由于本蒟蒻比较差，每一份代码都是交$AC$了才来的，所以不用怀疑它的真实行，当然代码和模板都不是luogu上面加强版，所以常数项恰好都是$1$，温馨提醒：不要乱用

多项式的逆元

理论推导

$$知识储备$$

$(a+b)\%c=(a\%c+b\%c)\%c$
$(a-b)\%c=(a\%c-b\%c+c)\%c$
$(a\ast b)\%c=(a\%c)\ast (b\%c)\%c$
也就是说如果你的算法只使用了加法、减法和乘法，你可以在所有的中间步骤取模，这和只对答案取模是等效的

同余的一些运算
$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}m),c\equiv d(\mathrm{m}\mathrm{o}\mathrm{d}m)=>a\ast c\equiv b\ast d(\mathrm{m}\mathrm{o}\mathrm{d}m)$
$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}m),c\equiv d(\mathrm{m}\mathrm{o}\mathrm{d}m)=>a\pm c\equiv b\pm d(\mathrm{m}\mathrm{o}\mathrm{d}m)$
$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}m)=>ka\equiv kb(\mathrm{m}\mathrm{o}\mathrm{d}km)$
$ac\equiv bc(\mathrm{m}\mathrm{o}\mathrm{d}m)=>a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}\frac{m}{(c,m)})$

---

接下来进入正题：
设给定的多项式为$A(x)$，求$A^{-1}(x)$满足：👇
$$A(x)\ast A^{-1}(x)\equiv 1(mod{x}^n)$$
$mod{x}^n$的意思即为舍去多项式里次数$\ge n$的项，只保留次数$\in [0,n-1]$的项
首先如果只有常数项，我们可以直接快速幂算出逆元（利用费马小定理）

假设已知$A(x)$在关于$mod{x}^{\lceil \frac{n}{2}\rceil }$意义下的逆元$B_0(x)$，满足
$$A(x)\ast B_0(x)\equiv 1(mod{x}^{\lceil \frac{n}{2}\rceil })$$
记要求的答案为$B(x)$，有
$$A(x)\ast B(x)\equiv 1(mod{x}^n)$$
两式相减，则有
$$A(x)\ast (B(x)-B_0(x))\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$$
因为$A(x)$肯定不为$0$，所以上述式子要成立的话只能寄希望于$(B(x)-B_0(x))$，所以化简得
$$B(x)-B_0(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$$
我们怎么把$mod{x}^{\lceil \frac{n}{2}\rceil }$升级到$mod{x}^n$，很简单 将式子平方一下
可以得到
$$(B(x)-B_0(x){)}^2\equiv 0(mod{x}^n)$$
展开这个式子：
$$B^2(x)+B_0^2(x)-2\ast B(x)\ast B_0(x)\equiv 0(mod{x}^n)$$
我们来证明一下为什么可以$mod{x}^n$，思考多项式乘法的过程，$c_i,a_i,b_i$都表示各自多项式中$x^i$的系数
$$c_i=\sum _{j=0}^i{a}_j\ast b_{i-j}\dots \dots ①$$
$$B(x)-B_0(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })\dots \dots ②$$
首先这两个式子的正确性不用证明了吧！我们把$B(x)-B_0(x)$看成一个整体$C(x)$，虽然不能保证$B,B_0$在$mod{x}^{\lceil \frac{n}{2}\rceil }$意义下的系数都为$0$，但是因为$②$我们可知$C(x)$在$mod{x}^{\lceil \frac{n}{2}\rceil }$意义下的系数都为$0$，那么平方就转换成了$C(x)\ast C(x)$，套用$①$
$$an{s}_i=\sum _{j=0}^i{c}_j\ast c_{i-j}$$
发现在$<n$的项至少是由其中一个$C(x)$中的$<\lceil \frac{n}{2}\rceil$项乘以其他项的结果，但是当$=n$时，是会遇到$c_{\lceil \frac{n}{2}\rceil }\ast c_{\lceil \frac{n}{2}\rceil }$，结合上面的公式$B(x)-B_0(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$，$x^{\lceil \frac{n}{2}\rceil }$是被舍掉了的，无法保证一定为$0$，
所以平方后的式子$[0,n-1]$次项都是$0$，但是$n$次项不确定，就成为一条分水岭，故把取模的界点设在$n$可以砍掉，而不设在$n-1$或者$n+1$等位置

