LibreOJ 6363 「地底蔷薇」

发表于 | 分类于 |

拉格朗日反演基本操作的一点改造？（

设 \(H(x)\) 为有标号有根无向连通图个数的 EGF。不会求这个的建议出门左转城市规划。

然后设 \(b_i\) 表示 \(i+1\) 个点的有标号点双个数，设 \(B(x)\) 是它的 EGF。
考虑一个有根连通图，删去其根应当形成若干连通块，而每个连通块中都有一部分点与根本在同一个点双。
考虑删去根再删去所有包含根的点双的边，那么这些点双被拆散以后会剩下一些以它们为根的连通图。
于是 \[ H(x) = x{\rm e}^{B(H(x))} \]

考虑求出 \(G(x) = \sum\limits_{i+1\in S} \frac{b_i x^i}{i!}\)。
可以试着对这个形式构造复合逆 \[ H^{-1}(x) = \frac{x}{ {\rm e}^{B(x)} } \]

反过来用复合逆表示 \(B\) \[ B(x) = \ln \frac x{H^{-1}(x)} \]

构造 \[ A(x) = \ln \frac{H(x)}x \]

根据扩展拉格朗日反演有 \[ [x^n] B(x) = [x^n] A(H^{-1}(x)) = \frac1n [x^{n-1}] A'(x) \left(\frac x{H(x)}\right)^n \]

有了 \(G\) 后，设 \(F\) 表示答案的 EGF，那么同理可得 \[ F(x) = x{\rm e}^{G(F(x))} \]

从而 \[ F^{-1}(x) = \frac x{ {\rm e}^{G(x)} } \\ [x^n] F(x) = \frac 1n[x^{n-1}] \left(\frac x{F^{-1}(x)}\right)^n = \frac 1n[x^{n-1}] {\rm e}^{nG(x)} \]

