導(dǎo)數(shù)卷積

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題目描述

題解

參考了一下標程的推導(dǎo)過程,因為這個推導(dǎo)對我這種數(shù)學(xué)弱渣真的有點難鴨.

[1]$f(x)$的$i$次導(dǎo)函數(shù):

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

[2]$f(x)$的$n?i?1$次導(dǎo)函數(shù):

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

對于積式$f^{(i)}(x)f^{(n?1?i)}(x)$結(jié)果中,次數(shù)為$d$的項$x^d$前的系數(shù)應(yīng)該是:

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

令$F_i=a_i?i!$

換種寫法:

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

對所有的積式求和得到:

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

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

在上個式子中必須要滿足$n?1+d?i\le n?1$, 即,$d\le i$ ,不然$F_{n?1+d?i}$將變成$0$,不過對答案沒有影響.

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

因此上個式子就是,

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

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

代碼

#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; }

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