---

大家最关心的时间复杂度，因为不停的$/2$用递归，就是$log$级别，所以$O(nlogn)$左右皆大欢喜

模板

void solve ( int deg, int *b ) {if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}solve ( ( deg + 1 ) >> 1, b );len = 1;l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}inv = qkpow ( len, mod - 2 );for ( int i = 0;i < len;i ++ )r[i] = ( r[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < n;i ++ )c[i] = 0;NTT ( c, 1 );NTT ( b, 1 );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1 );for ( int i = deg;i < n;i ++ )b[i] = 0;
}

例题：[luogu P4238]【模板】多项式乘法逆

题目

给定一个多项式 $F(x)$ ，请求出一个多项式 $G(x)$， 满足 $F(x)\ast G(x)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$系数对$998244353$ 取模。

输入格式
首先输入一个整数 $n$， 表示输入多项式的项数
接着输入$n$ 个整数，第 $i$ 个整数 $a_i$代表 $F(x)$次数为$i-1$的项的系数。

输出格式
输出 $n$个数字，第$i$个整数 $b_i$，代表 $G(x)$次数为 $i-1$ 的项的系数。 保证有解。

输入输出样例
输入
5
1 6 3 4 9
输出
1 998244347 33 998244169 1020
说明/提示
$1\le n\le 10^5,0\le a_i\le 10^9$

code

$998244353$的原根是$3$，所以就直接用了

#include <cstdio>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int n, l, pi, ni, len, inv;
int a[MAXN], b[MAXN], c[MAXN], r[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f ) {for ( int i = 0;i < len;i ++ )if ( i < r[i] )swap ( c[i], c[r[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {//求多项式a的逆元b if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );len = 1;l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}inv = qkpow ( len, mod - 2 );for ( int i = 0;i < len;i ++ )r[i] = ( r[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < n;i ++ )c[i] = 0;NTT ( c, 1 );NTT ( b, 1 );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1 );for ( int i = deg;i < n;i ++ )b[i] = 0;
}signed main() {pi = 3;ni = mod / pi + 1;scanf ( "%lld", &n );for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );polyinv ( n, a, b );for ( int i = 0;i < n;i ++ )printf ( "%lld ", b[i] );return 0;
}

多项式的除法/取模

理论推导

给定一个$n-1$次的多项式$A(x)$和$m-1$次的多项式$B(x)$，求$D(x),R(x)$满足
$$A(x)=D(x)\ast B(x)+R(x)\dots \dots ①$$
其中自然$D(x)$的最高次为$n-m$，$R(x)$最高次$<m-1$
或者$$A(x)\equiv R(x)(modB(x))$$
因为有余数$R(x)$难以对其进行处理，所以考虑去掉这个玩意儿带来的不良影响
定义反转操作，即将各项系数反转 dalao让我们这么搞，只能跟着

$$A^R(x)=x^{n-1}A(\frac{1}{x})=\sum _{i=0}^{n-1}{a}_{n-i-1}{x}^i\dots \dots ②$$
证明很简单，把中间的那一步$A(\frac{1}{x})$展开再与外面的$x^{n-1}$相乘，你会发现是一样的，好下一步
把$\frac{1}{x}$代入$①$，再同乘各多项式的$x^{n-1}$得到
$$x^{n-1}A(\frac{1}{x})=x^{n-1}\ast D(\frac{1}{x})\ast B(\frac{1}{x})+x^{n-1}\ast R(\frac{1}{x})$$
我们把$x^{n-1}$拆分成$x^{n-m}\ast x^{m-1}$，因为它们分别对应了$D(x),B(x)$的最高次，再把$x^{n-1}$拆分成$x^{n-m+1}\ast x^{m-2}$，因为$x^{m-2}$是$R(x)$能达到的最高次，我们就强买强卖要求$R(x)$达到最高次，大不了系数为0，即
$$x^{n-1}A(\frac{1}{x})=x^{n-m}D(\frac{1}{x})\ast x^{m-1}B(\frac{1}{x})+x^{n-m+1}\ast x^{m-2}R(\frac{1}{x})$$
由$②$可转换为
$$A^R(x)=D^R(x)B^R(x)+x^{n-m+1}{R}^R(x)$$
由于$D^R(x)$是$n-m$次的，所以在$mod{x}^{n-m+1}$意义下，就可以得到$$A^R(x)\equiv D^R(x)\ast B^R(x)(mod{x}^{n-m+1})$$，猥琐目的达到
所以我们可以用多项式求逆得到$D^R(x)$，反转即为$D(x)$，然后再带入求$R(x)$即可

