【Learning】多项式的一些东西

FFT

坑

---

NTT

将\(FFT\)中的单位复数根改成原根即可。

卡常版NTT模版

struct Mul
{	int Len;

	int wn[N], Lim;

	int rev[N];

	inline void getReverse(int * a)
	{
		static int rev[N];
		rev[0] = 0;
		for (register int i = 0; i < Len; i++)
		{
			rev[i] = (rev[i>>1] >> 1) | ((i&1) ? (Len >> 1) : 0);
			if (i < rev[i]) swap(a[i], a[rev[i]]);
		}
	}

	void NTT(int * a, int opt)
	{
		getReverse(a);

		for (register int i = 0; i < Len; i += 2)
		{
			int x = a[i], y = 1LL * wn[0] * a[i + 1] % mod;
			a[i] = Plus(x, y), a[i + 1] = Minus(x, y);
		}
		for (register int i = 0; i < Len; i += 4)
		{
			int x1 = a[i], y1 = 1LL * wn[0] * a[i + 2] % mod;
			int x2 = a[i + 1], y2 = 1LL * wn[Lim / 2] * a[i + 3] % mod;
			a[i] = Plus(x1, y1), a[i + 2] = Minus(x1, y1);
			a[i + 1] = Plus(x2, y2), a[i + 3] = Minus(x2, y2);
		}

		for (register int i = 8; i <= Len; i <<= 1)
		{
			int M = i >> 1;
			for (register int j = 0; j < Len; j += i)
			{
				for (register int k = 0; k < M; k += 4)
				{
					int x1 = a[j + k], y1 = 1LL * wn[Lim / M * k] * a[j + k + M] % mod;
					int x2 = a[j + k + 1], y2 = 1LL * wn[Lim / M * (k+1)] * a[j + k + M + 1] % mod;
					int x3 = a[j + k + 2], y3 = 1LL * wn[Lim / M * (k+2)] * a[j + k + M + 2] % mod;
					int x4 = a[j + k + 3], y4 = 1LL * wn[Lim / M * (k+3)] * a[j + k + M + 3] % mod;
					a[j + k] = Plus(x1, y1), a[j + k + M] = Minus(x1, y1);
					a[j + k + 1] = Plus(x2, y2), a[j + k + M + 1] = Minus(x2, y2);
					a[j + k + 2] = Plus(x3, y3), a[j + k + M + 2] = Minus(x3, y3);
					a[j + k + 3] = Plus(x4, y4), a[j + k + M + 3] = Minus(x4, y4);
				}
			}
		}

		if (opt == -1)
		{
			for (register int i = 0; i < Len; i++)
				a[i] = 1LL * a[i] * inv[Len] % mod;
			for(register int i = 1; i < Len; i++)
				if(i < Len-i) swap(a[i], a[Len-i]);
		}
	}

	inline void init()
	{
		Lim = Len;
		wn[0] = 1; wn[1] = power(7, (mod-1) / (Len << 1), mod);
		for (int i = 2; i < (Len << 1); i++) wn[i] = 1LL * wn[i-1] * wn[1] % mod;
	}

	inline int getLen(int l)
	{
		for (Len = 1; Len <= l; Len <<= 1);
		return Len;
	}
};

---

多项式求逆

也就是求

\[A(x)B(x) = 1 \pmod{x^n}
\]

假设我们已知

\[A(x)G(x) = 1 \pmod{x^{\lceil{n \over 2}\rceil}}
\]

两式相减得

\[A(x)(B(x) - G(x)) = 0 \pmod{x^{\lceil{n \over 2}\rceil}}
\]

\[B(x) - G(x) = 0 \pmod{x^{\lceil{n \over 2}\rceil}}
\]

\[B^2(x) + G^2(x) - 2B(x)G(x) = 0 \pmod{x^n}
\]

两边同时乘上\(A(x)\)

\[A(x)B^2(x) + A(x)G^2(x) = 2A(x)B(x)G(x)
\]

因为\(A(x)B(x) = 1\), 所以可以得到

\[B(x) + A(x)G^2(x) = 2G(x)
\]

\[B(x) = 2G(x) - A(x)G^2(x)
\]

就这样解决了

时间复杂度为\(T(n) = T(\frac{n}{2}) +O(n \log n)\)

代码

inline void getInv(int * A, int * B, int len)
{
	static int tmp1[N], tmp2[N];

	B[0] = power(A[0], mod-2, mod);

	for (register int k = 2; k <= len; k <<= 1)
	{
		int Len = k << 1;

		cpy(tmp1, A, k, Len);
		cpy(tmp2, B, (k >> 1), Len);

