为什么要做这题呢，当然是有用啊qwq
首先我们考虑非常经典的式子：
\[x^{\overline{n}}=\sum_i \left[^n_i\right] x^i\]
然后上倍增：
\[x^{\overline{2n}}=x^{\overline{n}}(x+n)^{\overline{n}}\]
相当于我们已经有了一个多项式\(f(x)\)，现在要求另一个多项式\(f(x+c)\)
\[f(x+c)=\sum_i f_i(x+c)^i\]
\[=\sum_i x_i\sum_j f_j\times C_i^j\times c^{i-j}\]
\[=\sum_i \frac{x_i}{i!} \sum_j j!\times f_j\times\frac{c^{i-j}}{(i-j)!}\]
发现此时\(\frac{c^{i-j}}{(i-j)!}\)不太好处理，因此我们把它和\(f\)都反过来做一遍卷积然后反回去即可
然后注意这是\(n\mod 2=0\)的情况，\(n=1\)是要暴力多乘上一个\((x+n)\)
复杂度为\(T(n)=T(n/2)+O(n\log n)=T(n\log n)\)
#include<cstdio>
#include<iostream>
#define RI register int
#define CI const int&
using namespace std;
const int N=1<<20,mod=167772161;
int n,lim,F[N],fact[N],inv[N];
inline int sum(CI x,CI y)
{
int t=x+y; return t>=mod?t-mod:t;
}
inline int sub(CI x,CI y)
{
int t=x-y; return t<0?t+mod:t;
}
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline void init(CI n)
{
RI i; for (fact[0]=i=1;i<=n;++i) fact[i]=1LL*fact[i-1]*i%mod;
for (inv[n]=quick_pow(fact[n]),i=n-1;~i;--i) inv[i]=1LL*inv[i+1]*(i+1)%mod;
}
namespace Poly
{
int rev[N],p;
inline void init(CI n)
{
for (lim=1,p=0;lim<=n;lim<<=1,++p);
for (RI i=0;i<lim;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<p-1);
}
inline void NTT(int *f,CI opt)
{
RI i,j,k; for (i=0;i<lim;++i) if (i<rev[i]) swap(f[i],f[rev[i]]);
for (i=1;i<lim;i<<=1)
{
int m=i<<1,D=quick_pow(3,~opt?(mod-1)/m:mod-1-(mod-1)/m);
for (j=0;j<lim;j+=m)
{
int W=1; for (k=0;k<i;++k,W=1LL*W*D%mod)
{
int x=f[j+k],y=1LL*f[i+j+k]*W%mod;
f[j+k]=sum(x,y); f[i+j+k]=sub(x,y);
}
}
}
if (!~opt)
{
int Inv=quick_pow(lim); for (i=0;i<lim;++i) f[i]=1LL*f[i]*Inv%mod;
}
}
inline void convolution(int *f,CI n,CI c,int *g)
{
static int A[N],B[N]; RI i; int bs; init(n<<1);
for (i=0;i<n;++i) A[n-1-i]=1LL*f[i]*fact[i]%mod;
for (bs=1,i=0;i<n;++i,bs=1LL*bs*c%mod) B[i]=1LL*bs*inv[i]%mod;
for (i=n;i<lim;++i) A[i]=B[i]=0; NTT(A,1); NTT(B,1);
for (i=0;i<lim;++i) A[i]=1LL*A[i]*B[i]%mod; NTT(A,-1);
for (i=0;i<n;++i) g[i]=1LL*A[n-1-i]*inv[i]%mod;
}
inline void solve(CI n,int *f)
{
if (!n) return (void)(f[0]=1); static int A[N],B[N];
RI i; int m=n>>1; solve(m,f); convolution(f,m+1,m,A);
for (i=0;i<=m;++i) B[i]=f[i]; for (i=m+1;i<lim;++i) A[i]=B[i]=0;
for (init(n),NTT(A,1),NTT(B,1),i=0;i<lim;++i) A[i]=1LL*A[i]*B[i]%mod;
NTT(A,-1); if (!(n&1)) for (i=0;i<=n;++i) f[i]=A[i]; else
for (i=0;i<=n;++i) f[i]=sum(i?A[i-1]:0,1LL*(n-1)*A[i]%mod);
//for (printf("%d\n",n),i=0;i<=n;++i) printf("%d%c",f[i]," \n"[i==n]);
}
};
int main()
{
//freopen("CODE.out","w",stdout);
scanf("%d",&n); init(n); Poly::solve(n,F);
for (RI i=0;i<=n;++i) printf("%d ",F[i]); return 0;
}