[NTT][斯特林数]BZOJ 5093: 图的价值

$Description$

一个带标号的图的价值定义为每个点度数的$p$次方的和。
给定$n$和$p$，请计算所有$n$个点的带标号的简单无向图的价值之和。

$Solution$

推式子：

$$\begin{eqnarray} ans&=&n2^{n-1\choose2}\sum_{d=0}^{n-1}d^p{n-1 \choose d}\\ x^p&=&\sum_{k=0}^pS(p,k)x^{k\downarrow}\\ &=&\sum_{k=0}^pS(p,k){x\choose k}k!\\ ans&=&n2^{n-1\choose2}\sum_{k=0}^pS(p,k)k!\sum_{d=0}^{n-1}{n-1 \choose d}{d\choose k}\\ &=&n2^{n-1\choose2}\sum_{k=0}^pS(p,k)k!{n-1 \choose k}2^{n-k-1}\\ S(p,k)&=&{1\over{p!}}\sum_{k=0}^p(-1)^{p-k}k^p{p\choose k}\\ &=&\sum_{k=0}^p{(-1)^{p-k}\over (p-k)!}{k^p\over k!} \end{eqnarray}$$
NTT就好了。
又忘了是循环卷积了QAQ

#include <bits/stdc++.h>
using namespace std;

const int G = 3;
const int N = 808080;
const int MOD = 998244353;
typedef long long ll;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int g, ig, num, L;
int n, p, m, ans;
int w[2][N];
int R[N];
int inv[N], fac[N], ifac[N];
int Sp[N], A[N], B[N], C[N];

inline int Pow(int a, int b) {
    int c = 1; 
    while (b) {
        if (b & 1) c = (ll)c * a % MOD;
        b >>= 1; a = (ll)a * a % MOD;
    }
    return c;
}
inline int Inv(int x) {
    return Pow(x, MOD - 2);
}
void Prep(int n) {
    g = Pow(G, (MOD - 1) / n);
    ig = Inv(g); num = n;
    w[0][0] = w[1][0] = 1;
    for (int i = 1; i <= n; i++) {
        w[0][i] = (ll)w[0][i - 1] * ig % MOD;
        w[1][i] = (ll)w[1][i - 1] * g % MOD;
    }
}
inline void FFT(int *a, int n, int r) {
    static int x, y, INV;
    for (int i = 0; i < n; i++)
        if (R[i] > i) swap(a[i], a[R[i]]);
    for (int i = 1; i < n; i <<= 1)
        for (int j = 0; j < n; j += (i << 1))
            for (int k = 0; k < i; k++) {
                x = a[j + k];
                y = (ll)a[j + k + i] * w[r][num / (i << 1) * k] % MOD;
                a[j + k] = (x + y) % MOD;
                a[j + k + i] = (x - y + MOD) % MOD;
            }
    if (!r) {
        INV = Inv(n);
        for (int i = 0; i < n; i++)
            a[i] = (ll)a[i] * INV % MOD;
    }
}
inline void Add(int &x, int a) {
    x = (x + a) % MOD;
}

int main(void) {
    freopen("1.in", "r", stdin);
    read(n); read(p);
    for (m = 1; m <= p; m <<= 1) ++L;
    m <<= 1; Prep(m);
    for (int i = 1; i < m; i++)
        R[i] = (R[i >> 1] >> 1) | ((i & 1) << L);
    inv[1] = 1;
    for (int i = 2; i < m; i++)
        inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;
    ifac[0] = fac[0] = 1;
    for (int i = 1; i < m; i++) {
        ifac[i] = (ll)ifac[i - 1] * inv[i] % MOD;
        fac[i] = (ll)fac[i - 1] * i % MOD;
    }
    for (int i = 0; i <= p; i++) {
        A[i] = (ll)Pow(i, p) * ifac[i] % MOD;
        if (i & 1) B[i] = MOD - ifac[i];
        else B[i] = ifac[i];
    }
    C[0] = 1;
    for (int i = 1; i <= p; i++)
        C[i] = (ll)C[i - 1] * (n - i) % MOD * inv[i] % MOD;
    FFT(A, m, 1); FFT(B, m, 1);
    for (int i = 0; i < m; i++) Sp[i] = (ll)A[i] * B[i] % MOD;
    FFT(Sp, m, 0);
    for (int i = 0; i <= p; i++)
        Add(ans, (ll)Sp[i] * fac[i] % MOD * C[i] % MOD * Pow(2, n - i - 1) % MOD);
    ans = (ll)ans * n % MOD * Pow(2, (ll)(n - 1) * (n - 2) / 2 % (MOD - 1)) % MOD;
    cout << ans << endl;
    return 0;
}