所以上面的操作真的是妙极了！！！

多项式牛顿迭代法

有一个关于多项式$f(x)$的方程$g(f(x))=0$
假设已经求出$f(x)$的前$n$项$f_0(x)$，则有
$$f(x)\equiv f_0(x)(mod{x}^n)$$
$$g(f_0(x))\equiv 0(mod{x}^n)$$
对$g(f(x))$在$f(x)$上进行泰勒展开
$$g(f(x))=g(f_0(x))+\frac{g^{\prime }(f_0(x))}{1!}(f(x)-f_0(x){)}^1+\frac{g^{\prime \prime }(f_0(x))}{2!}(f(x)-f_0(x){)}^2+\dots \dots$$
与上面内容结合$f(x)-f_0(x)$的前$n$项系数为$0$，于是$$g(f(x))\equiv g(f_0(x))+g^{\prime }(f_0(x))(f(x)-f_0(x))\equiv 0(mod{x}^{2n})$$

为什么从第三项开始就可以省略了呢？？里面有清晰的证明

即$$f(x)\equiv f_0(x)-\frac{g(f_0(x))}{g^{\prime }(f_0(x))}(mod{x}^{2n})$$

---

$$g(f(x))\equiv lnf(x)-A(x)$$

这个函数的零点为$e^A(x)$，即$A(x)$的$exp$
简单来说，多项式牛顿迭代法就是用来解关于多项式的方程的
——摘自某dalao

用这种方法也可以推多项式求逆，但是更多的是后面的一些基础操作，先卖个关子

模板

void polydiv ( int n, int m,int *d, int *a, int *b, int *r ) {for ( int i = 0;i < n;i ++ )//进行反转操作 A[i] = a[n - i - 1];for ( int i = 0;i < m;i ++ )B[i] = b[m - i - 1];polyinv ( n - m + 1, B, Binv );//根据推导的公式求出B在mod x^(n-m+1)意义下的逆len = 1;l = 0;while ( len < ( n << 1 ) - m ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( A, 1, len );NTT ( Binv, 1, len );for ( int i = 0;i < len;i ++ )//求反转的商D d[i] = A[i] * Binv[i] % mod;NTT ( d, -1, len ); for ( int i = 0, j = n - m;i < j;i ++, j -- )//把商反转成正常 swap ( d[i], d[j] );//接下来搞r memset ( B, 0, sizeof ( B ) );memset ( Binv, 0, sizeof ( Binv ) );memset ( rev, 0, sizeof ( rev ) );len = 1;l = 0;while ( len < n ) {l ++;len <<= 1;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) ); for ( int i = 0;i < m;i ++ )B[i] = b[i];for ( int i = 0;i < n - m + 1;i ++ )Binv[i] = d[i];NTT ( B, 1, len );NTT ( Binv, 1, len );for ( int i = 0;i < len;i ++ )r[i] = B[i] * Binv[i] % mod;NTT ( r, -1, len );for ( int i = 0;i < m;i ++ )r[i] = ( a[i] - r[i] + mod ) % mod;
}

例题：[luoguP4512]【模板】多项式除法

题目

给定一个 $n$ 次多项式 $F(x)$ 和一个 $m$ 次多项式 $G(x)$ ，请求出多项式 $Q(x),R(x)$，满足以下条件：
$Q(x)$ 次数为 $n-m$，$R(x)$ 次数小于 $m$
$F(x)=Q(x)\ast G(x)+R(x)$
所有的运算在模 $998244353$ 意义下进行。

输入格式
第一行两个整数$n$，$m$，意义如上。
第二行 $n+1$个整数，从低到高表示 $F(x)$ 的各个系数。
第三行 $m$ 个整数，从低到高表示 $G(x)$ 的各个系数。

输出格式
第一行 $n-m+1$ 个整数，从低到高表示 $Q(x)$ 的各个系数。
第二行 $m$ 个整数，从低到高表示 $R(x)$ 的各个系数。
如果 $R(x)$ 不足 $m-1$ 次，多余的项系数补 $0$。

