- Partition NumberReference p r ( n ) p_r(n) p r ( n ) n n n r r r
1.你可以 DP p r ( n ) = { 1 n = 1   o r   r = 1 p n ( n ) n < r 1 + p n − 1 ( n ) n = r p r − 1 ( n ) + p r ( n − r ) n > r p_r(n)=\begin{cases}
1&n=1\;or\;r=1\\
p_n(n)&n<r\\
1+p_{n-1}(n)&n=r\\
p_{r-1}(n)+p_{r}(n-r)&n>r
\end{cases} p r ( n ) = ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 1 p n ( n ) 1 + p n − 1 ( n ) p r − 1 ( n ) + p r ( n − r ) n = 1 o r r = 1 n < r n = r n > r n , r > 1 n,r>1 n , r > 1
性質：把 n n n r r r p r ( n ) p_r(n) p r ( n )
另外，把 n n n r r r n n n r r r
2.我會生成函數！ G ( x ) = ∑ n = 0 ∞ p r ( n ) x n G(x)=\sum\limits_{n=0}^\infty p_r(n)x^n G ( x ) = n = 0 ∑ ∞ p r ( n ) x n p r ( n ) p_r(n) p r ( n ) x x x G ( x ) = ∑ n = 0 ∞ p r ( n ) x n = ∏ k = 1 n ( 1 + x k + x 2 k + ⋯ + x ⌊ n k ⌋ k ) G(x)=\sum\limits_{n=0}^\infty p_r(n)x^n=\prod\limits_{k=1}^{n}(1+x^k+x^{2k}+\cdots+x^{\lfloor\frac{n}{k}\rfloor k}) G ( x ) = n = 0 ∑ ∞ p r ( n ) x n = k = 1 ∏ n ( 1 + x k + x 2 k + ⋯ + x ⌊ k n ⌋ k ) ∑ n = 0 ∞ x k n = 1 1 − x k \sum\limits_{n=0}^\infty x^{kn}=\frac{1}{1-x^k} n = 0 ∑ ∞ x k n = 1 − x k 1 G ( x ) = ∏ k = 1 n 1 1 − x k G(x)=\prod\limits_{k=1}^n\frac{1}{1-x^k} G ( x ) = k = 1 ∏ n 1 − x k 1 ln \ln ln ∏ \prod ∏ ln G ( x ) = ∑ k = 1 n − ln ( 1 − x k ) \ln G(x)=\sum\limits_{k=1}^n-\ln(1-x^k) ln G ( x ) = k = 1 ∑ n − ln ( 1 − x k ) ln ( 1 − x ) \ln(1-x) ln ( 1 − x ) ln G ( x ) = ∑ k = 1 n ∑ r = 0 ∞ x r k r \ln G(x)=\sum\limits_{k=1}^n\sum\limits_{r=0}^\infty\frac{x^{rk}}{r} ln G ( x ) = k = 1 ∑ n r = 0 ∑ ∞ r x r k A ( x ) = ln G ( x ) A(x)=\ln G(x) A ( x ) = ln G ( x ) G ( x ) = e A ( x ) G(x)=e^{A(x)} G ( x ) = e A ( x ) ( m o d x n + 1 ) \pmod{x^{n+1}} ( m o d x n + 1 ) 快去秒，loj上有
3.我會廣義五邊形數！ 你真的會嘛 參考百蠹百磕
Ex.多項式求逆Reference F ( x ) F(x) F ( x ) f ( x ) = F − 1 ( x ) f(x)=F^{-1}(x) f ( x ) = F − 1 ( x ) f i + 1 ( x ) ≡ 2 f i ( x ) − F ( x ) f i 2 ( x ) ( m o d x 2 i + 1 ) f_{i+1}(x)\equiv2f_i(x)-F(x){f_i}^2(x)\pmod{x^{2^{i+1}}} f i + 1 ( x ) ≡ 2 f i ( x ) − F ( x ) f i 2 ( x ) ( m o d x 2 i + 1 )
Ex.分治FFT F ( n ) = a F ( b ) + c + ∑ i = 0 n − 1 d F ( i ) G ( n − i ) F(n)=aF(b)+c+\sum\limits_{i=0}^{n-1}dF(i)G(n-i) F ( n ) = a F ( b ) + c + i = 0 ∑ n − 1 d F ( i ) G ( n − i ) O ( n log 2 n ) O(n\log^2n) O ( n log 2 n )
