Solution -「ZROI 1762」规划

标签：--   init   max   otherwise   ``   连接   有根树   ==   生成   

# $\mathcal{Description}$

  [Link](http://zhengruioi.com/problem/1761).

  （貌似未氪金玩家不能看题的嗷……如果有不良影响私信我就好√）

  把 $n$ 个有标号的点划分为若干个集合，接着从每个集合里选一个关键点，然后把所有关键点连成一棵有根树。大小为 $k$ 的集合的权值为 $v_k$，方案的权值为集合权值之积。对于 $t\in[1,N]$，分别求 $n=t$ 时所有方案的贡献之和，答案对 $998244353$ 取模。

  $N\le10^5$。

# $\mathcal{Solution}$

  首先，有标号有根树的 EGF 为：
$$
F(x)=\sum_{1\le i}\frac{i^{i-1}}{i!}x^i
$$
  而本题中，树上的一个结点实则为选出的关键点，也即对应着一个点集。不难写出点集的 EGF：
$$
G(x)=\sum_{1\le i}\frac{i\cdot v_i}{i!}x^i
$$
  其中 $i\cdot v_i$ 表示大小为 $i$ 的集合的权值为 $v_i$，且要从 $i$ 个点里选一个关键点，所以再乘上 $i$。

  那么，答案的 EGF 为 $F(G(x))$。直接求复合逆过不了（而且我不会 qwq），考虑 $F(x)$ 的性质：
$$
F(x)=xe^{F(x)}
$$
  组合意义：生成一片有根树森林，用一个真正的根连接每棵树根形成最终的树。

  代入 $G(x)$：
$$
F(G(x))=G(x)e^{F(G(x))}
$$
  相当于要求 $f(u,x)=u-G(x)e^u=0$ 的解 $u$。直接 Newton 迭代，令
$$
u_n-G(x)e^u\equiv0\pmod{x^n}
$$
  根据迭代公式，有
$$
\begin{aligned}u_{2n}&\equiv u_n-\frac{f(u_n,x)}{f_u‘(u_n,x)}\\&\equiv u_n-\frac{u_n-G(x)e^{u_n}}{1-G(x)e^{u_n}}\pmod{x^{2n}}\end{aligned}
$$
  $\mathcal O(n\log n)$ 求出来就行。~~别人说卡常，反正我写出来过得挺悠闲 owo。~~