输入输出样例
输入
5 1
1 9 2 6 0 8
1 7
输出
237340659 335104102 649004347 448191342 855638018
760903695
说明/提示
对于所有数据，$1\le m<n\le 10^5$ ，给出的系数均属于 $[0,998244353)\cap \mathbb{Z}$

code

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int l, pi, ni, len, inv;
int a[MAXN], b[MAXN], c[MAXN], r[MAXN], d[MAXN], rev[MAXN];
int A[MAXN], B[MAXN], Binv[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f, int len ) {for ( int i = 0;i < len;i ++ )if ( i < rev[i] )swap ( c[i], c[rev[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}inv = qkpow ( len, mod - 2 ); if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );len = 1;l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}inv = qkpow ( len, mod - 2 );for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < len;i ++ )c[i] = 0;NTT ( c, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1, len );for ( int i = deg;i < len;i ++ )b[i] = 0;
}void polydiv ( int n, int m,int *d, int *a, int *b, int *r ) {for ( int i = 0;i < n;i ++ )//进行反转操作 A[i] = a[n - i - 1];for ( int i = 0;i < m;i ++ )B[i] = b[m - i - 1];polyinv ( n - m + 1, B, Binv );//根据推导的公式求出B在mod x^(n-m+1)意义下的逆len = 1;l = 0;while ( len < ( n << 1 ) - m ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( A, 1, len );NTT ( Binv, 1, len );for ( int i = 0;i < len;i ++ )//求反转的商D d[i] = A[i] * Binv[i] % mod;NTT ( d, -1, len ); for ( int i = 0, j = n - m;i < j;i ++, j -- )//把商反转成正常 swap ( d[i], d[j] );//接下来搞r memset ( B, 0, sizeof ( B ) );memset ( Binv, 0, sizeof ( Binv ) );memset ( rev, 0, sizeof ( rev ) );len = 1;l = 0;while ( len < n ) {l ++;len <<= 1;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) ); for ( int i = 0;i < m;i ++ )B[i] = b[i];for ( int i = 0;i < n - m + 1;i ++ )Binv[i] = d[i];NTT ( B, 1, len );NTT ( Binv, 1, len );for ( int i = 0;i < len;i ++ )r[i] = B[i] * Binv[i] % mod;NTT ( r, -1, len );for ( int i = 0;i < m;i ++ )r[i] = ( a[i] - r[i] + mod ) % mod;
}signed main() {pi = 3;ni = mod / pi + 1;int n, m;scanf ( "%lld %lld", &n, &m );n ++;m ++; for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );for ( int i = 0;i < m;i ++ )scanf ( "%lld", &b[i] );polydiv ( n, m, d, a, b, r );for ( int i = 0;i < n - m + 1;i ++ )printf ( "%lld ", d[i] );printf ( "\n" );for ( int i = 0;i < m - 1;i ++ )printf ( "%lld ", r[i] );return 0;
}

多项式对数ln

理论推导

对于一个给定的多项式$A(x)$，求
$$lnA(x)$$

---

直接运用求导和积分完成，需要保证$A(x)$的常数项为1

对于$f(x),x$点的导数：$f^{\prime }(x)=\frac{f(x+△)-f(x)}{△}$

$$lnA(x)=\int ln(A(x){)}^{\prime }=\int \frac{A(x{)}^{\prime }}{A(x)}$$

---

反正对于小孩子而言，我是难以理解求导和积分的，可能这里会留个坑，等学到后面，再慢慢跟他们玩吧，如果有dalao能教教我（可私信），我一定感激死你

模板

void polyln ( int n, int *a, int *b ) {polyinv ( n, a, b );for ( int i = 1;i < n;i ++ )tmp[i - 1] = a[i] * i % mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( tmp, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = tmp[i] * b[i] % mod;NTT ( b, -1, len );for ( int i = n - 1;i; i -- )b[i] = b[i - 1] * qkpow ( i, mod - 2 ) % mod;b[0] = 0; for ( int i = n;i < len;i ++ )b[i] = 0;
} 

例题

题目

luogu
给出 $n-1$ 次多项式 $A(x)$，求一个 $\mathrm{m}\mathrm{o}\mathrm{d}{x}^n$下的多项式$B(x)$，满足 $B(x)\equiv \ln A(x)$.
在 $\text{mod }998244353$ 下进行，且 $a_i\in [0,998244353]\cap \mathbb{Z}$

