牛顿迭代，多项式求逆，除法，开方，exp，ln，求幂

牛顿迭代

若

\[G(F_0(x))\equiv 0(mod\ x^{2^t}) \]

牛顿迭代

\[F(x)\equiv F_0(x)-\frac{G(F_0(x))}{G'(F_0(x))}(mod\ x^{2^{t+1}}) \]

以下多数都可以牛顿迭代公式一步得到

多项式求逆

给定$A(x)$求满足$A(x)*B(x)=1$的$B(x)$

写成

\[A(x)*B(x)=1(mod \ x^n) \]

我们会求$$A(x)*B(x)=1(mod \ x^1)$$

然后我们考虑求$$A(x)*B(x)=1(mod \ x^t)$$

\[(A(x)*B(x)-1)^2=0(mod \ x^{2t}) \]

\[A(x)(2B(x)-A(x)*B^2(x))=1(mod \ x^{2t}) \]

把$2B(x)-A(x)*B^2(x)$当作新的$B$倍增算

从$mod \ x^1$倍增到大于等于$n$

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e5 + 5);
const int Zsy(998244353);
const int Phi(998244352);
const int G(3);

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

IL int Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; x = x * x % Zsy, y >>= 1)
		if(y & 1) ret = ret * x % Zsy;
	return ret;
}

int N, r[_], l, A[_], B[_];

IL void NTT(RG int *P, RG int opt){
	for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);
	for(RG int i = 1; i < N; i <<= 1){
		RG int wn = Pow(G, Phi / (i << 1));
		if(opt == -1) wn = Pow(wn, Zsy - 2);
		for(RG int j = 0, p = i << 1; j < N; j += p)
			for(RG int w = 1, k = 0; k < i; w = 1LL * w * wn % Zsy, ++k){
				RG int X = P[k + j], Y = 1LL * w * P[k + j + i] % Zsy;
				P[k + j] = (X + Y) % Zsy, P[k + j + i] = (X - Y + Zsy) % Zsy;
			}
	}
}

IL void Mul(RG int *a, RG int *b, RG int len){
	for(l = 0, N = 1, len += len - 1; N <= len; N <<= 1) ++l;
	for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for(RG int i = 0; i < N; ++i) A[i] = B[i] = 0;
	for(RG int i = 0; i <= len >> 1; ++i) A[i] = a[i], B[i] = b[i];
	NTT(A, 1), NTT(B, 1);
	for(RG int i = 0; i < N; ++i) A[i] = 1LL * A[i] * B[i] % Zsy * B[i] % Zsy;
	NTT(A, -1);
	RG int inv = Pow(N, Zsy - 2);
	for(RG int i = 0; i <= len; ++i) A[i] = 1LL * A[i] * inv % Zsy;
}

int n, a[_], b[_], c[_], m;

int main(RG int argc, RG char *argv[]){
	n = Input();
	for(RG int i = 0; i < n; ++i) a[i] = Input() % Zsy;
	for(m = 1; m < n; m <<= 1);
	b[0] = Pow(a[0], Zsy - 2);
	for(RG int t = 2; t <= m; t <<= 1){
		for(RG int i = 0; i < t; ++i) c[i] = b[i];
		Mul(a, b, t);
		for(RG int i = 0; i < t; ++i)
			b[i] = ((c[i] + c[i]) % Zsy - A[i] + Zsy) % Zsy;
	}
	for(RG int i = 0; i < n; ++i) printf("%d ", b[i]);
	return puts(""), 0;
}

多项式除法

已知$n$次多项式$A(x)$和$m$次多项式$B(x)$，$n>m$
求$A(x)$除以$B(x)$的商$C(x)$及余式$D(x)$

也就是

\[A(x)=B(x)C(x)+D(x) \]

\[A(\frac{1}{x})=B(\frac{1}{x})C(\frac{1}{x})+D(\frac{1}{x}) \]

\[x^nA(\frac{1}{x})=(x^mB(\frac{1}{x}))(x^{n-m}C(\frac{1}{x}))+x^{n-m+1}(x^{n-1}D(\frac{1}{x})) \]

