Luogu P5408 模板第一类斯特林数·行

Posted 2021-03-22 cjjsb

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 ight] 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_isum_j f_j imes C_i^j imes c^{i-j}]

[=sum_i frac{x_i}{i!} sum_j j! imes f_j imesfrac{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(nmod 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(nlog n)=T(nlog 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),i=0;i<=n;++i) printf("%d%c",f[i]," 
"[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;
}