對於至少是 F ( n ) = ∑ i = 0 n − 1 F ( i ) G ( n − i ) F(n)=\sum\limits_{i=0}^{n-1}F(i)G(n-i) F ( n ) = i = 0 ∑ n − 1 F ( i ) G ( n − i ) A ( x ) = ∑ i = 0 n F ( i ) x i A(x)=\sum\limits_{i=0}^nF(i)x^i A ( x ) = i = 0 ∑ n F ( i ) x i B ( x ) = ∑ i = 1 n G ( i ) x i B(x)=\sum\limits_{i=1}^nG(i)x^i B ( x ) = i = 1 ∑ n G ( i ) x i i i i A ( x ) B ( x ) = ∑ n = 1 ∞ x n ∑ i = 0 n − 1 F ( i ) G ( n − i ) = ∑ n = 1 ∞ F ( n ) x n = A ( x ) − F ( 0 ) A(x)B(x)=\sum\limits_{n=1}^\infty x^n\sum\limits_{i=0}^{n-1}F(i)G(n-i)=\sum\limits_{n=1}^\infty F(n)x^n=A(x)-F(0) A ( x ) B ( x ) = n = 1 ∑ ∞ x n i = 0 ∑ n − 1 F ( i ) G ( n − i ) = n = 1 ∑ ∞ F ( n ) x n = A ( x ) − F ( 0 ) A ( x ) ≡ ( 1 − B ( x ) ) − 1 F ( 0 ) ( m o d 2 t ) A(x)\equiv(1-B(x))^{-1}F(0)\pmod{2^t} A ( x ) ≡ ( 1 − B ( x ) ) − 1 F ( 0 ) ( m o d 2 t ) O ( n log n ) O(n\log n) O ( n log n )
Ex.多項式積分&求導
Ex.多項式求ln ln \ln ln ln ( 1 − x ) = − ∑ n = 1 ∞ x n n \ln(1-x)=-\sum\limits_{n=1}^\infty\frac{x^n}{n} ln ( 1 − x ) = − n = 1 ∑ ∞ n x n ln \ln ln ln F = ∫ F ′ F = ∫ F ′ F − 1 \ln F=\int\frac{F'}{F}=\int F'F^{-1} ln F = ∫ F F ′ = ∫ F ′ F − 1 F F F 1 1 1 ln \ln ln F F F
Ex.多項式牛頓迭代 G ( x ) G(x) G ( x ) F ( x ) F(x) F ( x ) G ( F ( x ) ) = 0 G(F(x))=0 G ( F ( x ) ) = 0
仍然考慮倍增。F 0 ( x ) F_0(x) F 0 ( x ) G ( F 0 ( x ) ) ≡ 0 ( m o d x t ) G(F_0(x))\equiv0\pmod{x^t} G ( F 0 ( x ) ) ≡ 0 ( m o d x t ) F ( x ) F(x) F ( x ) G ( F ( x ) ) ≡ 0 ( m o d x 2 i ) G(F(x))\equiv0\pmod{x^{2i}} G ( F ( x ) ) ≡ 0 ( m o d x 2 i ) G ( F ) = ∑ n = 0 ∞ G ( n ) ( F ) ⊗ ( F − F 0 ) n n ! G(F)=\sum\limits_{n=0}^\infty\frac{G^{(n)}(F)\otimes(F-F_0)^n}{n!} G ( F ) = n = 0 ∑ ∞ n ! G ( n ) ( F ) ⊗ ( F − F 0 ) n F F F F ( x ) F(x) F ( x ) F ≡ F 0 ( m o d x t ) F\equiv F_0\pmod{x^t} F ≡ F 0 ( m o d x t ) k > 1 k>1 k > 1 ( F − F 0 ) k ≡ 0 ( m o d x 2 t ) \left(F-F_0\right)^k\equiv0\pmod{x^{2t}} ( F − F 0 ) k ≡ 0 ( m o d x 2 t ) G ( F ) ≡ G ( F 0 ) + G ′ ( F 0 ) ⊗ ( F − F 0 ) ( m o d x 2 t ) G(F)\equiv G(F_0)+G'(F_0)\otimes(F-F_0)\pmod{x^{2t}} G ( F ) ≡ G ( F 0 ) + G ′ ( F 0 ) ⊗ ( F − F 0 ) ( m o d x 2 t ) x x x F ≡ F 0 − G ( F 0 ) G ′ ( F 0 ) ( m o d x 2 t ) F\equiv F_0-\frac{G(F_0)}{G'(F_0)}\pmod{x^{2t}} F ≡ F 0 − G ′ ( F 0 ) G ( F 0 ) ( m o d x 2 t ) G ′ ( F 0 ) G'(F_0) G ′ ( F 0 )
