P5219 无聊的水题 I
计有标号树,容易想到 \(\text{Prufer}\) 序列,那么对于度数的限制即使,每一个数的出现次数要小于等于\(m - 1\),且一定要有等于的,容斥一下,用小于等于 \(m-1\) 答案减去小于等于 \(m - 2\) 即可。
对于一种数 \(i\),设其\(EGF\)为
那么答案即为
\[(n - 2)![x^{n - 2}]F^n(x) \]多项式快速幂即可,时间复杂度\(O(n\log n)\)
Code
#include<cstdio>
#include<cstring>
#include<iostream>
#define IN inline
#define LL long long
using namespace std;
const int N = 1e5 + 5, P = 998244353, G = 3;
int n, m, a[N << 2], b[N << 2], inv[N << 2];
IN int read() {
int t = 0,res = 0; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) t |= (ch == '-');
for (; isdigit(ch); ch = getchar()) res = (res << 3) + (res << 1) + (ch ^ 48);
return t ? -res : res;
}
LL fpow(LL x, LL y) {
LL res = 1;
for (; x; x >>= 1, y = y * y % P)
if (x & 1) res = res * y % P;
return res;
}
const int invG = fpow(P - 2, G);
namespace Poly{
int rev[N << 2], insF[N << 2], sginv[N << 2];
void NTT(int *f, int len, int fl) {
static int W[N << 2] = {1};
if (len == 1) return;
for (int i = 0; i < len; i++)
if (i < rev[i]) swap(f[i], f[rev[i]]);
for (int l = 1; l < len; l <<= 1) {
int I = fpow((P - 1) / (l << 1), fl == 1 ? G : invG);
for (int i = 1; i < l; i++) W[i] = (LL)W[i - 1] * I % P;
for (int i = 0; i < len; i += (l << 1))
for (int j = 0; j < l; j++) {
int x = f[i | j], y = (LL)f[i | j | l] * W[j] % P;
f[i | j] = (x + y >= P ? x + y - P : x + y);
f[i | j | l] = (x - y < 0 ? x - y + P : x - y);
}
}
if (fl == -1) {
int IV = fpow(P - 2, len);
for (int i = 0; i < len; i++) f[i] = (LL)f[i] * IV % P;
}
}
void Mulpoly(int *f, int *g, int lenF, int lenG) {
int len = 1, bit = 0;
while (len <= lenF + lenG) len <<= 1, bit++;
for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
for (int i = lenF + 1; i < len; i++) f[i] = 0;
for (int i = lenG + 1; i < len; i++) g[i] = 0;
NTT(f, len, 1), NTT(g, len, 1);
for (int i = 0; i < len; i++) f[i] = (LL)f[i] * g[i] % P;
NTT(f, len, -1);
for (int i = lenF + lenG + 1; i < len; i++) f[i] = 0;
}
void Invpoly(int *f, int lenF, int *g) { // [0, lenF);
static int s2[N << 2], s1[N << 2];
int limit = 1; while (limit < lenF) limit <<= 1;
g[0] = fpow(P - 2, f[0]);
for (int len = 2, bit = 1; len <= limit; len <<= 1, bit++) {
for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
for (int i = 0; i < (len >> 1); i++) s2[i] = g[i];
for (int i = (len >> 1); i < len; i++) s2[i] = 0;
for (int i = 0; i < len; i++) s1[i] = f[i];
NTT(s2, len, 1), NTT(s1, len, 1);
for (int i = 0; i < len; i++) s1[i] = (LL)s2[i] * s1[i] % P;
NTT(s1, len, -1);
for (int i = 0; i < (len >> 1); i++) s1[i] = 0;
NTT(s1, len, 1);
for (int i = 0; i < len; i++) s1[i] = (LL)s1[i] * s2[i] % P;
NTT(s1, len, -1);
for (int i = (len >> 1); i < len; i++) g[i] = (s1[i] == 0 ? 0 : P - s1[i]);
}
for (int i = lenF; i < limit; i++) g[i] = 0;
for (int i = 0; i < limit; i++) s2[i] = s1[i] = 0;
}
void Getsginv(int len) {
sginv[0] = sginv[1] = 1;
for (int i = 2; i <= len; i++) sginv[i] = (LL)sginv[P % i] * (P - P / i) % P;
}
void Dpoly(int *f, int lenF) {
for (int i = 0; i < lenF; i++) f[i] = (LL)f[i + 1] * (i + 1) % P;
f[lenF] = 0;
}
void Jfpoly(int *f, int lenF) {
for (int i = lenF; i >= 0; i--) f[i + 1] = (LL)f[i] * sginv[i + 1] % P;
f[0] = 0;
}
void Lnpoly(int *f, int lenF) { // [0, lenF)
static int Inv_f[N << 2];
Invpoly(f, lenF, Inv_f), Dpoly(f, lenF - 1), Mulpoly(f, Inv_f, lenF - 1, lenF - 1), Jfpoly(f, lenF - 1);
for (int i = lenF; i <= (lenF << 2); i++) f[i] = 0;
}
void Exppoly(int *f, int lenF, int *g) { // [0, lenF)
static int s[N << 2];
int limit = 1; while (limit < lenF) limit <<= 1;
g[0] = 1;
for (int len = 2; len <= limit; len <<= 1) {
for (int i = (len >> 1); i < len; i++) g[i] = 0;
for (int i = 0; i < (len >> 1); i++) s[i] = g[i];
for (int i = (len >> 1); i < len; i++) s[i] = 0;
Lnpoly(s, len);
for (int i = 0; i < len; i++) s[i] = (f[i] - s[i] + P) % P;
s[0] = (s[0] + 1) % P, Mulpoly(g, s, len - 1, len - 1);
}
for (int i = lenF; i < limit; i++) g[i] = 0;
for (int i = 0; i < limit; i++) s[i] = 0;
}
}
int solve(int x) {
if (!x) return 0;
memset(a, 0, sizeof a), memset(b, 0, sizeof b);
for (int i = 0; i <= x; i++) a[i] = inv[i];
Poly::Lnpoly(a, n - 1);
for (int i = 0; i < n - 1; i++) a[i] = (LL)n * a[i] % P;
Poly::Exppoly(a, n - 1, b);
return b[n - 2];
}
int main() {
n = read(), m = read(), Poly::Getsginv(n << 1), inv[0] = inv[1] = 1;
for (int i = 2; i <= m; i++) inv[i] = (LL)inv[i - 1] * Poly::sginv[i] % P;
int fac = 1;
for (int i = 2; i <= n - 2; i++) fac = (LL)fac * i % P;
printf("%d\n", (LL)fac * (solve(m - 1) - solve(m - 2) + P) % P);
}