Codeforces 438E The Child and Binary Tree

发表于 | 分类于 |

首先把 \(c\) 看做一个集合，然后定义 \(C(x) = \sum\limits_{i \in c} x^i\)。
再设 \(f_m\) 为总权值和为 \(m\) 时的答案。

根据某些 DP 套路，可以得到 \[ f_m = 1 + \sum\limits_{i \in c}\sum\limits_{j = 0}^{m - i} f_j f_{m-i-j} \]

令 \(F(x)\) 为 \(f\) 的生成函数，则根据上式有 \[ \begin{align*} F(x) &= CF^2(x) + 1 \\ CF^2(x) - F(x) + 1 &= 0 \\ F(x) &= \frac{1 \pm \sqrt{1 - 4C}}{2C} \end{align*} \]

由于 \(0 \not\in c,f_0 = 1\) 所以 \(C(0) = 0,F(0) = 1\)。
据此可知应取负号，即 \[ F(x) = \frac{1 - \sqrt{1 - 4C}}{2C} \]

然后就发现似乎就可做了，但是 \(C\) 的常数项为 \(0\)，有点麻烦。
考虑用类似分母有理化的方法对分子有理化试试。
易得 \[ F(x) = \frac2{1+\sqrt{1-4C}} \]

然后就完全可做了。
贴个多项式开根和多项求逆就做完了。

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
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <utility>
#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 = 1 << 18;
const int mod = 998244353;
const int G = 3;
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;
}
struct poly
{
    int a[N + 5];
    inline const int &operator[](int x) const
    {
        return a[x];
    }
    inline int &operator[](int x)
    {
        return a[x];
    }
    inline void clear(int x = 0)
	{
        memset(a + x,0,(N - x + 1) << 2);
    }
} f;
int m,len,k,n,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(int len)
{
    for(n = 1;n < len;n <<= 1);
    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;
}
inline void ntt(poly &a,int type,int n)
{
    type == -1 && (reverse(a.a + 1,a.a + n),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;
}
inline void mul(poly &a,const poly &b,int n)
{
    static poly x,y;
    int lim = 1;
    x.clear(),y.clear();
    for(;lim < (n << 1);lim <<= 1);
    x = a,y = b;
    x.clear(n),y.clear(n);
    ntt(x,1,lim),ntt(y,1,lim);
    for(register int i = 0;i < lim;++i)
        x[i] = (long long)x[i] * y[i] % mod;
    ntt(x,-1,lim);
    x.clear(n),a = x;
}
inline poly inverse(const poly &f,int n)
{
    static int s[30];
    static poly g,h,q;
    int lim = 1,top = 0;
    g.clear();
    for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
    g[0] = fpow(f[0],mod - 2);
    for(;top;--top)
    {
        n = s[top];
        for(;lim < (n << 1);lim <<= 1);
        q = g,h = f,h.clear(n);
        ntt(g,1,lim),ntt(h,1,lim);
        for(register int i = 0;i < lim;++i)
            g[i] = (long long)g[i] * g[i] % mod * h[i] % mod;
        ntt(g,-1,lim);
        for(register int i = 0;i < n;++i)
            g[i] = dec(add(q[i],q[i]),g[i]);
        g.clear(n);
    }
    return g;
}
inline void derivative(poly &f,int n)
{
    for(register int i = 1;i < n;++i)
        f[i - 1] = (long long)f[i] * i % mod;
    f[n - 1] = 0;
}
inline void integral(poly &f,int n)
{
    for(register int i = n - 1;~i;--i)
        f[i + 1] = (long long)f[i] * inv[i + 1] % mod;
    f[0] = 0;
}
inline poly ln(const poly &f,int n)
{
    static poly g;
    g = f,derivative(g,n),mul(g,inverse(f,n),n),integral(g,n);
    return g;
}
inline poly exp(const poly &f,int n)
{
    static int s[30];
    static poly g,h;
    int lim = 1,top = 0;
    g.clear();
    for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
    g[0] = 1;
    for(;top;--top)
    {
        n = s[top];
        for(;lim < (n << 1);lim <<= 1);
        h = g,g = ln(g,n);
        for(register int i = 0;i < n;++i)
            g[i] = dec(f[i],g[i]);
        g[0] = add(g[0],1);
        ntt(g,1,lim),ntt(h,1,lim);
        for(register int i = 0;i < lim;++i) 
            g[i] = (long long)g[i] * h[i] % mod;
        ntt(g,-1,lim);
        g.clear(n);
    }
    return g;
}
inline poly power(const poly &f,int k,int n)
{
    static poly g;
    g = ln(f,n);
    for(register int i = 0;i < n;++i)
        g[i] = (long long)g[i] * k % mod;
    g = exp(g,n);
    return g;
}
namespace Mod_sqrt
{
    typedef pair<int,int> cp;
    int w;
    inline cp operator*(const cp &a,const cp &b)
    {
        return cp(((long long)a.first * b.first % mod + (long long)a.second * b.second % mod * w % mod) % mod,((long long)a.first * b.second % mod + (long long)a.second * b.first % mod) % mod);
    }
    inline cp pow(cp a,int b)
    {
        cp ret(1,0);
        for(;b;b >>= 1)
            (b & 1) && (ret = ret * a,1),a = a * a;
        return ret;
    }
    inline int mod_sqrt(int x)
    {
        int y = rand() % mod;
        for(;fpow(w = ((long long)y * y % mod - x + mod) % mod,mod - 1 >> 1) <= 1;y = rand() % mod);
        cp ret = pow(cp(y,1),mod + 1 >> 1);
        return min(ret.first,mod - ret.first);
    }
}
using Mod_sqrt::mod_sqrt;
inline poly sqrt(const poly &f,int n)
{
    static int s[30];
    static poly g,h;
    int top = 0;
    g.clear();
    for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
    g[0] = mod_sqrt(f[0]);
    for(;top;--top)
    {
        n = s[top];
        for(register int i = 0;i < n;++i)
            h[i] = add(g[i],g[i]);
        h = inverse(h,n),mul(g,g,n);
        for(register int i = 0;i < n;++i)
            g[i] = add(g[i],f[i]);
        mul(g,h,n);
    }
    return g;
}
int main()
{
    scanf("%d%d",&m,&len),init(++len << 1);
    int x;
    for(;m;--m)
        scanf("%d",&x),x < len && (f[x] = 1);
    for(register int i = 0;i < len;++i)
        f[i] = mod - 4LL * f[i] % mod;
    f[0] = add(f[0],1),f = sqrt(f,len),f[0] = add(f[0],1),f = inverse(f,len);
    for(register int i = 0;i < len;++i)
        f[i] = 2LL * f[i] % mod;
    for(register int i = 1;i < len;++i)
        printf("%d\n",f[i]);
}