「多項式除法」
阿新 • • 發佈:2020-12-30
「多項式除法」
前置知識
基本問題
給定一個 \(n\) 次多項式 \(F(x)\) 和一個 \(m\) 次多項式 \(G(x)\),求 \(A(x)\) 和 \(B(x)\) 滿足
\[F(x)=A(x)G(x)+B(x) \]
顯然 \(A(x)\) 的次數為 \(n - m\),\(B(x)\) 的次數 \(\leq m - 1\) 。
新定義
\[A_R(x)=x^nA(\frac{1}{x}) \]容易發現 \(A_R(x)\) 的係數是 \(A(x)\) 的係數 \(reverse\) 得到的。
簡單證明
\[A(x)=\sum^{n}_{i=0}a_ix^i \]證畢
兩邊同乘 \(x^n\)
\[x^nF(\frac{1}{x})=x^{n-m}A(\frac{1}{x})x^mG(\frac{1}{x})+x^{n-m+1}x^{m-1}B(\frac{1}{x}) \]然後通過 \(reverse\) 操作得到 \(A(x)\)
\[B(x)=F(x)-A(x)G(x) \]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, m; int f[maxn], g[maxn]; int fr[maxn], gr[maxn], ngr[maxn]; int rev[maxn], res[maxn]; int A[maxn], B[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 Division (register int * f, register int * g) { // 除法 Poly_Inv (n - m + 1, gr, ngr); register int len = 1, bit = 0; while (len <= 2 * n - m + 1) len <<= 1, bit ++; for (register int i = 0; i < len; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); NTT (len, fr, 3), NTT (len, ngr, 3); for (register int i = 0; i < len; i ++) A[i] = 1ll * fr[i] * ngr[i] % mod; NTT (len, A, inv3); register int inv = qpow (len, mod - 2); for (register int i = n - m + 1; i < len; i ++) A[i] = 0; for (register int i = n - m; i >= 0; i --) printf ("%d ", A[i] = 1ll * A[i] * inv % mod); putchar ('\n'); reverse (A, A + n - m + 1); } inline void Remainder (register int * f, register int * g) { // 取膜 register int len = 1, bit = 0; while (len <= n) len <<= 1, bit ++; for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); NTT (len, A, 3), NTT (len, g, 3); for (register int i = 0; i < len; i ++) res[i] = 1ll * A[i] * g[i] % mod; NTT (len, res, inv3); register int inv = qpow (len, mod - 2); for (register int i = 0; i <= m - 1; i ++) printf ("%lld ", ((f[i] - 1ll * res[i] * inv % mod) % mod + mod) % mod); putchar ('\n'); } int main () { n = read(), m = read(); for (register int i = 0; i <= n; i ++) f[i] = fr[n - i] = read(); for (register int i = 0; i <= m; i ++) g[i] = gr[m - i] = read(); Division (f, g), Remainder(f, g); return 0; }