输入格式
第一行一个整数 $n$
下一行有 $n$ 个整数，依次表示多项式的系数 $a_0,a_1,\cdots ,a_{n-1}$
输出格式
输出 $n$ 个整数，表示答案多项式中的系数 $a_0,a_1,\cdots ,a_{n-1}$

输入输出样例
输入
6
1 927384623 878326372 3882 273455637 998233543
输出
0 927384623 817976920 427326948 149643566 610586717
说明/提示
对于 $100\%$ 的数据，$n\le 10^5$

code

#include <cstdio>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int n, pi, ni, inv;
int a[MAXN], b[MAXN], c[MAXN], rev[MAXN], tmp[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f, int len ) {for ( int i = 0;i < len;i ++ )if ( i < rev[i] )swap ( c[i], c[rev[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}inv = qkpow ( len, mod - 2 );if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );int len = 1, l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < len;i ++ )c[i] = 0;NTT ( c, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1, len );for ( int i = deg;i < len;i ++ )b[i] = 0;
}void polyln ( int n, int *a, int *b ) {polyinv ( n, a, b );for ( int i = 1;i < n;i ++ )tmp[i - 1] = a[i] * i % mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( tmp, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = tmp[i] * b[i] % mod;NTT ( b, -1, len );for ( int i = n - 1;i; i -- )b[i] = b[i - 1] * qkpow ( i, mod - 2 ) % mod;b[0] = 0; for ( int i = n;i < len;i ++ )b[i] = 0;
} signed main() {pi = 3;ni = mod / pi + 1;scanf ( "%lld", &n );for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );polyln ( n, a, b );for ( int i = 0;i < n;i ++ )printf ( "%lld ", b[i] );return 0;
}

多项式开根sqrt

理论推导

对于一个给定的多项式$A(x)$，求$B(x)$满足
$$B^2(x)\equiv A(x)(mod{x}^n)$$

---

如果$n=1$就直接是常数项开方
$1.$可以直接套牛顿迭代
$$B(x)=B_0-\frac{B_0^2(x)-A(x)}{2B_0(x)}=\frac{B_0(x)+\frac{A(x)}{B_0(x)}}{2}$$

---

$2.$也可以用逆元平方的思想
设$B^{\prime }(x)$满足$B^{\prime 2}(x)\equiv A(x)(mod{x}^{\lceil \frac{n}{2}\rceil })$，且我们已经知道它了
两式相减$$B^2(x)-B^{\prime 2}(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$$
因式分解$$(B(x)-B^{\prime }(x))(B(x)+B^{\prime }(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$$
肯定只能是$$B(x)-B^{\prime }(x)\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$$不可能非负数系数开个根还能开出负数来，那你可以被飞签高等学府了
故伎重演，用求逆元的方式平方来一下，这里就直接打开了$$B^2(x)+B^{\prime 2}(x)-2\ast B(x)\ast B^{\prime }(x)\equiv 0(mod{x}^n)$$
结合$B^2(x)$需要满足的条件，替换成$A(x)$
$$A(x)+B^{\prime 2}(x)-2\ast B(x)\ast B^{\prime }(x)\equiv 0(mod{x}^n)$$
用$B^{\prime }(x),A(x)$去表示$B(x)$
$$B(x)\equiv \frac{A(x)+B^{\prime }(x{)}^2}{2\ast B^{\prime }(x)}$$

---

最后发现是孪生兄弟， 御用递归求解$O(nlogn)$

同时发现，只要常数项是二次项剩余且有逆元，这个多项式就可以开方
——摘自某dalao

模板

代码中常数项刚好是$1$，不是加强版，我太弱了
$$B(x)\equiv \frac{\frac{A(x)}{B^{\prime }(x)}+B^{\prime }(x)}{2}$$

void polysqrt ( int n, int *a, int *b ) {if ( n == 1 ) {b[0] = 1;return;}polysqrt ( ( n + 1 ) >> 1, a, b );static int tmp[MAXN], binv[MAXN];polyinv ( n, b, binv );//b就是公式的B'(x) tmp就是公式的A(x)int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < n;i ++ )tmp[i] = a[i];for ( int i = n;i < len;i ++ )tmp[i] = 0;NTT ( tmp, 1, len );NTT ( b, 1, len );NTT ( binv, 1, len );for ( int i = 0;i < len;i ++ )//求解B(x) 用b表示b[i] = ( tmp[i] * binv[i] % mod + b[i] ) % mod * inv2 % mod;NTT ( b, -1, len );for ( int i = n;i < len;i ++ )b[i] = binv[i] = 0;for ( int i = 0;i < n;i ++ )binv[i] = 0;
}

