導(dǎo)數(shù)卷積

$$\begin{gathered} 有f(x)=\sum _{i=0}^{n?1}{a}_i\times x^i \\ 求g(x)=\sum _{i=0}^{n?1}{f}^{(i)}(x)f^{(n?i?1)}(x) \\ \sum _{i=0}^{n?1}\sum _{j=0}^{n?1}{a}_{i+j}\frac{(i+j)!}{j!}\sum _{k=0}^{n?1}{a}_{k+(n?i?1)}\frac{(k+(n?i?1))!}{k!} \\ 設(shè)F(n)=a_n\times n! \\ 我們單獨(dú)求解g(x)的第m項(xiàng)，有g_m=\sum _{i=0}^{n?1}\sum _{j=1}^d(F(i+j)?F(n?i?1+d?j))\times (\frac{1}{j!}?\frac{1}{(d?j)!}) \\ 設(shè)H(n)=\frac{1}{n!} \end{gathered}$$

這里就可以看成是$(\sum _{i=0}^{n?1}F(i)\times F(n?1+d?i))\times (\sum _{i=0}^{d}H(i)\times H(d?i))$，也就是先對(duì)左右兩邊做卷積，然后相乘即可。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = mod + 1 >> 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex &a, Complex &b) {return Complex((1ll * a.r * b.r % mod + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans = Complex(1, 0);while (n) {if (n & 1) {ans = ans * a;}a = a * a;n >>= 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n == 0) {return 0;}if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {return -1;}uniform_int_distribution<int> r(0, mod - 1);int a = r(e);while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {a = r(e);}I2 = (1ll * a * a % mod - n + mod) % mod;int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;if(x > y) swap(x, y);return x;} }const int N = 1e6 + 10;int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * a * ans % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);} }void get_inv(int n) {inv[1] = 1;for (int i = 2; i <= n; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;} }void NTT(int *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}for (int mid = 1; mid < lim; mid <<= 1) {int wn = quick_pow(3, (mod - 1) / (mid << 1));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {int w = 1;for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;}}}if (rev == -1) {int inv = quick_pow(lim, mod - 2);reverse(f + 1, f + lim);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * inv % mod;}} }void polyinv(int *f, int *g, int n) {if (n == 1) {g[0] = quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n + 1 >> 1);for (int i = 0; i < n; i++) {t[i] = f[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;g[i] = 1ll * g[i] * cur % mod;t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }void polysqrt(int *f, int *g, int n) {if (n == 1) {g[0] = Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n + 1 >> 1);polyinv(g, b, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i < n; i++) {t[i] = f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] = t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }void derivative(int *a, int *b, int n) {for (int i = 0; i < n; i++) {b[i] = 1ll * a[i + 1] * (i + 1) % mod;} }void integrate(int *a, int n) {for (int i = n - 1; i >= 1; i--) {a[i] = 1ll * a[i - 1] * inv[i] % mod;}a[0] = 0; }void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * b[i] % mod;b[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}integrate(g, n); }void polyexp(int *f, int *g, int n) {if (n == 1) {g[0] = 1;return ;}polyexp(f, g, n + 1 >> 1);int lim = 1;while (lim < 2 * n) {lim <<= 1;}polyln(g, d, n);for (int i = 0; i < n; i++) {t[i] = (f[i] - d[i] + mod) % mod;}t[0] = (t[0] + 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * t[i] % mod;t[i] = d[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }/*b存放多項(xiàng)式逆，c存放多項(xiàng)式開根，d存放多項(xiàng)式對(duì)數(shù)ln，e存放多項(xiàng)式指數(shù)exp，t作為中間轉(zhuǎn)移數(shù)組,如果要用到polyln，得提前調(diào)用get_inv(n)先預(yù)先得到我們想要得到的逆元范圍。 */int A[N], B[N], f[N], invf[N], n;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);f[0] = invf[0] = 1;for (int i = 1; i < N; i++) {f[i] = 1ll * f[i - 1] * i % mod;}invf[N - 1] = quick_pow(f[N - 1], mod - 2);for (int i = N - 2; i >= 0; i--) {invf[i] = 1ll * invf[i + 1] * (i + 1) % mod;}scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &A[i]);A[i] = 1ll * A[i] * f[i] % mod;B[i] = invf[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(A, lim, 1);NTT(B, lim, 1);for (int i = 0; i < lim; i++) {A[i] = 1ll * A[i] * A[i] % mod;B[i] = 1ll * B[i] * B[i] % mod;}NTT(A, lim, -1);NTT(B, lim, -1);for (int i = 0; i < n; i++) {printf("%lld%c", 1ll * A[n - 1 + i] * B[i] % mod, i + 1 == n ? '\n' : ' ');}return 0; } 創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎(jiǎng)勵(lì)來(lái)咯，堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎(jiǎng)

總結(jié)

以上是生活随笔為你收集整理的导数卷积 （NTT）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。