[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：

ansxpansS(p,k)=======n(n−12)∑d=0n−1dp(n−1d)∑k=0pS(p,k)xk↓∑k=0pS(p,k)(xk)k!n(n−12)∑k=0pS(p,k)k!∑d=0n−1(n−1d)(dk)n(n−12)∑k=0pS(p,k)k!∑d=0n−1(n−1k)2n−k−11p!∑k=0p(−1)p−kkp(pk)∑k=0p(−1)p−k(p−k)!kpk!

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;}