有标号的DAG计数系列问题
II
设 \(f_i\) 表示 \(i\) 个点的答案
那么枚举至少 \(j\) 个点的出度为 \(0\)
\[\sum_{j=0}^{i}(-1)^j\binom{i}{j}f_{i-j}2^{(i-j)j}=0
\]
所以
\[f_i=\sum_{j=1}^{i}(-1)^{j+1}\binom{i}{j}f_{i-j}2^{(i-j)j}
\]
即
\[\frac{f_i}{i!}=\sum_{j=0}^{i-1}\frac{f_j}{j!}\frac{(-1)^{i-j+1}}{(i-j)!}2^{(i-j)j}
\]
现在的难处在于 \(2^{(i-j)j}\) 的拆分
可以发现
\[2^{(i-j)j}=\frac{2^{\frac{i^2}{2}}}{2^{\frac{(i-j)^2}{2}}2^{\frac{j^2}{2}}}
\]
所以
\[\frac{f_i}{i!\sqrt{2}^{i^2}}=\sum_{j=0}^{i-1}\frac{f_j}{j!\sqrt{2}^{j^2}}\frac{(-1)^{i-j+1}}{(i-j)!\sqrt{2}^{(i-j)^2}}
\]
求出 \(2\) 在模 \(998244353\) 意义下的二次剩余然后多项式求逆即可
二次剩余没有学QwQ直接暴力预处理出来即可
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn(4e5 + 5);
const int mod(998244353);
const int res2(116195171);
const int invres2(557219762);
inline void Inc(int &x, int y) {
if ((x += y) >= mod) x -= mod;
}
inline int Pow(ll x, int y) {
register ll ret = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ret = ret * x % mod;
return ret;
}
int a[maxn], b[maxn], w[2][maxn], deg, r[maxn], l;
inline void Init(int n) {
register int i, k, wn, iwn;
for (deg = 1, l = 0; deg < n; deg <<= 1) ++l;
for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
for (i = 1; i < deg; i <<= 1) {
w[0][0] = w[1][0] = 1;
wn = Pow(3, (mod - 1) / (i << 1)), iwn = Pow(wn, mod - 2);
for (k = 1; k < i; ++k) {
w[0][deg / i * k] = 1LL * w[0][deg / i * (k - 1)] * wn % mod;
w[1][deg / i * k] = 1LL * w[1][deg / i * (k - 1)] * iwn % mod;
}
}
}
inline void NTT(int *p, int opt) {
register int i, j, k, t, wn, x, y;
for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
for (i = 1; i < deg; i <<= 1)
for(t = i << 1, j = 0; j < deg; j += t)
for (k = 0; k < i; ++k) {
wn = w[opt == -1][deg / i * k];
x = p[j + k], y = 1LL * wn * p[i + j + k] % mod;
p[j + k] = x + y, p[i + j + k] = x - y;
if (p[j + k] >= mod) p[j + k] -= mod;
if (p[i + j + k] < 0) p[i + j + k] += mod;
}
if (opt == -1) {
wn = Pow(deg, mod - 2);
for (i = 0; i < deg; ++i) p[i] = 1LL * p[i] * wn % mod;
}
}
int n, f[maxn], g[maxn], fac[maxn], ifac[maxn];
void Inv(int *p, int *q, int len) {
if (len == 1) {
q[0] = Pow(p[0], mod - 2);
return;
}
Inv(p, q, len >> 1);
register int i, tmp = len << 1;
for (i = 0; i < len; ++i) a[i] = p[i], b[i] = q[i];
Init(tmp), NTT(a, 1), NTT(b, 1);
for (i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * b[i] % mod * b[i] % mod;
NTT(a, -1);
for (i = 0; i < len; ++i) q[i] = (2LL * q[i] + mod - a[i]) % mod;
for (i = 0; i < tmp; ++i) a[i] = b[i] = 0;
}
int main() {
register int i, len;
scanf("%d", &n), f[0] = fac[0] = ifac[0] = ifac[1] = 1;
for (i = 2; i <= n; ++i) ifac[i] = 1LL * (mod - mod / i) * ifac[mod % i] % mod;
for (i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod, ifac[i] = 1LL * ifac[i - 1] * ifac[i] % mod;
for (i = 0; i <= n; ++i) {
f[i] = 1LL * ifac[i] * Pow(invres2, 1LL * i * i % (mod - 1)) % mod;
if (i & 1) f[i] = mod - f[i];
}
for (len = 1; len <= n; len <<= 1);
Inv(f, g, len);
printf("%lld\n", 1LL * fac[n] * Pow(res2, 1LL * n * n % (mod - 1)) % mod * g[n] % mod);
return 0;
}
IV
设 \(f_i\) 表示不要求连通的 \(i\) 个点 的 \(DAG\) 的方案数
直接 \(II\) 那样求出来
设 \(g_i\) 表示 \(i\) 个点的答案
