(mathcal{Description})
link.
求包含 (n) 个点的边双连通图的个数。
(nle10^5)。
(mathcal{Solution})
类似于这道题，仍令 (D(x)) 为有根无向连通图的 ( ext{EGF})，(B(x)) 为边双连通图的 ( ext{EGF})，考虑用 (B) 表示 (D)。显然，根仅存在于一个边双连通分量中。我们可以在这个连通分量的任意一个点上挂一个有根连通图，这显然不会影响当前的边双，( ext{EGF}) 为 (nD(x))（(n) 为根所在边双大小。）事实上可以挂许多图，所以处根所在边双以外的 ( ext{EGF}) 为 (expleft(nD(x) ight))。枚举根所在边双大小，就有：
[D(x)=sum_{i=1}^{+infty}frac{b_ix^iexpleft(iD(x)
ight)}{i!}=Bleft(xexp D(x)
ight)
]
令 (F(x)=xexp D(x))，(F^{-1}(x)) 为其复合逆。将 (F^{-1}(x)) 代入上式得：
[B(x)=Dleft(F^{-1}(x)
ight)
]
扩展拉格朗日反演：
[[x^n]B(x)=frac{1}n[x^{n-1}]D'(x)left(frac{x}{F(x)}
ight)^n
]
重新展开 (F)，化简得：
[[x^n]B(x)=frac{1}n[x^{n-1}]D'(x)expleft(-nD(x)
ight)
]
求出 (D)，继而求出 ([x^n]B(x))。复杂度 (mathcal O(nlog n))。
(mathcal{Code})
#include <cmath>
#include <cstdio>
const int MAXN = 1 << 18, MOD = 998244353;
int n, fac[MAXN + 5], ifac[MAXN + 5], inv[MAXN + 5];
int F[MAXN + 5], G[MAXN + 5], H[MAXN + 5], T[MAXN + 5];
inline int qkpow ( int a, int b, const int p = MOD ) {
int ret = 1;
for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = 1ll * ret * ( b & 1 ? a : 1 ) % p;
return ret;
}
namespace Poly {
const int G = 3;
inline void NTT ( const int n, int* A, const int tp ) {
static int lstn = -1, rev[MAXN + 5] {};
if ( lstn ^ n ) {
int lgn = log ( n ) / log ( 2 ) + 0.5;
for ( int i = 0; i < n; ++ i ) rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << lgn >> 1 );
lstn = n;
}
for ( int i = 0; i < n; ++ i ) if ( i < rev[i] ) A[i] ^= A[rev[i]] ^= A[i] ^= A[rev[i]];
for ( int i = 2, stp = 1; i <= n; i <<= 1, stp <<= 1 ) {
int w = qkpow ( G, ( MOD - 1 ) / i );
if ( ! ~ tp ) w = qkpow ( w, MOD - 2 );
for ( int j = 0; j < n; j += i ) {
for ( int k = j, r = 1; k < j + stp; ++ k, r = 1ll * r * w % MOD ) {
int ev = A[k], ov = 1ll * r * A[k + stp] % MOD;
A[k] = ( ev + ov ) % MOD, A[k + stp] = ( ev - ov + MOD ) % MOD;
}
}
}
if ( ! ~ tp ) for ( int i = 0; i < n; ++ i ) A[i] = 1ll * A[i] * inv[n] % MOD;
}
inline void polyDer ( const int n, const int* A, int* R ) {
for ( int i = 1; i < n; ++ i ) R[i - 1] = 1ll * i * A[i] % MOD;
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] = 1ll * inv[i + 1] * A[i] % MOD;
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] = qkpow ( 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] = 1ll * tmp[0][i] * tmp[1][i] % MOD * tmp[1][i] % MOD;
NTT ( n << 1, tmp[0], -1 );
for ( int i = 0; i < n; ++ i ) R[i] = ( 2ll * R[i] % MOD - tmp[0][i] + MOD ) % MOD;
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] {};
polyDer ( 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] = 1ll * tmp[0][i] * tmp[1][i] % MOD;
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] = ( A[0] + 1 - tmp[0] + MOD ) % MOD;
for ( int i = 1; i < n; ++ i ) tmp[i] = ( A[i] - tmp[i] + MOD ) % MOD;
NTT ( n << 1, tmp, 1 ), NTT ( n << 1, R, 1 );
for ( int i = 0; i < n << 1; ++ i ) R[i] = 1ll * R[i] * tmp[i] % MOD;
NTT ( n << 1, R, -1 );
for ( int i = n; i < n << 1; ++ i ) R[i] = tmp[i] = 0;
}
} // namespace Poly.
inline void init () {
int len = MAXN >> 1;
inv[1] = fac[0] = ifac[0] = fac[1] = ifac[1] = 1;
for ( int i = 2; i <= MAXN; ++ i ) {
fac[i] = 1ll * i * fac[i - 1] % MOD;
inv[i] = 1ll * ( MOD - MOD / i ) * inv[MOD % i] % MOD;
ifac[i] = 1ll * inv[i] * ifac[i - 1] % MOD;
}
for ( int i = 0; i < len; ++ i ) F[i] = 1ll * qkpow ( 2, ( i * ( i - 1ll ) >> 1 ) % ( MOD - 1 ) ) * ifac[i] % MOD;
Poly::polyLn ( len, F, G );
for ( int i = 0; i < len; ++ i ) G[i] = 1ll * G[i] * i % MOD;
Poly::polyDer ( len, G, H ), Poly::NTT ( MAXN, H, 1 );
}
inline void solve () {
int len = MAXN >> 1;
if ( n == 1 ) return void ( puts ( "1" ) );
for ( int i = 0; i < MAXN; ++ i ) F[i] = T[i] = 0;
for ( int i = 0; i < len; ++ i ) F[i] = 1ll * ( MOD - n ) % MOD * G[i] % MOD;
Poly::polyExp ( len, F, T ), Poly::NTT ( MAXN, T, 1 );
for ( int i = 0; i < MAXN; ++ i ) F[i] = 1ll * T[i] * H[i] % MOD;
Poly::NTT ( MAXN, F, -1 );
printf ( "%d
", int ( 1ll * inv[n] * fac[n - 1] % MOD * F[n - 1] % MOD ) );
}
int main () {
init ();
for ( int i = 1; i <= 5; ++ i ) scanf ( "%d", &n ), solve ();
return 0;
}