「多项式除法」
前置知识
基本问题
给定一个 (n) 次多项式 (F(x)) 和一个 (m) 次多项式 (G(x))，求 (A(x)) 和 (B(x)) 满足
[F(x)=A(x)G(x)+B(x) ]
显然 (A(x)) 的次数为 (n - m)，(B(x)) 的次数 (leq m - 1) 。
新定义
[A_R(x)=x^nA(frac{1}{x})
]
容易发现 (A_R(x)) 的系数是 (A(x)) 的系数 (reverse) 得到的。
简单证明
[A(x)=sum^{n}_{i=0}a_ix^i
]
[A(frac{1}{x})=sum^{n}_{i=0}a_ix^{-i}
]
[A_R(x)=x^nA(frac{1}{x})=sum^{n}_{i=0}a_ix^{n-i}=sum^{n}_{i=0}a_{n-i}x^i
]
[A_R(x)=sum^{n}_{i=0}a_{n-i}x^i
]
证毕
[ecause F(x)=A(x)G(x)+B(x)
]
[F(frac{1}{x})=A(frac{1}{x})G(frac{1}{x})+B(frac{1}{x})
]
两边同乘 (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})
]
[F_R(x)=A_R(x)G_R(x)+x^{n-m+1}B_R(x)
]
[F_R(x)equiv A_R(x)G_R(x)(mod;x^{n-m+1})
]
[A_R(x)equiv F_R(x)G_R^{-1}(x)(mod;x^{n-m+1})
]
然后通过 (reverse) 操作得到 (A(x))
[B(x)=F(x)-A(x)G(x)
]
代码
#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 ('
');
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 ('
');
}
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;
}