方法一
钦定第一个连通块的一个点
\[g_i=f_i-\sum_{j=1}^{i-1}g_jf_{i-j}\binom{i-1}{j-1}
\]
组合数拆整理得到
\[\frac{f_i}{(i-1)!}=\sum_{j=1}^{i}\frac{g_j}{(i-1)!}\frac{f_{i-j}}{(i-j)!}
\]
直接多项式求逆后卷起来就好了
方法二
设
\[F(x)=\sum_{i=0}^{\infty}f_i\frac{x^i}{i!}
\]
\[G(x)=\sum_{i=0}^{\infty}g_i\frac{x^i}{i!}
\]
那么
\[F(x)=e^{G(x)},G(x)=ln F(x)
\]
方法一
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn(4e5 + 5);
const int mod(998244353);
const int res2(116195171);
const int invres2(557219762);
inline void Inc(int &x, int y) {
if ((x += y) >= mod) x -= mod;
}
inline int Pow(ll x, int y) {
register ll ret = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ret = ret * x % mod;
return ret;
}
int a[maxn], b[maxn], w[2][maxn], deg, r[maxn], l;
inline void Init(int n) {
register int i, k, wn, iwn;
for (deg = 1, l = 0; deg < n; deg <<= 1) ++l;
for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
for (i = 1; i < deg; i <<= 1) {
w[0][0] = w[1][0] = 1;
wn = Pow(3, (mod - 1) / (i << 1)), iwn = Pow(wn, mod - 2);
for (k = 1; k < i; ++k) {
w[0][deg / i * k] = 1LL * w[0][deg / i * (k - 1)] * wn % mod;
w[1][deg / i * k] = 1LL * w[1][deg / i * (k - 1)] * iwn % mod;
}
}
}
inline void NTT(int *p, int opt) {
register int i, j, k, t, wn, x, y;
for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
for (i = 1; i < deg; i <<= 1)
for(t = i << 1, j = 0; j < deg; j += t)
for (k = 0; k < i; ++k) {
wn = w[opt == -1][deg / i * k];
x = p[j + k], y = 1LL * wn * p[i + j + k] % mod;
p[j + k] = x + y, p[i + j + k] = x - y;
if (p[j + k] >= mod) p[j + k] -= mod;
if (p[i + j + k] < 0) p[i + j + k] += mod;
}
if (opt == -1) {
wn = Pow(deg, mod - 2);
for (i = 0; i < deg; ++i) p[i] = 1LL * p[i] * wn % mod;
}
}
int n, f[maxn], g[maxn], h[maxn], fac[maxn], ifac[maxn];
void Inv(int *p, int *q, int len) {
if (len == 1) {
q[0] = Pow(p[0], mod - 2);
return;
}
Inv(p, q, len >> 1);
register int i, tmp = len << 1;
for (i = 0; i < len; ++i) a[i] = p[i], b[i] = q[i];
Init(tmp), NTT(a, 1), NTT(b, 1);
for (i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * b[i] % mod * b[i] % mod;
NTT(a, -1);
for (i = 0; i < len; ++i) q[i] = (2LL * q[i] + mod - a[i]) % mod;
for (i = 0; i < tmp; ++i) a[i] = b[i] = 0;
}
int main() {
register int i, len;
scanf("%d", &n), f[0] = fac[0] = ifac[0] = ifac[1] = 1;
for (i = 2; i <= n; ++i) ifac[i] = 1LL * (mod - mod / i) * ifac[mod % i] % mod;
for (i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod, ifac[i] = 1LL * ifac[i - 1] * ifac[i] % mod;
for (i = 0; i <= n; ++i) {
f[i] = 1LL * ifac[i] * Pow(invres2, 1LL * i * i % (mod - 1)) % mod;
if (i & 1) f[i] = mod - f[i];
}
for (len = 1; len <= n; len <<= 1);
Inv(f, g, len);
for (i = 0; i <= n; ++i) g[i] = 1LL * g[i] * Pow(res2, 1LL * i * i % (mod - 1)) % mod;
for (memset(f, 0, sizeof(f)), i = n + 1; i < len; ++i) g[i] = 0;
for (i = 1; i <= n; ++i) f[i] = 1LL * g[i] * i % mod;
Inv(g, h, len), len <<= 1;
for (i = 0; i < len; ++i) a[i] = f[i], b[i] = h[i];
Init(len), NTT(a, 1), NTT(b, 1);
for (i = 0; i < len; ++i) a[i] = 1LL * a[i] * b[i] % mod;
NTT(a, -1);
printf("%lld\n", 1LL * a[n] * fac[n - 1] % mod);
return 0;
}
