题目传送门

题目大意

给出两个次数分别为\(n,m\)的下降幂多项式，表示为:

求出一个下降幂多项式\(H(x)\)使得对于\(\forall x,H(x)=F(x)G(x)\)。

\(n,m\le 10^5\)

思路

有一个思路特别\(\texttt{naive}\)但是码起来会死人的方法，就是两个下降幂多项式转换成普通多项式，然后乘起来再转换成下降幂多项式。时间复杂度\(\Theta(n\log ^2n)\)，常数估计也贼大，估计是通过不了这道题的。

以下内容借鉴了\(\texttt{command-block}\)的题解。

我们发现我们其实用与多项式乘法相同的做法，我们把下降幂多项式转换成点值表示法然后乘起来再转换成下降幂多项式。问题就是我们如何把一个下降幂多项式转换成点值表示法。

我们假设要转换的下降幂多项式为\(F\)，我们假设点值的\(\texttt{EGF}\)为\(F_1\)。

我们发现如果\(F(x)=x^{\underline {n}}\)，那我们可以得到\(F_1(x)\)为:

而我们根据定义可以得到:

又因为\(F(x)=\sum_{i=0}^{\infty} F[i]x^{\underline {i}}\)，所以我们可以得到:

于是，我们就可以直接普通多项式乘法\(\Theta(n\log n)\)求出一个下降幂多项式的点值的\(\texttt{EGF}\)，求出来之后直接乘起来。那要从点值\(\texttt{EGF}\)转换成下降幂多项式很显然直接乘上\(e^{-x}\)，这个东西用泰勒展开（显然）就是:

\(\texttt{Code}\)

#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline", "no-stack-protector", "unroll-loops")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"

#include <bits/stdc++.h>
using namespace std;

#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 1600005

int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
    int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
    return res;
}
int inv (int x){return qkpow (x,mod - 2);}

typedef vector <int> poly;

int up,w[MAXN],rev[MAXN];

void init_ntt (){
    int lim = 1 << 19;
    for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 18);
    int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
    for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
    for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}

void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
    static unsigned long long d[MAXN];
    for (Int i = 0,z = 19 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
    for (Int i = 1;i < lim;i <<= 1)
        for (Int j = 0;j < lim;j += i << 1)
            for (Int k = 0;k < i;++ k){
                int x = 1ll * d[i + j + k] * w[i + k] % mod;
                d[i + j + k] = d[j + k] + mod - x,d[j + k] += x;
            }
    for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
    if (type == -1){
        reverse (a.begin() + 1,a.begin() + lim);
        for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
    }
#undef G
#undef Gi 
}

poly operator * (poly a,poly b){
    int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
    a.resize (lim),b.resize (lim);
    ntt (a,lim,1),ntt (b,lim,1);
    for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
    ntt (a,lim,-1),a.resize (up + 1);
    return a;
}

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

poly I,Iv,A,B;
int n,m,fac[MAXN],ifac[MAXN];

signed main(){
    init_ntt (),read (n,m);up = n + m;A.resize (n + 1),B.resize (m + 1);
    for (Int i = 0;i <= n;++ i) read (A[i]);
    for (Int i = 0;i <= m;++ i) read (B[i]);
    fac[0] = 1;for (Int i = 1;i <= up;++ i) fac[i] = mul (fac[i - 1],i);
    ifac[up] = inv (fac[up]);for (Int i = up;i;-- i) ifac[i - 1] = mul (ifac[i],i);
    I.resize (up + 1);for (Int i = 0;i <= up;++ i) I[i] = ifac[i];
    A = A * I,B = B * I;
    for (Int i = 0;i <= up;++ i) A[i] = mul (mul (A[i],B[i]),fac[i]);
    Iv.resize (up + 1);for (Int i = 0;i <= up;++ i) Iv[i] = (i & 1 ? mod - ifac[i] : ifac[i]);
    A = A * Iv;
    for (Int i = 0;i <= up;++ i) write (A[i]),putchar (' ');
    putchar ('\n');
    return 0;
}

P.S.

本来没有卡过去，结果直接一波卡常就卡成\(\text {rank 3}\)了（至少在这个时候还是\(\text{rank 3}\)）。。。（雾