「多項式指數函式」
阿新 • • 發佈:2020-12-30
「多項式指數函式」
前置知識
導數
微積分
基本問題
給定一個 \(n\) 次多項式 \(A(x)\),求 \(B(x)\) 滿足
\[B(x)\equiv e^{A(x)}(mod\;x^n) \]
兩遍同時用 \(ln\) 取對數
\[ln^{B(x)}\equiv A(x)(mod\;x^n) \]\[ln^{B(x)} - A(x)\equiv 0(mod\;x^n) \]套用牛頓迭代,設
\[G(x)=ln^{B(x)} - A(x) \]\[B(x)=B_0(x) - \frac{G(x)}{G'(x)} \]由於 \(A(x)\) 是常數,所以直接消掉不影響求導
\[G'(x)=\frac{1}{B(x)} \]最後遞迴求解即可
程式碼
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> typedef long long ll; typedef unsigned long long ull; using namespace std; const int maxn = 3e5 + 50, INF = 0x3f3f3f3f, mod = 998244353, inv3 = 332748118; inline int read () { register int x = 0, w = 1; register char ch = getchar (); for (; ch < '0' || ch > '9'; ch = getchar ()) if (ch == '-') w = -1; for (; ch >= '0' && ch <= '9'; ch = getchar ()) x = x * 10 + ch - '0'; return x * w; } inline void write (register int x) { if (x / 10) write (x / 10); putchar (x % 10 + '0'); } int n; int f[maxn]; int f0[maxn], rf0[maxn], nf0[maxn], lnf0[maxn]; int res[maxn], tmp[maxn], rev[maxn]; inline int qpow (register int a, register int b, register int ans = 1) { for (; b; b >>= 1, a = 1ll * a * a % mod) if (b & 1) ans = 1ll * ans * a % mod; return ans; } inline void NTT (register int len, register int * a, register int opt) { for (register int i = 1; i < len; i ++) if (i < rev[i]) swap (a[i], a[rev[i]]); for (register int d = 1; d < len; d <<= 1) { register int w1 = qpow (opt, (mod - 1) / (d << 1)); for (register int i = 0; i < len; i += d << 1) { register int w = 1; for (register int j = 0; j < d; j ++, w = 1ll * w * w1 % mod) { register int x = a[i + j], y = 1ll * w * a[i + j + d] % mod; a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod; } } } } inline void Poly_Inv (register int d, register int * a, register int * b) { if (d == 1) return b[0] = qpow (a[0], mod - 2), void (); Poly_Inv ((d + 1) >> 1, a, b); register int len = 1, bit = 0; while (len < (d << 1)) len <<= 1, bit ++; for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); for (register int i = 0; i < d; i ++) res[i] = a[i]; NTT (len, res, 3), NTT (len, b, 3); for (register int i = 0; i < len; i ++) b[i] = ((2ll * b[i] % mod - 1ll * res[i] * b[i] % mod * b[i] % mod) % mod + mod) % mod; NTT (len, b, inv3); register int inv = qpow (len, mod - 2); for (register int i = 0; i < d; i ++) b[i] = 1ll * b[i] * inv % mod; for (register int i = d; i < len; i ++) b[i] = 0; } inline void Poly_Ln (register int d, register int * f0, register int * lnf0) { register int len = 1, bit = 0; while (len < (d << 1)) len <<= 1, bit ++; for (register int i = 0; i < len; i ++) rf0[i] = nf0[i] = lnf0[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); for (register int i = 0; i < d; i ++) rf0[i] = 1ll * f0[i + 1] * (i + 1) % mod; Poly_Inv (d, f0, nf0); NTT (len, rf0, 3), NTT (len, nf0, 3); for (register int i = 0; i < len; i ++) lnf0[i] = 1ll * rf0[i] * nf0[i] % mod; NTT (len, lnf0, inv3); register int inv = qpow (len, mod - 2); for (register int i = 0; i < d; i ++) lnf0[i] = 1ll * lnf0[i] * inv % mod; for (register int i = d; i < len; i ++) lnf0[i] = 0; for (register int i = d - 1; i > 0; i --) lnf0[i] = 1ll * lnf0[i - 1] * qpow (i, mod - 2) % mod; lnf0[0] = 0; } inline void Poly_Exp (register int d, register int * f, register int * f0) { if (d == 1) return f0[0] = 1, void (); Poly_Exp ((d + 1) >> 1, f, f0); Poly_Ln (d, f0, lnf0); register int len = 1, bit = 0; while (len < (d << 1)) len <<= 1, bit ++; for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); for (register int i = 0; i < d; i ++) res[i] = (f[i] - lnf0[i] + mod) % mod; res[0] ++; NTT (len, f0, 3), NTT (len, res, 3); for (register int i = 0; i < len; i ++) f0[i] = 1ll * f0[i] * res[i] % mod; NTT (len, f0, inv3); register int inv = qpow (len, mod - 2); for (register int i = 0; i < d; i ++) f0[i] = 1ll * f0[i] * inv % mod; for (register int i = d; i < len; i ++) f0[i] = 0; } int main () { n = read() - 1; for (register int i = 0; i <= n; i ++) f[i] = read(); Poly_Exp (n + 1, f, f0); for (register int i = 0; i <= n; i ++) printf ("%d ", f0[i]); putchar ('\n'); return 0; }