Ex.牛頓迭代 x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n )
Ex.多項式exp e x = ∑ n = 0 ∞ x n n ! e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!} e x = n = 0 ∑ ∞ n ! x n e F ( x ) = exp F ( x ) = ∑ n = 0 ∞ F n ( x ) n ! e^{F(x)}=\exp{F(x)}=\sum\limits_{n=0}^\infty\frac{F^n(x)}{n!} e F ( x ) = exp F ( x ) = n = 0 ∑ ∞ n ! F n ( x ) F F F 0 0 0 e e e e 0 e^0 e 0 exp \exp exp F F F exp ln F = F \exp\ln F=F exp ln F = F ln exp F = F \ln\exp F=F ln exp F = F ln G = F \ln G=F ln G = F ln G − F = 0 \ln G-F=0 ln G − F = 0 A ( G ) = ln G − F A(G)=\ln G-F A ( G ) = ln G − F G ≡ G 0 − A ( G 0 ) A ′ ( G 0 ) G\equiv G_0-\frac{A(G_0)}{A'(G_0)} G ≡ G 0 − A ′ ( G 0 ) A ( G 0 ) G ≡ G 0 − ln G 0 − F G 0 − 1 ≡ G 0 ( 1 − ln G 0 + F ) G\equiv G_0-\frac{\ln G_0-F}{{G_0}^{-1}}\equiv G_0(1-\ln G_0+F) G ≡ G 0 − G 0 − 1 ln G 0 − F ≡ G 0 ( 1 − ln G 0 + F ) O ( n log n ) O(n\log n) O ( n log n )
另一種做法？G ( x ) = exp F ( x ) G(x)=\exp F(x) G ( x ) = exp F ( x ) G ′ ( x ) = F ′ ( x ) exp F ( x ) G'(x)=F'(x)\exp F(x) G ′ ( x ) = F ′ ( x ) exp F ( x ) G ′ ( x ) = F ′ ( x ) G ( x ) G'(x)=F'(x)G(x) G ′ ( x ) = F ′ ( x ) G ( x ) G ( x ) = ∫ F ′ ( x ) G ( x ) G(x)=\int F'(x)G(x) G ( x ) = ∫ F ′ ( x ) G ( x ) O ( n log 2 n ) O(n\log^2 n) O ( n log 2 n )
我發現本來只是打算補一下用到的東西結果快把多項式補完了Ex.多項式快速冪 O ( n log 2 n ) O(n\log^2n) O ( n log 2 n ) G = F k G=F^k G = F k ln G = k ln F \ln G=k\ln F ln G = k ln F ln F = ln G k \ln F=\frac{\ln G}{k} ln F = k ln G exp \exp exp O ( n log n ) O(n\log n) O ( n log n ) 1 1 1 1 1 1
開 k > 0 k>0 k > 0
Ex.多項式開根
Ex.多項式帶餘除法 F ( x ) = G ( x ) Q ( x ) + R ( x ) F(x)=G(x)Q(x)+R(x) F ( x ) = G ( x ) Q ( x ) + R ( x ) F ( 1 x ) = G ( 1 x ) Q ( 1 x ) + R ( 1 x ) F(\frac{1}{x})=G(\frac{1}{x})Q(\frac{1}{x})+R(\frac{1}{x}) F ( x 1 ) = G ( x 1 ) Q ( x 1 ) + R ( x 1 ) x n F ( 1 x ) = x n G ( 1 x ) Q ( 1 x ) + x n R ( 1 x ) x^nF(\frac{1}{x})= x^nG(\frac{1}{x})Q(\frac{1}{x})+x^nR(\frac{1}{x}) x n F ( x 1 ) = x n G ( x 1 ) Q ( x 1 ) + x n R ( x 1 ) F ‾ ( x ) = G ‾ ( x ) Q ‾ ( x ) + x n − m + 1 R ( x ) \overline{F}(x)=\overline{G}(x)\overline{Q}(x)+x^{n-m+1}R(x) F ( x ) = G ( x ) Q ( x ) + x n − m + 1 R ( x ) F ‾ ( x ) ≡ G ‾ ( x ) Q ‾ ( x ) ( m o d x n − m + 1 ) \overline{F}(x)\equiv\overline{G}(x)\overline{Q}(x)\pmod{x^{n-m+1}} F ( x ) ≡ G ( x ) Q ( x ) ( m o d x n − m + 1 )
Ex.拉格朗日插值法Reference L ( x ) = ∑ j = 0 n ( y i ∏ i ≠ j x − x i x j − x i ) \frak{L}(x)=\sum\limits_{j=0}^n\left(y_i\prod\limits_{i\ne j}\frac{x-x_i}{x_j-x_i}\right) L ( x ) = j = 0 ∑ n ⎝ ⎛ y i i ̸ = j ∏ x j − x i x − x i ⎠ ⎞ x x x O ( n 2 ) O(n^2) O ( n 2 ) x i x_i x i O ( n ) O(n) O ( n )
Ex.重心拉格朗日插值法 ω j = ∏ i ≠ j ( x j − x i ) \omega_j=\prod\limits_{i\ne j}(x_j-x_i) ω j = i ̸ = j ∏ ( x j − x i ) L ( x ) = ( ∏ i = 0 n ( x − x i ) ) ( ∑ j = 0 n ω j ⋅ y j x − x j ) \frak{L}(x)=\left(\prod\limits_{i=0}^n(x-x_i)\right)\left(\sum\limits_{j=0}^n\frac{\omega_j\cdot y_j}{x-x_j}\right) L ( x ) = ( i = 0 ∏ n ( x − x i ) ) ( j = 0 ∑ n x − x j ω j ⋅ y j ) O ( n ) O(n) O ( n )
Ex.多項式多點求值 A ( a i )   ,   i ∈ [ 1 , n ] A(a_i)\;,\;i\in[1,n] A ( a i ) , i ∈ [ 1 , n ] B ( x ) = ∏ i x − x i B(x)=\prod_i x-x_i B ( x ) = ∏ i x − x i A 0 ≡ A ( m o d B 0 ) A_0\equiv A\pmod{B_0} A 0 ≡ A ( m o d B 0 ) A 1 ≡ A ( m o d B 1 ) A_1\equiv A\pmod{B_1} A 1 ≡ A ( m o d B 1 ) B 0 B_0 B 0 B 1 B_1 B 1 B B B A 0 A_0 A 0 A 1 A_1 A 1 A A A O ( n log 2 n ) O(n\log^2 n) O ( n log 2 n )
Ex.多項式快速插值 ∏ j 不 等 於 i ( x i − x j ) = lim x → x i ∏ 所 有 的 j x − x j x − x i = ( ∏ 所 有 的 j x − x j ) ∣ x = x i ′ \prod\limits_{j不等於i}(x_i-x_j)=\lim\limits_{x\to x_i}\frac{\prod\limits_{所有的j}x-x_j}{x-x_i}=\left(\prod\limits_{所有的j}x-x_j\right)|'_{x=x_i} j 不 等 於 i ∏ ( x i − x j ) = x → x i lim x − x i 所 有 的 j ∏ x − x j = ( 所 有 的 j ∏ x − x j ) ∣ x = x i ′ ∏ x − x j \prod x-x_j ∏ x − x j ∏ L R \prod\limits_{L}^R L ∏ R ∏ L M ⊗ ∏ M + 1 R \prod\limits_L^M\otimes\prod\limits_{M+1}^R L ∏ M ⊗ M + 1 ∏ R O ( n log 2 n ) O(n\log^2n) O ( n log 2 n ) x i x_i x i O ( n log 2 n ) O(n\log^2n) O ( n log 2 n )