对$x^{n-m+1}$取模

\[A^R(x)=B^R(x)C^R(x)(mod \ x^{n-m+1}) \]

$R$表示把系数倒置，即$swap$前后
那么可以多项式求逆，再相乘得到$C^R(x)$
然后去掉$R$带入原式相乘再相减得到$D(x)$

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int mod(998244353);
const int phi(998244352);
const int maxn(8e5 + 5);
const int g(3);

IL int Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; y >>= 1, x = x * x % mod)
		if(y & 1) ret = ret * x % mod;
	return ret;
}

int x[maxn], y[maxn], z[maxn], s[maxn], c[maxn], n, m, a[maxn], b[maxn], r[maxn], len;

IL void NTT(RG int *p, RG int m, RG int opt){
	RG int l = 0, n = 1;
	for(; n < m; n <<= 1) ++l;
	for(RG int i = 0; i < n; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for(RG int i = 0; i < n; ++i) if(r[i] < i) swap(p[r[i]], p[i]);
	for(RG int i = 1; i < n; i <<= 1){
		RG int wn = Pow(g, phi / (i << 1));
		if(opt == -1) wn = Pow(wn, mod - 2);
		for(RG int j = 0, t = i << 1; j < n; j += t)
			for(RG int k = 0, w = 1; k < i; ++k, w = 1LL * w * wn % mod){
				RG int x = p[j + k], y = 1LL * w * p[j + k + i] % mod;
				p[j + k] = (x + y) % mod, p[j + k + i] = (x - y + mod) % mod;
			}
		}
	if(opt == -1){
		RG int inv = Pow(n, mod - 2);
		for(RG int i = 0; i < n; ++i) p[i] = 1LL * p[i] * inv % mod;
	}
}

IL void Inv(RG int *p, RG int *q, RG int num){
	if(num == 1){
		q[0] = Pow(p[0], mod - 2);
		return;
	}
	Inv(p, q, num >> 1);
	for(RG int i = 0; i < num; ++i) a[i] = p[i], b[i] = q[i];
	RG int l = num << 1;
	NTT(a, l, 1), NTT(b, l, 1);
	for(RG int i = 0; i < l; ++i) a[i] = 1LL * a[i] * b[i] % mod * b[i] % mod;
	NTT(a, l, -1);
	for(RG int i = 0; i < num; ++i) q[i] = ((2 * q[i] - a[i]) % mod + mod) % mod;
	for(RG int i = 0; i < l; ++i) a[i] = b[i] = 0;
}

int main(){
	n = Input(), m = Input();
	for(RG int i = 0; i <= n; ++i) y[i] = Input();
	for(RG int i = 0; i <= m; ++i) x[i] = Input();
	reverse(x, x + m + 1), reverse(y, y + n + 1);
	for(RG int i = 0; i <= n - m; ++i) s[i] = x[i];
	for(len = 1; len <= n - m; len <<= 1);
	Inv(s, z, len);
	for(len = 1; len <= (n - m) << 1; len <<= 1);
	for(RG int i = 0; i <= n - m; ++i) a[i] = y[i];
	for(RG int i = 0; i <= n - m; ++i) b[i] = z[i];
	len <<= 1;
	NTT(a, len, 1), NTT(b, len, 1);
	for(RG int i = 0; i < len; ++i) b[i] = 1LL * b[i] * a[i] % mod;
	NTT(b, len, -1), reverse(b, b + n - m + 1);
	for(RG int i = 0; i <= n - m; ++i) c[i] = b[i], printf("%d ", c[i]);
	puts(""), reverse(y, y + n + 1), reverse(x, x + m + 1);
	for(RG int i = 0; i < len; ++i) a[i] = b[i] = 0;
	for(len = 1; len <= n; len <<= 1);
	for(RG int i = 0; i <= n; ++i) a[i] = x[i];
	for(RG int i = 0; i <= n - m; ++i) b[i] = c[i];
	NTT(a, len, 1), NTT(b, len, 1);
	for(RG int i = 0; i < len; ++i) a[i] = 1LL * a[i] * b[i] % mod;
	NTT(a, len, -1);
	for(RG int i = 0; i < m; ++i) y[i] = (y[i] - a[i] + mod) % mod, printf("%d ", y[i]);
	return 0;
}

