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 ) ) ≡ 0 ( m o d x 2 t ) G(F_0(x))\equiv 0(mod\ x^{2^t}) G(F0(x))≡0(mod x2t)
牛顿迭代
F ( x ) ≡ F 0 ( x ) − G ( F 0 ( x ) ) G ′ ( F 0 ( x ) ) ( m o d x 2 t + 1 ) F(x)\equiv F_0(x)-\frac{G(F_0(x))}{G'(F_0(x))}(mod\ x^{2^{t+1}}) F(x)≡F0(x)−G′(F0(x))G(F0(x))(mod x2t+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 ) A(x) A(x)求满足 A ( x ) ∗ B ( x ) = 1 A(x)*B(x)=1 A(x)∗B(x)=1的 B ( x ) B(x) B(x)
写成
A ( x ) ∗ B ( x ) = 1 ( m o d x n ) A(x)*B(x)=1(mod \ x^n) A(x)∗B(x)=1(mod xn)
我们会求 A ( x ) ∗ B ( x ) = 1 ( m o d x 1 ) A(x)*B(x)=1(mod \ x^1) A(x)∗B(x)=1(mod x1)
然后我们考虑求 A ( x ) ∗ B ( x ) = 1 ( m o d x t ) A(x)*B(x)=1(mod \ x^t) A(x)∗B(x)=1(mod xt)
( A ( x ) ∗ B ( x ) − 1 ) 2 = 0 ( m o d x 2 t ) (A(x)*B(x)-1)^2=0(mod \ x^{2t}) (A(x)∗B(x)−1)2=0(mod x2t)
A ( x ) ( 2 B ( x ) − A ( x ) ∗ B 2 ( x ) ) = 1 ( m o d x 2 t ) A(x)(2B(x)-A(x)*B^2(x))=1(mod \ x^{2t}) A(x)(2B(x)−A(x)∗B2(x))=1(mod x2t)
把 2 B ( x ) − A ( x ) ∗ B 2 ( x ) 2B(x)-A(x)*B^2(x) 2B(x)−A(x)∗B2(x)当作新的 B B B倍增算
从 m o d x 1 mod \ x^1 mod x1倍增到大于等于 n n n
# include 多项式除法
已知 n n n次多项式 A ( x ) A(x) A(x)和 m m m次多项式 B ( x ) B(x) B(x)， n > m n>m n>m
求 A ( x ) A(x) A(x)除以 B ( x ) B(x) B(x)的商 C ( x ) C(x) C(x)及余式 D ( x ) D(x) D(x)
也就是
A ( x ) = B ( x ) C ( x ) + D ( x ) A(x)=B(x)C(x)+D(x) A(x)=B(x)C(x)+D(x)
A ( 1 x ) = B ( 1 x ) C ( 1 x ) + D ( 1 x ) A(\frac{1}{x})=B(\frac{1}{x})C(\frac{1}{x})+D(\frac{1}{x}) A(x1)=B(x1)C(x1)+D(x1)
x n A ( 1 x ) = ( x m B ( 1 x ) ) ( x n − m C ( 1 x ) ) + x n − m + 1 ( x n − 1 D ( 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})) xnA(x1)=(xmB(x1))(xn−mC(x1))+xn−m+1(xn−1D(x1))
对 x n − m + 1 x^{n-m+1} xn−m+1取模
A R ( x ) = B R ( x ) C R ( x ) ( m o d x n − m + 1 ) A^R(x)=B^R(x)C^R(x)(mod \ x^{n-m+1}) AR(x)=BR(x)CR(x)(mod xn−m+1)
R R R表示把系数倒置，即 s w a p swap swap前后
那么可以多项式求逆，再相乘得到 C R ( x ) C^R(x) CR(x)
然后去掉 R R R带入原式相乘再相减得到 D ( x ) D(x) D(x)
# include 多项式开方
给定 B ( x ) B(x) B(x)，求 A ( x ) A(x) A(x)使 A 2 ( x ) = B ( x ) A^2(x)=B(x) A2(x)=B(x)
