LibreOJ 6268.分拆数

欧拉 NB!

首先显然地 \(\sum\limits_{i=0}^{\infty} f(i)x^i = \prod\limits_{i=1}^{\infty} \frac1{1-x^i}\)
然后根据欧拉五边形数定理,有 \(\prod\limits_{i=1}^{\infty} (1-x^i)=1+\sum\limits_{i=1}^{\infty} (-1)^i x^{\frac{i(3i-1)}2} (1+x^i)\)
然后多项式求逆即可。

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
#include <cstdio>
#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 = 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[](const int &x) const
{
return a[x];
}
inline int &operator[](const int &x)
{
return a[x];
}
inline void clear(const int &x = 0)
{
memset(a + x,0,sizeof (a + x));
}
} a;
int len,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;
}
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,k = n >> 1;w <= n;w <<= 1,m <<= 1,k >>= 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;
}
void mul(poly &a,const poly &b,int n)
{
static poly x,y;
x = a,y = b;
x.clear(n >> 1),y.clear(n >> 1);
ntt(x,1,n),ntt(y,1,n);
for(register int i = 0;i < n;++i)
x[i] = (long long)x[i] * y[i] % mod;
ntt(x,-1,n);
a = x;
}
poly inverse(const poly &f,int n)
{
static int s[30];
static poly g,h,q;
int lim = 1,top = 0;
g.clear();
while(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;
for(register int i = n;i < lim;++i)
h[i] = 0;
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]);
for(register int i = n;i < lim;++i)
g[i] = 0;
}
return g;
}
int main()
{
scanf("%d",&len),init((len + 1) << 1),a[0] = 1;
for(register int i = 1;i * (3 * i - 1) / 2 <= len;++i)
a[i * (3 * i - 1) / 2] = (i & 1) ? mod - 1 : 1,
i * (3 * i + 1) / 2 <= len && (a[i * (3 * i + 1) / 2] = (i & 1) ? mod - 1 : 1);
a = inverse(a,len + 1);
for(register int i = 1;i <= len;++i)
printf("%d\n",a[i]);
}