## $\mathcal{Code}$

```cpp
/* Clearink */

#include <cmath>
#include <cstdio>
#include <algorithm>

inline char fgc () {
	static char buf[1 << 17], *p = buf, *q = buf;
	return p == q && ( q = buf + fread ( p = buf, 1, 1 << 17, stdin ), p == q ) ?
		EOF : *p++;
}

inline int rint () {
	int x = 0, s = fgc ();
	for ( ; s < ‘0‘ || ‘9‘ < s; s = fgc () );
	for ( ; ‘0‘ <= s && s <= ‘9‘; s = fgc () ) x = x * 10 + ( s ^ ‘0‘ );
	return x;
}

template<typename Tp>
inline void wint ( Tp x ) {
	if ( x < 0 ) putchar ( ‘-‘ ), x = -x;
	if ( 9 < x ) wint ( x / 10 );
	putchar ( x % 10 ^ ‘0‘ );
}

const int MOD = 998244353, MAXN = 1 << 18;
int n, fac[MAXN + 5], ifac[MAXN + 5], G[MAXN + 5], H[MAXN + 5];

inline int mul ( const long long a, const int b ) { return a * b % MOD; }
inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline int mpow ( int a, int b ) {
	int ret = 1;
	for ( ; b; a = mul ( a, a ), b >>= 1 ) ret = mul ( ret, b & 1 ? a : 1 );
	return ret;
}

inline void init ( const int n ) {
	fac[0] = 1;
	for ( int i = 1; i <= n; ++i ) fac[i] = mul ( i, fac[i - 1] );
	ifac[n] = mpow ( fac[n], MOD - 2 );
	for ( int i = n - 1; ~i; --i ) ifac[i] = mul ( i + 1, ifac[i + 1] );
}

namespace PolyOper {

/*
	In operations like `func(n,A,m,B,R)`, R could be A or B.
	Otherwise, R can NOT be A.
*/

const int G = 3;
int inv[MAXN + 5], omega[19][MAXN + 5];

inline void init () {
	inv[1] = 1;
	for ( int i = 2; i <= MAXN; ++i ) {
		inv[i] = mul ( MOD - MOD / i, inv[MOD % i] );
	}
	for ( int i = 0; i < 19; ++i ) {
		int* omegaI = omega[i];
		omegaI[0] = 1;
		int& o1 = omegaI[1] = mpow ( G, ( MOD - 1 ) / ( 1 << i << 1 ) );
		for ( int j = 2; j < 1 << i; ++j ) omegaI[j] = mul ( omegaI[j - 1], o1 );
	}
}

inline void NTT ( const int n, int* A, const int type ) {
	static int rev[MAXN + 5], las = -1;
	if ( las != n ) {
		las = n, rev[0] = 1;
		int lgn = log ( n ) / log ( 2 ) + 0.5;
		for ( int i = 1; i < n; ++i ) {
			rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << lgn >> 1 );
		}
	}
	for ( int i = 1; i < n; ++i ) {
		if ( i < rev[i] ) {
			A[i] ^= A[rev[i]] ^= A[i] ^= A[rev[i]];
		}
	}
	for ( int i = 0, stp = 1; stp < n; ++i, stp <<= 1 ) {
		const int* omegaI = omega[i];
		for ( int j = 0; j < n; j += stp << 1 ) {
			for ( int k = j; k < j + stp; ++k ) {
				int ev = A[k], ov = mul ( omegaI[k - j], A[k + stp] );
				A[k] = add ( ev, ov ), A[k + stp] = sub ( ev, ov );
			}
		}
	}
	if ( type == -1 ) {
		for ( int invn = MOD - ( MOD - 1 ) / n, i = 0; i < n; ++i ) {
			A[i] = mul ( A[i], invn );
		}
		std::reverse ( A + 1, A + n );
	}
}

inline void polyConv ( const int n, const int* A,
	const int m, const int* B, int* R ) {
	static int tmp[2][MAXN + 5];
	int len = 1, lg = 0;
	for ( ; len < n + m - 1; len <<= 1, ++lg );
	for ( int i = 0; i < len; ++i ) {
		tmp[0][i] = i < n ? A[i] : 0, tmp[1][i] = i < m ? B[i] : 0;
	}
	NTT ( len, tmp[0], 1 ), NTT ( len, tmp[1], 1 );
	for ( int i = 0; i < len; ++i ) tmp[0][i] = mul ( tmp[0][i], tmp[1][i] );
	NTT ( len, tmp[0], -1 );
	for ( int i = 0; i < n + m - 1; ++i ) R[i] = tmp[0][i];
}

inline void polyDir ( const int n, const int* A, int* R ) {
	for ( int i = 1; i < n; ++i ) R[i - 1] = mul ( i, A[i] );
	R[n - 1] = 0;
}

inline void polyInt ( const int n, const int* A, int* R ) {
	for ( int i = n - 1; ~i; --i ) R[i + 1] = mul ( inv[i + 1], A[i] );
	R[0] = 0;
}

inline void polyInv ( const int n, const int* A, int* R ) {
	static int tmp[2][MAXN + 5] {};
	if ( n == 1 ) return void ( R[0] = mpow ( A[0], MOD - 2 ) );
	polyInv ( n >> 1, A, R );
	for ( int i = 0; i < n; ++i ) tmp[0][i] = A[i], tmp[1][i] = R[i];
	NTT ( n << 1, tmp[0], 1 ), NTT ( n << 1, tmp[1], 1 );
	for ( int i = 0; i < n << 1; ++i ) {
		tmp[0][i] = mul ( mul ( tmp[0][i], tmp[1][i] ), tmp[1][i] );
	}
	NTT ( n << 1, tmp[0], -1 );
	for ( int i = 0; i < n; ++ i ) R[i] = sub ( mul ( 2, R[i] ), tmp[0][i] );
	for ( int i = 0; i < n << 1; ++ i ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyLn ( const int n, const int* A, int* R ) {
	static int tmp[2][MAXN + 5] {};
	polyDir ( n, A, tmp[0] ), polyInv ( n, A, tmp[1] );
	NTT ( n << 1, tmp[0], 1 ), NTT ( n << 1, tmp[1], 1 );
	for ( int i = 0; i < n << 1; ++i ) tmp[0][i] = mul ( tmp[0][i], tmp[1][i] );
	NTT ( n << 1, tmp[0], -1 ), polyInt ( n << 1, tmp[0], R );
	for ( int i = 0; i < n << 1; ++i ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyExp ( const int n, const int* A, int* R ) {
	static int tmp[MAXN + 5] {};
	if ( n == 1 ) return void ( R[0] = 1 );
	polyExp ( n >> 1, A, R ), polyLn ( n, R, tmp );
	tmp[0] = sub ( add ( A[0], 1 ), tmp[0] );
	for ( int i = 1; i < n; ++i ) tmp[i] = sub ( A[i], tmp[i] );
	NTT ( n << 1, tmp, 1 ), NTT ( n << 1, R, 1 );
	for ( int i = 0; i < n << 1; ++i ) R[i] = mul ( R[i], tmp[i] );
	NTT ( n << 1, R, -1 );
	for ( int i = n; i < n << 1; ++i ) R[i] = tmp[i] = 0;
}

} // namespace PolyOper.

inline void solve ( const int n, const int* A, int* R ) {
	if ( n == 1 ) return void ( R[0] = A[0] );
	solve ( n >> 1, A, R );
	static int tmp[2][MAXN + 5] {};
	int len = 1; for ( ; len <= n; len <<= 1 );
	PolyOper::polyExp ( n, R, tmp[0] );
	PolyOper::polyConv ( n, G, n, tmp[0], tmp[0] );
	for ( int i = 0; i < n; ++i ) tmp[0][i] = sub ( 0, tmp[0][i] );
	tmp[0][0] = add ( tmp[0][0], 1 );
	PolyOper::polyInv ( n, tmp[0], tmp[1] );
	tmp[0][0] = sub ( tmp[0][0], 1 );
	for ( int i = 0; i < n; ++i ) tmp[0][i] = add ( tmp[0][i], R[i] );
	PolyOper::polyConv ( n, tmp[0], n, tmp[1], tmp[0] );
	for ( int i = 0; i < n; ++i ) R[i] = sub ( R[i], tmp[0][i] );
	for ( int i = 0; i < len; ++i ) tmp[0][i] = tmp[1][i] = 0;
}

int main () {
	PolyOper::init (), init ( n = rint () );
	for ( int i = 1; i <= n; ++i ) G[i] = mul ( ifac[i - 1], rint () );
	int len = 1; for ( ; len <= n; len <<= 1 );
	solve ( len, G, H );
	for ( int i = 1; i <= n; ++i ) {
		wint ( mul ( fac[i], H[i] ) );
		putchar ( i ^ n ? ‘ ‘ : ‘\n‘ );
	}
	return 0;
}

Solution -「ZROI 1762」规划

标签：--   init   max   otherwise   ``   连接   有根树   ==   生成   

原文地址：https://www.cnblogs.com/rainybunny/p/14391236.html