多项式开方

给定$B(x)$，求$A(x)$使$A^2(x)=B(x)$
写成

\[A^2(x)=B(x) (mod \ x^n) \]

然后

\[A^2(x)=B(x)(mod \ x^1) \]

是可以求的，好像是什么二次剩余
~~留坑在这里以后补~~

考虑求

\[A^2(x)=B(x)(mod \ x^{2t}) \]

令

\[C^2(x)=B(x)(mod \ x^t) \]

那么

\[(A^2(x)-C^2(x))^2=0(mod \ x^{2t}) \]

\[A^4(x)-2A^2(x)C^2(x)+C^$(x)=0(mod \ x^{2t}) \]

\[A^2(x)=\frac{A^4(x)+C^4(x)}{2C^2(x)}(mod \ x^{2t}) \]

\[2A^2(x)=\frac{A^4(x)+2C^2(x)A^2(x)+C^4(x)}{2C^2(x)}(mod \ x^{2t}) \]

\[A^2(x)=\frac{(A^2(x)+C^2(x))^2}{(2C(x))^2}(mod \ x^{2t}) \]

\[A(x)=\frac{B(x)+C^2(x)}{2C(x)}(mod \ x^{2t}) \]

同样的还是倍增求

多项式ln

设$B(x)=ln(A(x))$
同时求导
就是

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

那么多项式求导，然后多项式求逆，然后多项式积分就好了

多项式exp

设$B(x)=e^{A(x)}$
那么

\[ln(B(x))=A(x) \]

牛顿迭代
得

\[B(x)(mod\ 2^{t+1})=B(x)(mod\ 2^t)(1-ln(B(x)(mod\ 2^t))+A(x)) \]

多项式求幂

设$B(x)=A^k(x)$

那么$ln(B(x))=kln(A(x))$

然后就是求$ln$然后$exp$就好了

模板！！！

COGS2189. [HZOI 2015] 帕秋莉的超级多项式

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
 
IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

const int maxn(8e5 + 5);
const int mod(998244353);
const int phi(998244352);
const int inv2(499122177);
const int g(3);

int a[maxn], b[maxn], c[maxn], d[maxn], r[maxn];

IL int Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; y >>= 1, x = x * x % mod)
		if(y & 1) ret = ret * x % mod;
	return ret;
}

IL void NTT(RG int *p, RG int m, RG int opt){
	RG int n = 1, l = 0;
	for(; n < m; n <<= 1) ++l;
	for(RG int i = 0; i < n; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for(RG int i = 0; i < n; ++i) if(r[i] < i) swap(p[i], p[r[i]]);
	for(RG int i = 1; i < n; i <<= 1){
		RG int t = i << 1, wn = Pow(g, phi / t);
		if(opt == -1) wn = Pow(wn, mod - 2);
		for(RG int j = 0; j < n; j += t)
			for(RG int k = 0, w = 1; k < i; ++k, w = 1LL * w * wn % mod){
				RG int x = p[k + j], y = 1LL * w * p[k + j + i] % mod;
				p[k + j] = (x + y) % mod, p[k + j + i] = (x - y + mod) % mod;
			}
	}
	if(opt == -1){
		RG int inv = Pow(n, mod - 2);
		for(RG int i = 0; i < n; ++i) p[i] = 1LL * p[i] * inv % mod;
	}
}

IL void Inv(RG int *p, RG int *q, RG int len){
	if(len == 1){
		q[0] = Pow(p[0], mod - 2);
		return;
	}
	Inv(p, q, len >> 1);
	for(RG int i = 0; i < len; ++i) a[i] = p[i], b[i] = q[i];
	RG int tmp = len << 1;
	NTT(a, tmp, 1), NTT(b, tmp, 1);
	for(RG int i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * b[i] % mod * b[i] % mod;
	NTT(a, tmp, -1);
	for(RG int i = 0; i < len; ++i) q[i] = ((2 * q[i] - a[i]) % mod + mod) % mod;
	for(RG int i = 0; i < tmp; ++i) a[i] = b[i] = 0;
}