写成
A 2 ( x ) = B ( x ) ( m o d x n ) A^2(x)=B(x) (mod \ x^n) A2(x)=B(x)(mod xn)
然后
A 2 ( x ) = B ( x ) ( m o d x 1 ) A^2(x)=B(x)(mod \ x^1) A2(x)=B(x)(mod x1)
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 ) ( m o d x 2 t ) A^2(x)=B(x)(mod \ x^{2t}) A2(x)=B(x)(mod x2t)
令
C 2 ( x ) = B ( x ) ( m o d x t ) C^2(x)=B(x)(mod \ x^t) C2(x)=B(x)(mod xt)
那么
( A 2 ( x ) − C 2 ( x ) ) 2 = 0 ( m o d x 2 t ) (A^2(x)-C^2(x))^2=0(mod \ x^{2t}) (A2(x)−C2(x))2=0(mod x2t)
A 4 ( x ) − 2 A 2 ( x ) C 2 ( x ) + C 4 ( x ) = 0 ( m o d x 2 t ) A^4(x)-2A^2(x)C^2(x)+C^4(x)=0(mod \ x^{2t}) A4(x)−2A2(x)C2(x)+C4(x)=0(mod x2t)
A 2 ( x ) = A 4 ( x ) + C 4 ( x ) 2 C 2 ( x ) ( m o d x 2 t ) A^2(x)=\frac{A^4(x)+C^4(x)}{2C^2(x)}(mod \ x^{2t}) A2(x)=2C2(x)A4(x)+C4(x)(mod x2t)
2 A 2 ( x ) = A 4 ( x ) + 2 C 2 ( x ) A 2 ( x ) + C 4 ( x ) 2 C 2 ( x ) ( m o d x 2 t ) 2A^2(x)=\frac{A^4(x)+2C^2(x)A^2(x)+C^4(x)}{2C^2(x)}(mod \ x^{2t}) 2A2(x)=2C2(x)A4(x)+2C2(x)A2(x)+C4(x)(mod x2t)
A 2 ( x ) = ( A 2 ( x ) + C 2 ( x ) ) 2 ( 2 C ( x ) ) 2 ( m o d x 2 t ) A^2(x)=\frac{(A^2(x)+C^2(x))^2}{(2C(x))^2}(mod \ x^{2t}) A2(x)=(2C(x))2(A2(x)+C2(x))2(mod x2t)
A ( x ) = B ( x ) + C 2 ( x ) 2 C ( x ) ( m o d x 2 t ) A(x)=\frac{B(x)+C^2(x)}{2C(x)}(mod \ x^{2t}) A(x)=2C(x)B(x)+C2(x)(mod x2t)
同样的还是倍增求
多项式ln
设 B ( x ) = l n ( A ( x ) ) B(x)=ln(A(x)) B(x)=ln(A(x))
同时求导
就是
B ′ ( x ) = A ′ ( x ) A ( x ) B'(x)=\frac{A'(x)}{A(x)} B′(x)=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 ) B(x)=e^{A(x)} B(x)=eA(x)
那么
l n ( B ( x ) ) = A ( x ) ln(B(x))=A(x) ln(B(x))=A(x)
牛顿迭代
得
B ( x ) ( m o d 2 t + 1 ) = B ( x ) ( m o d 2 t ) ( 1 − l n ( B ( x ) ( m o d 2 t ) ) + A ( x ) ) B(x)(mod\ 2^{t+1})=B(x)(mod\ 2^t)(1-ln(B(x)(mod\ 2^t))+A(x)) B(x)(mod 2t+1)=B(x)(mod 2t)(1−ln(B(x)(mod 2t))+A(x))
多项式求幂
设 B ( x ) = A k ( x ) B(x)=A^k(x) B(x)=Ak(x)
那么 l n ( B ( x ) ) = k l n ( A ( x ) ) ln(B(x))=kln(A(x)) ln(B(x))=kln(A(x))
然后就是求 l n ln ln然后 e x p exp exp就好了
模板！！！
COGS2189. [HZOI 2015] 帕秋莉的超级多项式
# include