复杂度 \(O(n \log n)\)（视 \(n,\sum\limits_{i\in S}i\) 同阶）。

代码： 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#define add(a,b) (a + b >= mod ? a + b - mod : a + b)
#define dec(a,b) (a < b ? a - b + mod : a - b)
using namespace std;
const int N = 1e5;
const int mod = 998244353;
inline int fpow(int a,int b)
{
    int ret = 1;
    for(;b;b >>= 1)
        (b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
    return ret;
}
int n,m;
int a[N + 5];
namespace Poly
{
    const int N = 1 << 18;
    const int G = 3;
    int lg2[N + 5];
    int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
    int rt[N + 5],irt[N + 5];
    inline void init()
    {
        for(register int i = 2;i <= N;++i)
            lg2[i] = lg2[i >> 1] + 1;
        int w = fpow(G,(mod - 1) / N);
        rt[N >> 1] = 1;
        for(register int i = (N >> 1) + 1;i <= N;++i)
            rt[i] = (long long)rt[i - 1] * w % mod;
        for(register int i = (N >> 1) - 1;i;--i)
            rt[i] = rt[i << 1];
        fac[0] = 1;
        for(register int i = 1;i <= N;++i)
            fac[i] = (long long)fac[i - 1] * i % mod;
        ifac[N] = fpow(fac[N],mod - 2);
        for(register int i = N;i;--i)
            ifac[i - 1] = (long long)ifac[i] * i % mod;
        for(register int i = 1;i <= N;++i)
            inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
    }
    struct poly
    {
        vector<int> a;
        inline poly(int x = 0)
        {
            x && (a.push_back(x),1);
        }
        inline poly(const vector<int> &o)
        {
            a = o,shrink();
        }
        inline poly(const poly &o)
        {
            a = o.a,shrink();
        }
        inline void shrink()
        {
            for(;!a.empty() && !a.back();a.pop_back());
        }
        inline int size() const
        {
            return a.size();
        }
        inline void resize(int x)
        {
            a.resize(x);
        }
        inline int operator[](int x) const
        {
            if(x < 0 || x >= size())
                return 0;
            return a[x];
        }
        inline void clear()
        {
            vector<int>().swap(a);
        }
        inline void ntt(int type = 1)
        {
            int n = size();
            type == -1 && (reverse(a.begin() + 1,a.end()),1);
            int lg = lg2[n] - 1;
            for(register int i = 0;i < n;++i)
                rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
                i < rev[i] && (swap(a[i],a[rev[i]]),1);
            for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
                for(register int i = 0;i < n;i += w)
                    for(register int j = 0;j < m;++j)
                    {
                        int t = (long long)rt[m | j] * a[i | j | m] % mod;
                        a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
                    }
            if(type == -1)
                for(register int i = 0;i < n;++i)
                    a[i] = (long long)a[i] * inv[n] % mod;
        }
        friend inline poly operator+(const poly &a,const poly &b)
        {
            vector<int> ret(max(a.size(),b.size()));
            for(register int i = 0;i < ret.size();++i)
                ret[i] = add(a[i],b[i]);
            return poly(ret);
        }
        friend inline poly operator-(const poly &a,const poly &b)
        {
            vector<int> ret(max(a.size(),b.size()));
            for(register int i = 0;i < ret.size();++i)
                ret[i] = dec(a[i],b[i]);
            return poly(ret);
        }
        friend inline poly operator*(poly a,poly b)
        {
            if(a.a.empty() || b.a.empty())
                return poly();
            int lim = 1,tot = a.size() + b.size() - 1;
            for(;lim < tot;lim <<= 1);
            a.resize(lim),b.resize(lim);
            a.ntt(),b.ntt();
            for(register int i = 0;i < lim;++i)
                a.a[i] = (long long)a[i] * b[i] % mod;
            a.ntt(-1),a.shrink();
            return a;
        }
        poly &operator+=(const poly &o)
        {
            resize(max(size(),o.size()));
            for(register int i = 0;i < o.size();++i)
                a[i] = add(a[i],o[i]);
            return *this;
        }
        poly &operator-=(const poly &o)
        {
            resize(max(size(),o.size()));
            for(register int i = 0;i < o.size();++i)
                a[i] = dec(a[i],o[i]);
            return *this;
        }
        poly &operator*=(poly o)
        {
            return (*this) = (*this) * o;
        }
        poly deriv() const
        {
            if(a.empty())
                return poly();
            vector<int> ret(size() - 1);
            for(register int i = 0;i < size() - 1;++i)
                ret[i] = (long long)(i + 1) * a[i + 1] % mod;
            return poly(ret);
        }
        poly integ() const
        {
            if(a.empty())
                return poly();
            vector<int> ret(size() + 1);
            for(register int i = 0;i < size();++i)
                ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
            return poly(ret);
        }
        inline poly modxn(int n) const
        {
            if(a.empty())
                return poly();
            n = min(n,size());
            return poly(vector<int>(a.begin(),a.begin() + n));
        }
        inline poly inver(int m) const
        {
            poly ret(fpow(a[0],mod - 2));
            for(register int k = 1;k < m;)
                k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
            return ret.modxn(m);
        }
        inline poly log(int m) const
        {
            return (deriv() * inver(m)).integ().modxn(m);
        }
        inline poly exp(int m) const
        {
            poly ret(1);
            for(register int k = 1;k < m;)
                k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
            return ret.modxn(m);
        }
        inline poly pow(int m,int k1,int k2 = -1) const
        {
            if(a.empty())
                return poly();
            if(k2 == -1)
                k2 = k1;
            int t = 0;
            for(;t < size() && !a[t];++t);
            if((long long)t * k1 >= m)
                return poly();
            poly ret;
            ret.resize(m);
            int u = fpow(a[t],mod - 2),v = fpow(a[t],k2);
            for(register int i = 0;i < m - t * k1;++i)
                ret.a[i] = (long long)operator[](i + t) * u % mod;
            ret = ret.log(m - t * k1);
            for(register int i = 0;i < ret.size();++i)
                ret.a[i] = (long long)ret[i] * k1 % mod;
            ret = ret.exp(m - t * k1),t *= k1,ret.resize(m);
            for(register int i = m - 1;i >= t;--i)
                ret.a[i] = (long long)ret[i - t] * v % mod;
            for(register int i = 0;i < t;++i)
                ret.a[i] = 0;
            return ret;
        }
    };
}
using Poly::init;
using Poly::poly;
poly G,H,A;
int ans;
int main()
{
    init();
    scanf("%d%d",&n,&m);
    H.resize(n + 2),G.resize(n + 1);
    for(register int i = 0;i < H.size();++i)
        H.a[i] = (long long)fpow(2,(long long)i * (i - 1) / 2 % (mod - 1)) * Poly::ifac[i] % mod;
    H = H.log(n + 2);
    for(register int i = 0;i < H.size();++i)
        H.a[i] = (long long)H[i] * i % mod;
    A.resize(n + 1);
    for(register int i = 0;i < A.size();++i)
        A.a[i] = H[i + 1];
    A = (A.deriv() * (H = A.inver(n + 1))).modxn(n + 1);
    for(register int i = 1;i <= m;++i)
    {
        scanf("%d",a + i);
        if(a[i] > 1 && a[i] <= n)
            --a[i],
            G.a[a[i]] = (long long)Poly::inv[a[i]] * (A.modxn(a[i]) * H.pow(a[i],a[i]))[a[i] - 1] % mod;
    }
    ans = (long long)Poly::inv[n] * (n * G).exp(n)[n - 1] % mod * Poly::fac[n - 1] % mod;
    printf("%d\n",ans);
}