);
		NTT(tmp5, 1, Len);
		for (register int j = 0; j < Len; j++)
			tmp4[j] = (tmp4[j] * tmp5[j]) % mod;
		NTT(tmp4, -1, Len);
		for (register int j = 0; j < (i << 1); j++)
			b[j] = (inv2 * b[j] + tmp4[j]) % mod;
	}
}

LL a[N], b[N], ans[N];

LL v[N];

int main()
{	int n, m;
	n = rd_int(), m = rd_int();
	m++;
	for (register int i = 1; i <= n; i++)
	{	int x;
		x = rd_int();
		v[x]++;
	}
	inv2 = power(2, mod - 2);
	getBit(m);
	a[0] = 1;
	for (register int i = 1; i < m; i++)
		a[i] = (4 * (mod - v[i])) % mod;
	Sqrt(a, b, (1 << Bit));
	b[0] = (b[0] + 1) % mod;
	get_Inv(b, ans, (1 << Bit));
	for (register int i = 1; i < m; i++)
		printf("%lld\n", (ans[i] * 2LL) % mod);
	return 0;
}
posted @ 2018-07-05 22:23  EZ_WYC  阅读(...)  评论(...编辑  收藏