知識前備
首先有複合函數的求導公式:G(x) = F(A(x)) → G'(x) = F'(A(x))A'(x)。
ln 函數的導數:
顯然由於 \({\rm e}^{\ln(x+\delta) - \ln(x)} = \dfrac{x + \delta}x\), 那麼, \(\ln(x+\delta) - \ln(x) = \ln(\dfrac{x+\delta}x)\), 於是就有:
有個通過洛必達法則得出的結論,我沒時間學,只好記了, 那就是 t 趨於 0 的時候, ln(1 + t) 趨近於 t。
那麼上式就是:\(\lim_{\delta \to 0} \dfrac{\dfrac{\delta}x}\delta = \dfrac 1x\)。
因此 ln 函數的導函數是 \(\dfrac 1x\)。
本題
給定多項式 A(x), 要求求多項式 B(x) 滿足 B(x) ≡ ln(A(x)) mod xn。
於是求導後積分即可。
多項式求導:axn 的導函數 anxn-1, 常數項滅亡。
void Dervt(int *a, int *b, int n) {
for(int i = 0; i < n - 1; ++i) b[i] = (LL)a[i + 1] * (i + 1) % mo;
b[n - 1] = 0;
}
多項式積分,準確的說是不定積分:axn 的反導函數是 axn+1/(n+1)。
void Integ(int *a, int *b, int n) {
for(int i = 1; i < n; ++i) b[i] = (LL)a[i - 1] * Inv(i) % mo;
b[0] = 0;
}
容我複習下求逆。
void poly_inv(int deg, int *a, int *b) {
if(deg == 1) { b[0] = ksm(a[0], mo - 2); return; }
poly_inv((deg + 1) >> 1, a, b);
int len = 1; while(len < (deg << 1)) len = len << 1;
for(int i = 0; i < deg; ++i) c[i] = a[i];
for(int i = deg; i < len; ++i) c[i] = 0;
for(int i = 1; i < len; ++i) rv[i] = (rv[i>>1]>>1) | (i&1?len>>1:0);
NTT(c, len, 1), NTT(b, len, 1);
for(int i = 0; i < len; ++i) b[i] = (LL)b[i] * (2ll - (LL)c[i] * b[i] % mo) % mo;
NTT(b, len, -1);
for(int i = deg; i < len; ++i) b[i] = 0;
}
#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
#define int long long
const int N = 3e5 + 233, mo = 998244353;
int ksm(int a, int b) {
int res = 1;
for(; b; b = b>>1, a = ((LL)a * a) % mo)
if(b & 1) res = ((LL)res * a) % mo;
return res % mo;
}
const int g = 3, ig = ksm(g, mo - 2);
void Dervt(int n, int *a, int *b) {
for(int i = 0; i < n - 1; ++i) b[i] = (LL)a[i + 1] * (i + 1ll) % mo;
b[n - 1] = 0;
}
void Integ(int n, int *a, int *b) {
for(int i = 1; i < n; ++i) b[i] = (LL)a[i - 1] * ksm(i, mo - 2) % mo;
b[0] = 0;
}
int rv[N], t[N], n;
void NTT(int *a, int n, int type) {
for(int i = 0; i < n; ++i) if(i < rv[i]) swap(a[i], a[rv[i]]);
for(int m = 2; m <= n; m = m << 1) {
int w = ksm(type == 1 ? g : ig, (mo - 1) / m);
for(int i = 0; i < n; i += m) {
int tmp = 1;
for(int j = 0; j < (m >> 1); ++j) {
int p = a[i + j] % mo, q = (LL)tmp * a[i + j + (m >> 1)] % mo;
a[i + j] = (p + q) % mo, a[i + j + (m >> 1)] = (p - q + mo) % mo;
tmp = (LL)tmp * w % mo;
}
}
}
if(type == -1) {
int Inv = ksm(n, mo - 2);
for(int i = 0; i < n; ++i) a[i] = (LL)a[i] * Inv % mo;
}
}
void poly_inv(int deg, int *a, int *b) {
if(deg == 1) { b[0] = ksm(a[0], mo - 2); return; }
poly_inv((deg + 1) >> 1, a, b);
int len = 1; while(len < (deg << 1)) len = len << 1;
for(int i = 0; i < deg; ++i) t[i] = a[i];
for(int i = deg; i < len; ++i) t[i] = 0;
for(int i = 1; i < len; ++i) rv[i] = (rv[i>>1]>>1) | (i&1 ? len>>1 : 0);
NTT(b, len, 1), NTT(t, len, 1);
for(int i = 0; i < len; ++i) b[i] = (LL)b[i] * (2ll - (LL)t[i] * b[i] % mo) % mo;
NTT(b, len, -1);
for(int i = deg; i < len; ++i) b[i] = 0;
}
int b[N], c[N];
void poly_ln(int *a, int n) {
poly_inv(n, a, b), Dervt(n, a, c);
int len = 1; while(len < (n << 1)) len = len << 1;
for(int i = n; i < len; ++i) b[i] = c[i] = 0;
for(int i = 1; i < len; ++i) rv[i] = (rv[i >> 1] >> 1) | (i & 1 ? len >> 1 : 0);
NTT(b, len, 1), NTT(c, len, 1);
for(int i = 0; i < len; ++i) b[i] = (LL)b[i] * c[i] % mo;
NTT(b, len, -1);
Integ(n, b, a);
}
int a[N];
signed main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i) scanf("%lld", &a[i]), a[i] %= mo;
poly_ln(a, n);
for(int i = 0; i < n; ++i) cout << (a[i] + mo) % mo << ' ';
return 0;
}