Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5396 第二类斯特林数·列

P5396 第二类斯特林数·列

考虑 EGF 。对于一个集合有 $F(x) = \sum _ {i = 1} {i \brace 1} \frac {x ^ i} {i!} = e ^ x - 1$ 。

那么对于 $k$ 个集合,有生成函数 $F(x) = \frac {(e ^ x - 1) ^ k} {k!}$ 。上多项式快速幂即可。

查看代码
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
#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, mod = 167772161, inv2 = mod + 1 >> 1;
int n, k, rev[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void ntt (int *x, int bit, int op)
{
int tot = 1 << bit;
for (int i = 1; i < tot; ++i)
if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
swap(x[rev[i]], x[i]);
for (int mid = 1; mid < tot; mid <<= 1)
{
int w1 = binpow(3, (mod - 1) / (mid << 1));
if (!~op) w1 = binpow(w1);
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
{
int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
x[i | j] = ((LL)p + q) % mod, x[i | j | mid] = ((LL)p - q) % mod;
}
}
if (~op) return;
int itot = binpow(tot);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
int bit = 0;
while (1 << bit < n + m - 1) ++bit;
int tot = 1 << bit;
for (int i = n; i < tot; ++i) f[i] = 0;
for (int i = m; i < tot; ++i) g[i] = 0;
ntt(f, bit, 1), ntt(g, bit, 1);
for (int i = 0; i < tot; ++i)
res[i] = (LL)f[i] * g[i] % mod;
ntt(res, bit, -1);
for (int i = nm; i < tot; ++i) res[i] = 0;
}
void PolyInv(int n, int *x, int *g)
{
if (n == 1) return void(g[0] = binpow(x[0]));
int m = n + 1 >> 1;
int bit = 0;
while (1 << bit < n + m + m - 2) ++bit;
int tot = 1 << bit;
PolyInv(m, x, g);
for (int i = m; i < tot; ++i) g[i] = 0;
static int A[N];
for (int i = 0; i < n; ++i) A[i] = x[i];
for (int i = n; i < tot; ++i) A[i] = 0;
ntt(g, bit, 1), ntt(A, bit, 1);
for (int i = 0; i < tot; ++i)
g[i] = (2 - (LL)g[i] * A[i]) % mod * g[i] % mod;
ntt(g, bit, -1);
for (int i = n; i < tot; ++i) g[i] = 0;
}
void PolyDerivate(int n, int *x, int *g)
{
for (int i = 1; i < n; ++i)
g[i - 1] = (LL)x[i] * i % mod;
g[n - 1] = 0;
}
void PolyIntegrate(int n, int *x, int *g)
{
for (int i = 1; i < n; ++i)
g[i] = (LL)x[i - 1] * binpow(i) % mod;
g[0] = 0;
}
void PolyLn(int n, int *x, int *g)
{
static int A[N], B[N];
PolyDerivate(n, x, A);
PolyInv(n, x, B);
PolyMul(n, A, n, B, n, A);
PolyIntegrate(n, A, g);
}
void PolyExp(int n, int *x, int *g)
{
if (n == 1) return void(g[0] = 1);
int m = n + 1 >> 1;
PolyExp(m, x, g);
for (int i = m; i < n; ++i) g[i] = 0;
static int A[N];
PolyLn(n, g, A);
for (int i = 0; i < n; ++i)
A[i] = ((LL)x[i] - A[i]) % mod;
++A[0];
PolyMul(n, A, m, g, n, g);
}
void PolyBinpow(int n, int *x, int k, int *g)
{
static int A[N], B[N];
int t = 0;
while (t < n && !x[t]) ++t;
if ((LL)k * t >= n)
{
for (int i = 0; i < n; ++i) g[i] = 0;
return;
}
int s = binpow(x[t]);
for (int i = 0; i < n - t; ++i)
A[i] = (LL)x[i + t] * s % mod;
PolyLn(n - t, A, B);
for (int i = 0; i < n - t; ++i)
B[i] = (LL)B[i] * k % mod;
PolyExp(n - t, B, g);
s = binpow(x[t], k);
for (int i = n - 1; i >= k * t; --i)
g[i] = (LL)g[i - k * t] * s % mod;
for (int i = 0; i < k * t; ++i) g[i] = 0;
}
int main ()
{
static int n, m, A[N], B[N];
read(n, m); ++n;
(++A[1]) %= mod;
PolyExp(n, A, B);
(--B[0]) %= mod;
PolyBinpow(n, B, m, A);
int k = 1;
for (int i = 2; i <= m; ++i)
k = (LL)k * i % mod;
k = binpow(k);
for (int i = 0, t = 1; i < n; t = (LL)t * ++i % mod)
write(((LL)A[i] * k % mod * t % mod + mod) % mod), putchar(' ');
return 0;
}