例题

题目

luogu
给定一个$n-1$次多项式$A(x)$，求一个在$\mathrm{m}\mathrm{o}\mathrm{d}{x}^n$
意义下的多项式$B(x)$，使得$B^2(x)\equiv A(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$若有多解，请取零次项系数较小的作为答案。

多项式的系数在$\mathrm{m}\mathrm{o}\mathrm{d}998244353$的意义下进行运算。

输入格式
第一行一个正整数$n$
接下来$n$个整数，依次表示多项式的系数$a_0,a_1,\dots ,a_{n-1}$
输出格式
输出$n$个整数，表示答案多项式的系数$b_0,b_1,\dots ,b_{n-1}$

输入输出样例
输入
3
1 2 1
输出
1 1 0

输入
7
1 8596489 489489 4894 1564 489 35789489
输出
1 503420421 924499237 13354513 217017417 707895465 411020414
说明/提示
对于$100\%$的数据：$n\le 10^5{a}_i\in [0,998244352]\cap \mathbb{Z}$

code

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int pi, ni, inv, inv2;
int rev[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f, int len ) {for ( int i = 0;i < len;i ++ )if ( i < rev[i] )swap ( c[i], c[rev[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}inv = qkpow ( len, mod - 2 );if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {static int c[MAXN];if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );int len = 1, l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < len;i ++ )c[i] = 0;NTT ( c, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1, len );for ( int i = deg;i < len;i ++ )b[i] = 0;
}void polysqrt ( int n, int *a, int *b ) {if ( n == 1 ) {b[0] = 1;return;}polysqrt ( ( n + 1 ) >> 1, a, b );static int tmp[MAXN], binv[MAXN];polyinv ( n, b, binv );int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < n;i ++ )tmp[i] = a[i];for ( int i = n;i < len;i ++ )tmp[i] = 0;NTT ( tmp, 1, len );NTT ( b, 1, len );NTT ( binv, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( tmp[i] * binv[i] % mod + b[i] ) % mod * inv2 % mod;NTT ( b, -1, len );for ( int i = n;i < len;i ++ )b[i] = binv[i] = 0;for ( int i = 0;i < n;i ++ )binv[i] = 0;
}int a[MAXN], b[MAXN];signed main() {pi = 3;ni = mod / pi + 1;inv2 = qkpow ( 2, mod - 2 );int n;scanf ( "%lld", &n );for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );polysqrt ( n, a, b );for ( int i = 0;i < n;i ++ )printf ( "%lld ", b[i] );return 0;
}

多项式指数exp

理论推导

对于一个给定的多项式$A(x)$，求$B(x)$满足$$B(x)\equiv {\text{e}}^{A(x)}(mod{x}^n)$$

---

取个对数，移个项$$ln(B(x))\equiv A(x)(mod{x}^n)——>ln(B(x))-A(x)\equiv 0(mod{x}^n)$$

---

对数求导公式：$(lnx{)}^{\prime }=\frac{1}{x}$别问我为什么，我也证不来，求教

接着套牛顿迭代$$B(x)\equiv B_0(x)-\frac{g(B_0(x))}{g^{\prime }(B_0(x))}\equiv B_0(x)-\frac{g(B_0(x))}{\frac{1}{B_0}}\equiv B_0(x)-\frac{ln{B}_0(x)-A(x)}{\frac{1}{B_0(x)}}$$$$\equiv B_0(x)-B_0(x)(ln{B}_0(x)-A(x))\equiv B_0(x)(1-ln{B}_0(x)+A(x))(mod{x}^{2n})$$