Ex.多項式複合逆：拉格朗日反演Reference f ( g ( x ) ) = x f(g(x))=x f ( g ( x ) ) = x f f f g g g f ( g ( x ) ) = g ( f ( x ) ) f(g(x))=g(f(x)) f ( g ( x ) ) = g ( f ( x ) ) [ x k ] f ( x ) [x^k]f(x) [ x k ] f ( x ) f ( x ) f(x) f ( x ) x k x^k x k g g g f f f [ x n ] f ( x ) = 1 x [ x n − 1 ] ( x g ( x ) ) n [x^n]f(x)=\frac{1}{x}[x^{n-1}]\left(\frac{x}{g(x)}\right)^n [ x n ] f ( x ) = x 1 [ x n − 1 ] ( g ( x ) x ) n
Ex.擴展拉格朗日反演 [ x n ] h ( f ( x ) ) = 1 x [ x n − 1 ] h ′ ( x ) ( x g ( x ) ) n [x^n]h(f(x))=\frac{1}{x}[x^{n-1}]h'(x)\left(\frac{x}{g(x)}\right)^n [ x n ] h ( f ( x ) ) = x 1 [ x n − 1 ] h ′ ( x ) ( g ( x ) x ) n
Ex.Berlekamp-MasseyReference f f f g g g F F F G G G ∀ n ,   F ⊗ G ≡ f n x n ( m o d x n + 1 ) \forall n,\;F\otimes G\equiv f_nx^n\pmod{x^{n+1}} ∀ n , F ⊗ G ≡ f n x n ( m o d x n + 1 ) h h h H H H H ⊗ F ≡ ( f n + ζ ) ⋅ x n ( m o d x n + 1 ) H\otimes F\equiv (f_n+\zeta)\cdot x^n\pmod{x^{n+1}} H ⊗ F ≡ ( f n + ζ ) ⋅ x n ( m o d x n + 1 ) e e e E E E x m E ⊗ F ≡ ( f n − m + ϵ ) ⋅ x n ( m o d x n − m + 1 ) x^mE\otimes F\equiv(f_{n-m}+\epsilon)\cdot x^{n}\pmod{x^{n-m+1}} x m E ⊗ F ≡ ( f n − m + ϵ ) ⋅ x n ( m o d x n − m + 1 ) x m E ⊗ F ≡ ( f n − m + ϵ ) ⋅ x n ( m o d x n + 1 ) x^mE\otimes F\equiv(f_{n-m}+\epsilon)\cdot x^{n}\pmod{x^{n+1}} x m E ⊗ F ≡ ( f n − m + ϵ ) ⋅ x n ( m o d x n + 1 ) H ⊗ F − ζ f n + m + ϵ ⋅ x m E ⊗ F ≡ f n x n ( m o d x n + 1 ) H\otimes F-\frac{\zeta}{f_{_{n+m}}+\epsilon}\cdot x^{m} E\otimes F\equiv f_nx^n\pmod{x^{n+1}} H ⊗ F − f n + m + ϵ ζ ⋅ x m E ⊗ F ≡ f n x n ( m o d x n + 1 ) ( H − ϵ f n + m + ϵ ⋅ x m E ) ⊗ F ≡ f n x n ( m o d x n + 1 ) (H-\frac{\epsilon}{f_{_{n+m}}+\epsilon}\cdot x^mE)\otimes F\equiv f_nx^n\pmod{x^{n+1}} ( H − f n + m + ϵ ϵ ⋅ x m E ) ⊗ F ≡ f n x n ( m o d x n + 1 )
哦，E E E
Ex.Cayley-HamiltonReference f n = ∑ i = 1 m a i f n − i f_n=\sum\limits_{i=1}^m a_if_{n-i} f n = i = 1 ∑ m a i f n − i f f f m m m f y f_y f y f m = ∑ i = 1 m a i f m − i f_m=\sum\limits_{i=1}^m a_if_{m-i} f m = i = 1 ∑ m a i f m − i f y = f y f_y=f_y f y = f y f y = x y f_y=x^y f y = x y x m − ∑ i = 1 m a i x m − i x^m-\sum\limits_{i=1}^ma_ix^{m-i} x m − i = 1 ∑ m a i x m − i f y f_y f y m m m f i f_i f i O ( m ) O(m) O ( m ) f y f_y f y
實際上上面全都是口胡的因爲這個多項式全家桶我一點也沒有打
- 將 n 個小球放入 m 個盒子中的方案數Reference