IL void Sqrt(RG int *p, RG int *q, RG int len){
	if(len == 1){
		q[0] = sqrt(p[0]); //???
		return;
	}
	Sqrt(p, q, len >> 1), Inv(q, c, len);	
	RG int tmp = len << 1;
	for(RG int i = 0; i < len; ++i) a[i] = p[i];
	NTT(a, tmp, 1), NTT(c, tmp, 1);
	for(RG int i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * c[i] % mod;
	NTT(a, tmp, -1);
	for(RG int i = 0; i < len; ++i) q[i] = 1LL * (q[i] + a[i]) % mod * inv2 % mod;
	for(RG int i = 0; i < tmp; ++i) a[i] = c[i] = 0;
}

IL void ICalc(RG int *p, RG int *q, RG int len){
	q[len - 1] = 0;
	for(RG int i = 1; i < len; ++i) q[i - 1] = 1LL * p[i] * i % mod;
}

IL void Calc(RG int *p, RG int *q, RG int len){
	q[0] = 0;
	for(RG int i = 1; i < len; ++i) q[i] = 1LL * Pow(i, mod - 2) * p[i - 1] % mod;
}

IL void Ln(RG int *p, RG int *q, RG int len){
	Inv(p, c, len), ICalc(p, a, len);
	RG int tmp = len << 1;
	NTT(c, tmp, 1), NTT(a, tmp, 1);
	for(RG int i = 0; i < tmp; ++i) c[i] = 1LL * c[i] * a[i] % mod;
	NTT(c, tmp, -1), Calc(c, q, len);
	for(RG int i = 0; i < tmp; ++i) a[i] = c[i] = 0;
}

IL void Exp(RG int *p, RG int *q, RG int len){
	if(len == 1){
		q[0] = 1;
		return;
	}
	Exp(p, q, len >> 1), Ln(q, b, len);
	for(RG int i = 0; i < len; ++i) b[i] = (mod - b[i] + p[i]) % mod, c[i] = q[i];
	b[0] = (b[0] + 1) % mod;
	RG int tmp = len << 1;
	NTT(b, tmp, 1), NTT(c, tmp, 1);
	for(RG int i = 0; i < tmp; ++i) b[i] = 1LL * b[i] * c[i] % mod;
	NTT(b, tmp, -1);
	for(RG int i = 0; i < len; ++i) q[i] = b[i];
	for(RG int i = 0; i < tmp; ++i) b[i] = c[i] = 0;
}

IL void CalcPow(RG int *p, RG int *q, RG int len, RG int y){
	Ln(p, d, len);
	for(RG int i = 0; i < len; ++i) d[i] = 1LL * d[i] * y % mod;
	Exp(d, q, len);
	for(RG int i = 0; i < len; ++i) d[i] = 0;
}

int f[maxn], h[maxn], n, k, len;

IL void Record(){
	for(RG int i = 0; i < len; ++i) f[i] = h[i], h[i] = 0;
}

int main(){
	freopen("polynomial.in", "r", stdin);
	freopen("polynomial.out", "w", stdout);
	n = Input() - 1, k = Input();
	for(RG int i = 0; i <= n; ++i) f[i] = Input();
	for(len = 1; len <= n; len <<= 1);
	Sqrt(f, h, len), Record();
	Inv(f, h, len), Record();
	Calc(f, h, n + 1), Record();
	Exp(f, h, len), Record();
	Inv(f, h, len), Record();
	f[0] = (f[0] + 1) % mod;
	Ln(f, h, len), Record();
	f[0] = (f[0] + 1) % mod;
	CalcPow(f, h, len, k), Record();
	ICalc(f, h, n + 1);
	for(RG int i = 0; i <= n; ++i) printf("%d ", h[i]);
    return 0;
}

posted @ 2018-03-25 08:02 Cyhlnj 阅读(918) 评论(0) 编辑收藏举报

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AFO

牛顿迭代，多项式求逆，除法，开方，exp，ln，求幂

牛顿迭代

多项式求逆

多项式除法

多项式开方

多项式ln

多项式exp

多项式求幂

模板！！！

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