		Calc.Len = Len;

		Calc.NTT(tmp1, 1);
		Calc.NTT(tmp2, 1);
		for (int i = 0; i < Len; i++)
			tmp1[i] = Minus(Plus(tmp2[i], tmp2[i]), 1LL * tmp1[i] * tmp2[i] % mod * tmp2[i] % mod);
		Calc.NTT(tmp1, -1);

		cpy(B, tmp1, k, Len);
	}

	return;
}

---

多项式开根

也就是求

\[B^2(x) = A(x) \pmod{x^n}
\]

假设我们已知

\[G^2(x) = A(x) \pmod{x^{\lceil{n \over 2}\rceil}}
\]

两式相减得

\[B^2(x) - G^2(x) = 0 \pmod{x^{\lceil{n \over 2}\rceil}}
\]

然后平方一下

\[B^4(x) + G^4(x) - 2B^2(x)G^2(x) = 0 \pmod{x^n}
\]

\[B^4(x) + G^4(x) = 2B^2(x)G^2(x) \pmod{x^n}
\]

配一下方

\[B^4(x) + G^4(x) + 2B^2(x)G^2(x) = 4B^2(x)G^2(x) \pmod{x^n}
\]

\[(B^2(x) + G^2(x))^2 = (2B(x)G(x))^2 \pmod{x^n}
\]

\[B^2(x) + G^2(x) = 2B(x)G(x) \pmod{x^n}
\]

因为\(A(x)B(x) = 1\), 所以可以得到

\[B(x) = {A(x) + G^2(x) \over {2G(x)}}
\]

为了方便实现，我们常把它化成

\[B(x) = {A(x) \over 2G(x)} + {G(x) \over 2}
\]

然后就解决了

时间复杂度为\(T(n) = T(\frac{n}{2}) + O(n \log n) = O(n \log n)\)

---

多项式求导

模拟即可

代码

inline void getDeri(int * a, int len)
{	for (register int i = 0; i < len; i++)
		a[i] = 1LL * a[i+1] * (i+1) % mod;
}

---

多项式积分

模拟即可

代码

inline void getInte(int * a, int len)
{	for (int i = len-1; i >= 1; i--)
		a[i] = 1LL * a[i-1] * inv[i] % mod;
	a[0] = 0;
}

---

多项式求\(\ln\)

已知多项式\(A(x)\), 求\(B(x) = \ln (A(x))\)

根据链式法则， 我们可以得到

\[B'(x) = \frac{\mathbb{d}(\ln(A(x)))}{\mathbb{d}A(x)} \frac{\mathbb{d}A(x)}{\mathbb{d}x} = \frac{A'(x)}{A(x)}
\]

所以

\[B(x) = \int \frac{A'(x)}{A(x)} \mathbb{d}x
\]

代码

void getLn(int * A, int len)
{
	static int tmp1[N], tmp2[N], tmp3[N];

	int Len = len << 1;

	cpy(tmp1, A, len, Len);
	cpy(tmp2, A, len, Len);

	getDeri(tmp1, len);
	getInv(tmp2, tmp3, len);

	Calc.Len = Len;

	Calc.NTT(tmp1, 1);
	Calc.NTT(tmp3, 1);
	for (int i = 0; i < Len; i++)
		tmp1[i] = 1LL * tmp1[i] * tmp3[i] % mod;
	Calc.NTT(tmp1, -1);

	memset(tmp1 + len, 0, 4 * (Len - len));
	getInte(tmp1, len);

	cpy(A, tmp1, len, Len);
}

时间复杂度还是\(T(n) = T(\frac{n}{2}) + O(n \log n) = O(n \log n)\)

---

牛顿迭代

我们要求\(f(x) = 0\)的根

那么可以使用泰勒公式来近似\(f(x) = 0\)的根

我们设当前的近似值是\(x_{n}\)， 我们想要得到的近似值是\(x_{n+1}\)

截取泰勒公式的线性部分： $$f(x_{n}) + f'(x_{n})(x_{n+1}-x_n) = 0$$

解方程得： $$x_{n+1} = x_n - \frac{f(x_{n})}{f'(x_{n})}$$

我们可以用这种方法来解多项式方程。

---

多项式\(\exp\)

我们要求的是 \(exp(A(x))\), 这相当于解一个方程： \(g(exp(A(x))) = \ln(exp(A(x))) - A(x) = 0\)

可以直接套用牛顿迭代法求解。