必须保证A(x)常数项为0，B(x)常数项为1，不知道为什么，求dalao教

模板

void polyexp ( int n, int *a, int *b ) {if ( n == 1 ) {b[0] = 1;return;}polyexp ( ( n + 1 ) >> 1, a, b );static int bln[MAXN];polyln ( n, b, bln );for ( int i = 0;i < n;i ++ )bln[i] = ( i == 0 ) + a[i] - bln[i] + mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( bln, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = b[i] * bln[i] % mod;NTT ( b, -1, len );for ( int i = n;i < len;i ++ )b[i] = bln[i] = 0;for ( int i = 0;i < n;i ++ )bln[i] = 0;
}

例题

luogu
给出 $n-1$ 次多项式 $A(x)$，求一个 $\mathrm{m}\mathrm{o}\mathrm{d}{x}^n$下的多项式 $B(x)$，满足 $B(x)\equiv {\text{e}}^{A(x)}$，系数对 $998244353$ 取模
输入格式
第一行一个整数 $n$

下一行有 $n$ 个整数，依次表示多项式的系数 $a_0,a_1,\cdots ,a_{n-1}$
输出格式
输出 $n$ 个整数，表示答案多项式中的系数 $a_0,a_1,\cdots ,a_{n-1}$

输入输出样例
输入
6
0 927384623 817976920 427326948 149643566 610586717
输出
1 927384623 878326372 3882 273455637 998233543
说明/提示
对于 $100\%$ 的数据，$n\le 10^5$

题目

code

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int pi, ni, inv;
int rev[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f, int len ) {for ( int i = 0;i < len;i ++ )if ( i < rev[i] )swap ( c[i], c[rev[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}inv = qkpow ( len, mod - 2 );if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {static int c[MAXN];if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );int len = 1, l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < len;i ++ )c[i] = 0;NTT ( c, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1, len );for ( int i = deg;i < len;i ++ )b[i] = 0;
}void polyln ( int n, int *a, int *b ) {static int tmp[MAXN];polyinv ( n, a, b );for ( int i = 1;i < n;i ++ )tmp[i - 1] = a[i] * i % mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( tmp, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = tmp[i] * b[i] % mod;NTT ( b, -1, len );for ( int i = n - 1;i; i -- )b[i] = b[i - 1] * qkpow ( i, mod - 2 ) % mod;b[0] = 0;for ( int i = n;i < len;i ++ )b[i] = 0;
} void polyexp ( int n, int *a, int *b ) {if ( n == 1 ) {b[0] = 1;return;}polyexp ( ( n + 1 ) >> 1, a, b );static int bln[MAXN];polyln ( n, b, bln );for ( int i = 0;i < n;i ++ )bln[i] = ( i == 0 ) + a[i] - bln[i] + mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( bln, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = b[i] * bln[i] % mod;NTT ( b, -1, len );for ( int i = n;i < len;i ++ )b[i] = bln[i] = 0;for ( int i = 0;i < n;i ++ )bln[i] = 0;
}int a[MAXN], b[MAXN];signed main() {pi = 3;ni = mod / pi + 1;int n;scanf ( "%lld", &n );for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );polyexp ( n, a, b );for ( int i = 0;i < n;i ++ )printf ( "%lld ", b[i] );return 0;
}

多项式快速幂

理论推导

通过上述多项式的基本操作后，我们直接暴力来一发拆公式

$$B(x)\equiv A^k(x)(mod{x}^n)$$
$==>$ $$B(x)\equiv e^{lnA(x{)}^k}$$
$==>$ $$B(x)\equiv e^{k\ast lnA(x)}$$
所以我们只用上述所讲的$ln,exp$即可完成快速幂

例题

题目

luogu
给定一个$n-1$次多项式$A(x)$，求一个在$\mathrm{m}\mathrm{o}\mathrm{d}{x}^n$
意义下的多项式$B(x)$，使得$B(x)\equiv A^k(x)(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$
多项式的系数在$\mathrm{m}\mathrm{o}\mathrm{d}998244353$的意义下进行运算。

输入格式
第一行两个正整数$n,k$。

接下来$n$个整数，依次表示多项式的系数$a_0,a_1,\dots ,a_{n-1}$
输出格式
输出$n$个整数，表示答案多项式的系数$b_0,b_1,\dots ,b_{n-1}$

输入输出样例
输入
9 18948465
1 2 3 4 5 6 7 8 9
输出
1 37896930 597086012 720637306 161940419 360472177 560327751 446560856 524295016

