一.一些约定.
对于接下来的所有操作，我们都是给出 F ( x ) F(x) F(x)，求 G ( x ) = T ( F ( x ) ) G(x)=T(F(x)) G(x)=T(F(x))，其中 T ( ) T() T()是操作函数，我们的目标是在 O ( n 2 ) O(n^{2}) O(n2)的时间复杂度内完成所有操作.
多项式
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fi,gi.
二.多项式求逆.
对于一个多项式 F ( x ) F(x) F(x)，我们要求 G ( x ) G(x) G(x)满足 F ( x ) G ( x ) = 1 F(x)G(x)=1 F(x)G(x)=1.
那么有：
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G(x)F(x)=1\\ \sum_{j=0}^{i}g_jf_{i-j}=0\\ g_i=-\frac{1}{f_0}\sum_{j=0}^{i-1}g_{j}f_{i-j}
G(x)F(x)=1j=0∑igjfi−j=0gi=−f01j=0∑i−1gjfi−j
其中初项
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g0=f01.
三.多项式ln.
对于一个常数项为 0 0 0的多项式 F ( x ) F(x) F(x)，我们要求 G ( x ) G(x) G(x)满足 G ( x ) = ln ( 1 + F ( x ) ) G(x)=\ln (1+F(x)) G(x)=ln(1+F(x)).
那么有：
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G(x)=\ln(1+F(x))\\ G'(x)=\frac{F'(x)}{1+F(x)}\\ G'(x)(1+F(x))=F'(x)\\ G'(x)=F'(x)-G'(x)F(x)\\ (i+1)g_{i+1}=(i+1)f_{i+1}-\sum_{j=0}^{i-1}(j+1)g_{j+1}f_{i-j}\\ g_{i+1}=f_{i+1}-\frac{1}{i+1}\sum_{j=0}^{i-1}(j+1)g_{j+1}f_{i-j}\\ g_i=f_i-\frac{1}{i}\sum_{j=1}^{i-1}jg_{j}f_{i-j}
G(x)=ln(1+F(x))G′(x)=1+F(x)F′(x)G′(x)(1+F(x))=F′(x)G′(x)=F′(x)−G′(x)F(x)(i+1)gi+1=(i+1)fi+1−j=0∑i−1(j+1)gj+1fi−jgi+1=fi+1−i+11j=0∑i−1(j+1)gj+1fi−jgi=fi−i1j=1∑i−1jgjfi−j
其中初项
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四.多项式exp.
对于一个常数项为 0 0 0的多项式 F ( x ) F(x) F(x)，我们要求 G ( x ) G(x) G(x)满足 G ( x ) = e F ( x ) G(x)=e^{F(x)} G(x)=eF(x).
那么有：
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G(x)=e^{F(x)}\\ G'(x)=e^{F(x)}F'(x)\\ G'(x)=F'(x)G(x)\\ (i+1)g_{i+1}=\sum_{j=0}^{i}(j+1)f_{j+1}g_{i-j}\\ g_{i+1}=\frac{1}{i+1}\sum_{j=0}^{i}(j+1)f_{j+1}g_{i-j}\\ g_{i}=\frac{1}{i}\sum_{j=1}^{i}jf_{j}g_{i-j}
G(x)=eF(x)G′(x)=eF(x)F′(x)G′(x)=F′(x)G(x)(i+1)gi+1=j=0∑i(j+1)fj+1gi−jgi+1=i+11j=0∑i(j+1)fj+1gi−jgi=i1j=1∑ijfjgi−j
其中初项
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g0=1.
五.多项式开方.
对于一个多项式 F ( x ) F(x) F(x)，我们要求 G ( x ) G(x) G(x)满足 G 2 ( x ) = F ( x ) G^{2}(x)=F(x) G2(x)=F(x).
那么有：
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G^{2}(x)=F(x)\\ \sum_{j=0}^{i}g_{j}g_{i-j}=f_i\\ 2g_{0}g_{i}=f_i-\sum_{j=1}^{i-1}g_{j}g_{i-j}\\ g_i=\frac{1}{2g_0}\left(f_i-\sum_{j=1}^{i-1}g_{j}g_{i-j}\right)
G2(x)=F(x)j=0∑igjgi−j=fi2g0gi=fi−j=1∑i−1gjgi−jgi=2g01(fi−j=1∑i−1gjgi−j)
其中初项
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g0=f0.
六.多项式求幂.
对于一个多项式 F ( x ) F(x) F(x)，我们要求 G ( x ) G(x) G(x)满足 G ( x ) = F k ( x ) G(x)=F^{k}(x) G(x)=Fk(x)，其中 k k k是一个常数.
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G(x)=F^{k}(x)\\ G'(x)=kF'(x)F^{k-1}(x)\\ G'(x)=kF'(x)\frac{G(x)}{F(x)}\\ G'(x)F(x)=kG(x)F'(x)\\ \sum_{j=0}^{i}(j+1)g_{j+1}f_{i-j}=k\sum_{j=0}^{i}g_{j}(i-j+1)f_{i-j+1}\\ (i+1)f_{0}g_{i+1}=k\sum_{j=0}^{i}g_{j}(i-j+1)f_{i-j+1}-\sum_{j=0}^{i-1}(j+1)g_{j+1}f_{i-j}\\ (i+1)f_{0}g_{i+1}=k\sum_{j=0}^{i}(i-j+1)g_{j}f_{i-j+1}-\sum_{j=1}^{i}jg_{j}f_{i-j+1}\\ (i+1)f_{0}g_{i+1}=k(i+1)g_0f_{i+1}+\sum_{j=1}^{i}g_{j}f_{i-j+1}(ki-kj+k-j)\\ g_{i+1}=\frac{1}{f_0}\left(kg_0f_{i+1}+\frac{1}{i+1}\sum_{j=1}^{i}g_jf_{i-j+1}(ki-kj+k-j)\right)\\ g_{i}=\frac{1}{f_0}\left(kg_0f_{i}+\frac{1}{i}\sum_{j=1}^{i-1}g_{j}f_{i-j}(ki-kj-j)\right)
G(x)=Fk(x)G′(x)=kF′(x)Fk−1(x)G′(x)=kF′(x)F(x)G(x)G′(x)F(x)=kG(x)F′(x)j=0∑i(j+1)gj+1fi−j=kj=0∑igj(i−j+1)fi−j+1(i+1)f0gi+1=kj=0∑igj(i−j+1)fi−j+1−j=0∑i−1(j+1)gj+1fi−j(i+1)f0gi+1=kj=0∑i(i−j+1)gjfi−j+1−j=1∑ijgjfi−j+1(i+1)f0gi+1=k(i+1)g0fi+1+j=1∑igjfi−j+1(ki−kj+k−j)gi+1=f01(kg0fi+1+i+11j=1∑igjfi−j+1(ki−kj+k−j))gi=f01(kg0fi+i1j=1∑i−1gjfi−j(ki−kj−j))
其中初项 g 0 = f 0 k g_0=f_0^{k} g0=f0k.
对于 f 0 = 0 f_0=0 f0=0的情况，我们可以先把最后面的几个 0 0 0去掉，算完之后再添加回去.