题意：第二类斯特林数(egin{Bmatrix} n \m end{Bmatrix})表示把(n)个不同元素划分成(m)个相同的集合（不能有空集）的方案数。
给定(n,k)，对于所有的整数(iin[0,n])，你要求出(egin{Bmatrix} i \k end{Bmatrix})。
由于答案会非常大，所以你的输出需要对(167772161)（(2^{25} imes 5+1)，是一个质数）取模。
我们可以先处理(k=1)的情况，考虑一个EGF：
[F(x)=sum_{i=1}^{infin}egin{Bmatrix}i\1end{Bmatrix}frac{x^i}{i!}
]
而显然(egin{Bmatrix}i\1end{Bmatrix}=1)，那么我们有：
[F(x)=sum_{i=1}^{infin}frac{x^i}{i!}=e^x-1
]
推广到(k)的情况就是：
[frac{F^k(x)}{k!}
]
因为EGF是关于排列的答案，而我们要求关于组合的答案，所以需要除以(k!)
于是我们可以直接愉快地多项式快速幂了。
Code
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 6e5;
const int p = 167772161;
using namespace std;
int n,k,G[N + 5][2],rev[N + 5],lg,maxn,a[N + 5],ans[N + 5];
int mypow(int a,int x){int s = 1;for (;x;x & 1 ? s = 1ll * s * a % p : 0,a = 1ll * a * a % p,x >>= 1);return s;}
void pre(int n)
{
maxn = 1,lg = 0;
while (maxn <= n)
maxn <<= 1,lg++;
for (int i = 0;i < maxn;i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
}
void clear(int *a,int n,int l = 0)
{
for (int i = l;i < n;i++)
a[i] = 0;
}
void ntt(int *a,int typ)
{
for (int i = 0;i < maxn;i++)
if (i < rev[i])
swap(a[i],a[rev[i]]);
for (int i = 1;i < maxn;i <<= 1)
for (int j = 0;j < maxn;j += i << 1)
for (int k = 0;k < i;k++)
{
int x = a[j + k],t = 1ll * G[k + i][typ] * a[j + k + i] % p;
a[j + k] = (x + t) % p;
a[j + k + i] = (x - t) % p;
}
if (!typ)
{
int inv = mypow(maxn,p - 2);
for (int i = 0;i < maxn;i++)
a[i] = 1ll * a[i] * inv % p;
}
}
int INVa[N + 5];
void INV(int *a,int *ans,int n)
{
if (n == 1)
{
ans[0] = mypow(a[0],p - 2);
return;
}
INV(a,ans,n + 1 >> 1);
pre(n * 2);
for (int i = 0;i < n;i++)
INVa[i] = a[i];
clear(INVa,maxn,n);
ntt(INVa,1);
ntt(ans,1);
for (int i = 0;i < maxn;i++)
ans[i] = (2ll * ans[i] % p - 1ll * INVa[i] * ans[i] % p * ans[i] % p) % p;
ntt(ans,0);
clear(ans,maxn,n);
}
int Lna[N + 5],Lnb[N + 5];
void DOV(int *a,int *f,int n)
{
for (int i = 1;i < n;i++)
f[i - 1] = 1ll * i * a[i] % p;
f[n - 1] = 0;
}
void DOVINV(int *a,int *f,int n)
{
f[0] = 0;
for (int i = 1;i < n;i++)
f[i] = 1ll * mypow(i,p - 2) * a[i - 1] % p;
}
void Ln(int *a,int *ans,int n)
{
DOV(a,Lna,n);
pre(n * 2);
clear(Lnb,maxn);
INV(a,Lnb,n);
pre(n * 2);
clear(Lna,maxn,n);
ntt(Lna,1);
ntt(Lnb,1);
for (int i = 0;i < maxn;i++)
Lna[i] = 1ll * Lna[i] * Lnb[i] % p;
ntt(Lna,0);
DOVINV(Lna,ans,n);
clear(ans,maxn,n);
}
int expa[N + 5],expb[N + 5];
void exp(int *a,int *ans,int n)
{
if (n == 1)
{
ans[0] = 1;
return;
}
exp(a,ans,n + 1 >> 1);
Ln(ans,expa,n);
pre(n * 2);
for (int i = 0;i < n;i++)
expb[i] = a[i];
clear(expb,maxn,n);
ntt(ans,1);
ntt(expa,1);
ntt(expb,1);
for (int i = 0;i < maxn;i++)
ans[i] = 1ll * ans[i] * ((1 - expa[i] + expb[i]) % p) % p;
ntt(ans,0);
clear(ans,maxn,n);
}
int pa[N + 5];
void mypow(int *a,int *ans,int n,int k)
{
Ln(a,pa,n);
for (int i = 0;i < n;i++)
pa[i] = 1ll * pa[i] * k % p;
exp(pa,ans,n);
}
int main()
{
scanf("%d%d",&n,&k);
pre(n * 2 + 2);
for (int i = 1;i < maxn;i <<= 1)
{
int g1 = mypow(3,(p - 1) / (i << 1)),g2 = mypow(mypow(3,p - 2),(p - 1) / (i << 1));
G[i][0] = G[i][1] = 1;
for (int j = 1;j < i;j++)
G[i + j][1] = 1ll * G[i + j - 1][1] * g1 % p,G[i + j][0] = 1ll * G[i + j - 1][0] * g2 % p;
}
int res = 1;
for (int i = 0;i < n;i++)
res = 1ll * res * (i + 1) % p,a[i] = mypow(res,p - 2);
mypow(a,ans,n,k);
for (int i = n;i >= k;i--)
ans[i] = ans[i - k];
for (int i = 0;i < k;i++)
ans[i] = 0;
res = 1;
for (int i = 1;i <= k;i++)
res = 1ll * res * i % p;
int sm = mypow(res,p - 2);
for (int i = k;i <= n;i++)
ans[i] = 1ll * ans[i] * res % p * sm % p,res = 1ll * res * (i + 1) % p;
for (int i = 0;i <= n;i++)
printf("%d ",(ans[i] + p) % p);
return 0;
}