LibreOJ 6268 分拆数

欧拉 NB！

首先显然地 ∞∑i=0f(i)xi=∞∏i=111−xi。
然后根据欧拉五边形数定理，有 ∞∏i=1(1−xi)=1+∞∑i=1(−1)ixi(3i−1)2(1+xi)。
然后多项式求逆即可。

代码：

#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]);
}