题意
略。（语文太差）
题解
首先一个结论：随机一个([0, 1])之间的实数序列，只用到各个位置相互之间的大小关系，每种关系出现的概率等同于随机一个([1, n])的排列。
原因是出现某些位置值相同的概率是无穷小，并且只有有限种情况出现相同的值，因此可以忽略。
然后在草稿纸上画一画，就知道如果(p_1, p_2, ldots, p_x in [n - x + 1, n])，那么(1, 2, ldots, x)和(x + 1, x + 2, ldots, n)之间必然没有边。
考虑dp。但是如果直接dp答案会有点难，所以设(f_n)表示长度为(n)的序列，不存在一个(x)，使得(1, 2, ldots, x)和(x + 1, x + 2, ldots, n)没有边的不同序列数，并规定(f_0 = 1)。
考虑补集转化，有
[f_n = n! - sum_{i = 1} ^ {n - 1} f_{n - i} i! \
2n! = sum_{i = 0} ^ {n} f_{n - i} i! + [n = 0] \
]
考虑其生成函数
[F(x) = sum_{i geq 0} f_i x ^ i
]
与一个辅助生成函数
[G(x) = sum_{i geq 0} i! x ^ i
]
则
[2G(x) = G(x)F(x) + 1 \
F(x) = frac{2G(x) - 1}{G(x)} \
]
这可以用多项式求逆求出。
考虑答案。设(H(x) = sum_{i geq 0} i f_i x ^ i)（此时(f_i)已求出），则有
[ans = [x ^ n] sum_{k geq 0} H(X) ^ k = [x ^ n] frac{1}{1 - H(x)}
]
再做一次多项式求逆即可。
复杂度(mathcal O(n log n))。
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector <int> poly;
const int N = 1 << 20, mod = 998244353, G = 3;
const ll infty = 1ll * mod * mod;
int power (int a, int b) {
int ret = 1;
if (b < 0) {
b += mod - 1;
}
for ( ; b; b >>= 1, a = 1ll * a * a % mod) {
if (b & 1) {
ret = 1ll * ret * a % mod;
}
}
return ret;
}
namespace {
void printp (poly a) {
if (!a.size()) {
return;
}
printf("%d", a[0]);
for (int i = 1; i < (int)a.size(); ++i) {
printf(" %d", a[i]);
}
putchar('
');
}
int adjust (int n) {
int ret = 1;
for ( ; ret < n; ret <<= 1);
return ret;
}
poly trans (int n, int *a) {
int m = adjust(n);
poly ret; ret.resize(m, 0);
for (int i = 0; i < n; ++i) {
ret[i] = a[i];
}
return ret;
}
void dnt (int n, poly &_a) {
static int rev[N], a[N], wi[N];
for (int i = 0; i < n; ++i) {
rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? n >> 1 : 0);
a[i] = _a[rev[i]];
}
for (int l = 2, _w; l <= n; l <<= 1) {
_w = power(G, (mod - 1) / l), wi[l >> 1] = 1;
for (int i = (l >> 1) + 1; i < l; ++i) {
wi[i] = 1ll * wi[i - 1] * _w % mod;
}
}
for (int l = 2, m = 1; l <= n; m = l, l <<= 1) {
for (int i = 0; i < n; i += l) {
int *u = a + i, *v = a + i + m, *w = wi + m;
for (int j = 0, x, y; j < m; ++u, ++v, ++w, ++j) {
x = *u, y = 1ll * (*v) * (*w) % mod;
*u = (x + y) % mod, *v = (x - y + mod) % mod;
}
}
}
for (int i = 0; i < n; ++i) {
_a[i] = a[i];
}
}
void idnt (int n, poly &_a) {
static int rev[N], a[N], wi[N];
for (int i = 0; i < n; ++i) {
rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? n >> 1 : 0);
a[i] = _a[rev[i]];
}
for (int l = 2, _w; l <= n; l <<= 1) {
_w = power(G, mod - 1 - (mod - 1) / l), wi[l >> 1] = 1;
for (int i = (l >> 1) + 1; i < l; ++i) {
wi[i] = 1ll * wi[i - 1] * _w % mod;
}
}
for (int l = 2, m = 1; l <= n; m = l, l <<= 1) {
for (int i = 0; i < n; i += l) {
int *u = a + i, *v = a + i + m, *w = wi + m;
for (int j = 0, x, y; j < m; ++u, ++v, ++w, ++j) {
x = *u, y = 1ll * (*v) * (*w) % mod;
*u = (x + y) % mod, *v = (x - y + mod) % mod;
}
}
}
int invn = power(n, mod - 2);
for (int i = 0; i < n; ++i) {
_a[i] = 1ll * a[i] * invn % mod;
}
}
poly conv (int n, poly a, poly b, int f = 0) {
a.resize(n, 0), b.resize(n, 0);
n <<= 1, a.resize(n, 0), b.resize(n, 0);
dnt(n, a), dnt(n, b);
for (int i = 0; i < n; ++i) {
a[i] = 1ll * a[i] * b[i] % mod;
if (f) {
a[i] = 1ll * a[i] * b[i] % mod;
}
}
idnt(n, a);
return a;
}
poly plu (int n, poly a, poly b) {
for (int i = 0; i < n; ++i) {
if ((a[i] += b[i]) >= mod) {
a[i] -= mod;
}
}
return a;
}
poly minus (int n, poly a, poly b) {
for (int i = 0; i < n; ++i) {
if ((a[i] -= b[i]) < 0) {
a[i] += mod;
}
}
return a;
}
poly inv (int n, poly f) {
if (n == 1) {
return (poly) {power(f[0], mod - 2)};
}
f.resize(n);
poly g = inv(n >> 1, f);
g.resize(n, 0);
f = conv(n, f, g, 1);
g = minus(n, plu(n, g, g), f);
return g.resize(n), g;
}
}
int fac[N], c[N];
poly A, B, C;
int solve (int n) {
fac[0] = 1;
for (int i = 1; i <= n; ++i) {
fac[i] = 1ll * fac[i - 1] * i % mod;
}
A = trans(n, fac);
B = plu(A.size(), A, A), B[0] = (B[0] - 1 + mod) % mod;
A = inv(A.size(), A);
B = conv(B.size(), B, A);
for (int i = 0; i < n; ++i) {
c[i] = (mod - 1ll * i * B[i] % mod) % mod;
}
c[0] = (c[0] + 1) % mod;
C = trans(n, c);
C = inv(C.size(), C);
return C[n - 1];
}
int n;
int main () {
cin >> n;
cout << solve(n + 1) << endl;
return 0;
}