导数卷积

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题目描述

题解

参考了一下标程的推导过程,因为这个推导对我这种数学弱渣真的有点难鸭.

[1]$f(x)$的$i$次导函数:

$f^{(i)}(x)=a_i\ast \frac{i!}{0!}+a_{i+1}\ast \frac{(i+1)!}{1!}\ast x^1+...+a_{n-1}\ast \frac{(n-1)!}{(n-1-i)!}\ast x^{n-1-i}$

[2]$f(x)$的$n-i-1$次导函数:

$f^{(n-i-1)}(x)=a_{n-i-1}\ast \frac{(n-1-i)!}{0!}+a_{n-i}\ast \frac{(n-i)!}{1!}\ast x^1+...+a_{n-1}\ast \frac{(n-1)!}{i!}\ast x^i$

对于积式$f^{(i)}(x)f^{(n-1-i)}(x)$结果中,次数为$d$的项$x^d$前的系数应该是:

${\sum }_{t=0}^d{a}_{i+t}\ast \frac{(i+t)!}{t!}\ast a_{n-i-1+(d-t)}\ast \frac{(n-i-1+(d-t))!}{(d-t)!}$

令$F_i=a_i\ast i!$

换种写法:

${\sum }_{t=0}^d{F}_{i+t}\ast \frac{1}{t!}\ast F_{n-1+d-(i+t)}\ast \frac{1}{(d-t)!}$

对所有的积式求和得到:

${\sum }_{i=0}^{n-1}{\sum }_{t=0}^d{F}_{i+t}\ast \frac{1}{t!}\ast F_{n-1+d-(i+t)}\ast \frac{1}{(d-t)!}$

$={\sum }_{i=0}^{n-1+d}{F}_i\ast F_{n-1+d-i}\ast {\sum }_{j=0}^d\frac{1}{j!(d-j)!}$

在上个式子中必须要满足$n-1+d-i\le n-1$, 即,$d\le i$ ,不然$F_{n-1+d-i}$将变成$0$,不过对答案没有影响.

$j$的取值范围是$j:[0,min(d,i)]$,也就是$j:[0,d]$.

因此上个式子就是,

$={\sum }_{i=0}^{n-1+d}{F}_i\ast F_{n-1+d-i}\ast \frac{2^d}{d!}=\frac{2^d}{d!}\ast {\sum }_{i=0}^{n-1+d}{F}_i\ast F_{n-1+d-i}$

对后面的部分${\sum }_{i=0}^{n-1+d}{F}_i\ast F_{n-1+d-i}$直接套用$NTT$来求就可以了.

代码

#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;
#define rep(i,a,b) for(int i = a;i <= b;++i)
typedef long long LL;
const int N = 1 << 20;
const int P = 998244353;
const int G = 3;
const int NUM = 20;LL  wn[NUM];
LL  a[N], b[N];LL quick_mod(LL a, LL b, LL m)
{LL ans = 1;a %= m;while(b){if(b & 1){ans = ans * a % m;b--;}b >>= 1;a = a * a % m;}return ans;
}void GetWn()
{for(int i = 0; i < NUM; i++){int t = 1 << i;wn[i] = quick_mod(G, (P - 1) / t, P);}
}
void Rader(LL a[], int len)
{int j = len >> 1;for(int i = 1; i < len - 1; i++){if(i < j) swap(a[i], a[j]);int k = len >> 1;while(j >= k){j -= k;k >>= 1;}if(j < k) j += k;}
}void NTT(LL a[], int len, int on)
{Rader(a, len);int id = 0;for(int h = 2; h <= len; h <<= 1){id++;for(int j = 0; j < len; j += h){LL w = 1;for(int k = j; k < j + h / 2; k++){LL u = a[k] % P;LL t = w * a[k + h / 2] % P;a[k] = (u + t) % P;a[k + h / 2] = (u - t + P) % P;w = w * wn[id] % P;}}}if(on == -1){for(int i = 1; i < len / 2; i++)swap(a[i], a[len - i]);LL inv = quick_mod(len, P - 2, P);for(int i = 0; i < len; i++)a[i] = a[i] * inv % P;}
}void Conv(LL a[], LL b[], int n)
{NTT(a, n, 1);NTT(b, n, 1);for(int i = 0; i < n; i++)a[i] = a[i] * b[i] % P;NTT(a, n, -1);
}
int n;
LL Fac[N],iFac[N];
int main()
{GetWn();Fac[0] = 1;rep(i,1,N-1) Fac[i] = Fac[i-1] * i % P;rep(i,0,N-1)iFac[i] = quick_mod(Fac[i],P-2,P);ios::sync_with_stdio(false);cin >> n;rep(i,0,n-1) {std::cin >> a[i];a[i] = b[i] = a[i] * Fac[i] % P;}int len = 1;while(len < n) len <<= 1;len <<= 1;Conv(a,b,len);rep(i,0,n-1) {std::cout << a[n-1+i] * quick_mod(2,i,P) % P * iFac[i] % P << " ";}return 0;
}