luogu P5219 无聊的水题 I

https://www.luogu.com.cn/problem/P5219

通过学习prufer序列我们知道度数最大为$M$表示在prefer序列中一个数出现次数最多的为$M-1$次
恰好为$M-1$不好求，可以差分一下
因为要考虑顺序，所以要用EGF
考虑对每个数构造EGF
$$F(x)=\sum _{i=0}^{m-1}\frac{x^i}{i!}$$
答案就是$$[x^{n-2}]F^n(x)$$
冲个多项式快速幂即可
注意用的是EGF，最后要$\ast (n-2)!$
code:

#include<bits/stdc++.h>
#define int long long
#define mod 998244353
#define G 3
#define N 800005
using namespace std;
int qpow(int x, int y){
    int ret = 1;
    for(; y; y >>= 1, x = x * x % mod) if(y & 1) ret = ret * x % mod;
    return ret;
}
int rev[N], G_inv, len_inv;
void ntt(int *a, int len, int o){
    len_inv = qpow(len, mod - 2);
    for(int i = 0; i <= len; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i&1) * len >> 1);
    for(int i = 0; i <= len; i ++) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int i = 2; i <= len; i <<= 1){
        int wn = qpow((o == 1)? G:G_inv, (mod - 1) / i);
        for(int j = 0, p = i / 2; j + i - 1 <= len; j += i){
            int w0 = 1;
            for(int k = j; k < j + p; k ++, w0 = w0 * wn % mod){
                int X = a[k], Y = w0 * a[k + p] % mod;
                a[k] = (X + Y) % mod;
                a[k + p] = (X - Y + mod) % mod;
            }
        }
    }
    if(o == -1)
        for(int i = 0; i <= len; i ++) a[i] = a[i] * len_inv % mod;
}
int c[N];
void inv(int *a, int *b, int sz){
    if(sz == 0) {b[0] = qpow(a[0], mod - 2); return;}
    inv(a, b, sz / 2);
    int len = 1;
    for(; len <= sz + sz; len <<= 1);
    for(int i = 0; i <= sz; i ++) c[i] = a[i];
    for(int i = sz + 1; i <= len; i ++) c[i] = 0;
    ntt(c, len, 1), ntt(b, len, 1);
    for(int i = 0; i <= len; i ++) b[i] = (b[i] * 2 % mod - b[i] * b[i] % mod * c[i] % mod + mod) % mod;
    ntt(b, len, -1);
    for(int i = sz + 1; i <= len; i ++) b[i] = 0;
}
void qiudao(int *a, int sz) {
    for(int i = 0; i < sz; i ++) a[i] = a[i + 1] * (i + 1) % mod;
    a[sz] = 0;
}
void jifen(int *a, int sz) {
    for(int i = sz; i >= 1; i --) a[i] = a[i - 1] * qpow(i, mod - 2) % mod;
    a[0] = 0;
}
int Ad[N], An[N];
void ln(int *A, int n) {
    for(int i = 0; i <= n; i ++) Ad[i] = A[i];
    qiudao(Ad, n);
    inv(A, An, n);
    int len = 1;
    for(; len <= n + n;) len <<= 1;
    ntt(Ad, len, 1), ntt(An, len, 1);
    for(int i = 0; i <= len; i ++) Ad[i] = Ad[i] * An[i] % mod;
    ntt(Ad, len, -1);
    jifen(Ad, n);
    for(int i = 0; i <= n; i ++) A[i] = Ad[i];
    for(int i = 0; i <= len; i ++) An[i] = Ad[i] = 0;
}
int fln[N];
void exp(int *a, int *b, int n) {
    if(n == 0) {b[0] = 1; return;}
    exp(a, b, n / 2);
    for(int i = 0; i <= n; i ++) fln[i] = b[i]; ln(fln, n);
    fln[0] = 1;
    for(int i = 1; i <= n; i ++) fln[i] = (a[i] - fln[i] + mod ) % mod;
    int len = 1;
    for(; len <= n + n;) len <<= 1;
    ntt(b, len, 1), ntt(fln, len, 1);
    for(int i = 0; i <= len; i ++) b[i] = b[i] * fln[i] % mod;
    ntt(b, len, -1);
    for(int i = 0; i <= len; i ++) fln[i] = 0;
}
int a[N], b[N], n, m, fac[N], ifac[N], ib[N];
void init(int n) {
    fac[0] = 1;
    for(int i = 1; i <= n; i ++) fac[i] = fac[i - 1] * i % mod;
    ifac[n] = qpow(fac[n], mod - 2);
    for(int i = n - 1; i >= 0; i --) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
int calc(int n, int m) {
    if(m <= 0) return 0;
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    for(int i = 0; i <= m; i ++) a[i] = ifac[i];
    ln(a, n);
    for(int i = 0; i <= n; i ++) a[i] = a[i] * n % mod;
    exp(a, b, n);
    return b[n - 2] * fac[n - 2] % mod;
}
signed main(){
    init(N - 10);
    G_inv = qpow(G, mod - 2);
    scanf("%lld%lld", &n, &m);
    printf("%lld", (calc(n, m - 1) - calc(n, m - 2) + mod) % mod);
    return 0;
}