输入
4 1
1 1 0 0
输出
1 1 0 0

输入
4 2
1 1 0 0
输出
1 2 1 0

输入
4 3
1 1 0 0
输出
1 3 3 1
说明/提示
对于$100\%$的数据：$n\le 10^{5}2\le k\le 10^{10^5}{a}_i\in [0,998244352]\cap \mathbb{Z}$

code

唯一的坑点🕳就是注意$k$很大，$longlong$装不下，必须快读取模

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define mod 998244353
#define MAXN 300005
#define int long long
int pi, ni, inv;
int rev[MAXN], result[MAXN];int qkpow ( int x, int y ) {int ans = 1;while ( y ) {if ( y & 1 )ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void NTT ( int *c, int f, int len ) {for ( int i = 0;i < len;i ++ )if ( i < rev[i] )swap ( c[i], c[rev[i]] );for ( int i = 1;i < len;i <<= 1 ) {int omega = qkpow ( f == 1 ? pi : ni, ( mod - 1 ) / ( i << 1 ) );for ( int j = 0;j < len;j += ( i << 1 ) ) {int w = 1;for ( int k = 0;k < i;k ++, w = w * omega % mod ) {int x = c[k + j], y = w * c[k + j + i] % mod;c[k + j] = ( x + y ) % mod;c[k + j + i] = ( x - y + mod ) % mod;}}}inv = qkpow ( len, mod - 2 );if ( f == -1 )for ( int i = 0;i < len;i ++ )c[i] = c[i] * inv % mod;
}void polyinv ( int deg, int *a, int *b ) {static int c[MAXN];if ( deg == 1 ) {b[0] = qkpow ( a[0], mod - 2 );return;}polyinv ( ( deg + 1 ) >> 1, a, b );int len = 1, l = 0;while ( len < deg * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );for ( int i = 0;i < deg;i ++ )c[i] = a[i];for ( int i = deg;i < len;i ++ )c[i] = 0;NTT ( c, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = ( 2 - b[i] * c[i] % mod + mod ) % mod * b[i] % mod; NTT ( b, -1, len );for ( int i = deg;i < len;i ++ )b[i] = 0;
}void polyln ( int n, int *a, int *b ) {static int tmp[MAXN];polyinv ( n, a, b );for ( int i = 1;i < n;i ++ )tmp[i - 1] = a[i] * i % mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( tmp, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = tmp[i] * b[i] % mod;NTT ( b, -1, len );for ( int i = n - 1;i; i -- )b[i] = b[i - 1] * qkpow ( i, mod - 2 ) % mod;b[0] = 0;for ( int i = n;i < len;i ++ )b[i] = tmp[i] = 0;for ( int i = 0;i < n;i ++ )tmp[i] = 0;
} void polyexp ( int n, int *a, int *b ) {if ( n == 1 ) {b[0] = 1;return;}polyexp ( ( n + 1 ) >> 1, a, b );static int bln[MAXN];polyln ( n, b, bln );for ( int i = 0;i < n;i ++ )bln[i] = ( ( i == 0 ) + a[i] - bln[i] + mod ) % mod;int len = 1, l = 0;while ( len < n * 2 - 1 ) {len <<= 1;l ++;}for ( int i = 0;i < len;i ++ )rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) );NTT ( bln, 1, len );NTT ( b, 1, len );for ( int i = 0;i < len;i ++ )b[i] = b[i] * bln[i] % mod;NTT ( b, -1, len );for ( int i = n;i < len;i ++ )b[i] = bln[i] = 0;for ( int i = 0;i < n;i ++ )bln[i] = 0;
}int a[MAXN], b[MAXN];void read ( int &x ) {x = 0;char s = getchar();while ( s < '0' || s > '9' )s = getchar();while ( s >= '0' && s <= '9' ) {x = ( ( x << 1 ) + ( x << 3 ) + ( s - '0' ) ) % mod;s = getchar();}
} signed main() {pi = 3;ni = mod / pi + 1;int n, k;scanf ( "%lld", &n );read ( k );//k太大了，必须要取模long long装不下 for ( int i = 0;i < n;i ++ )scanf ( "%lld", &a[i] );polyln ( n, a, b );for ( int i = 0;i < n;i ++ )b[i] = b[i] * k % mod;polyexp ( n, b, result );for ( int i = 0;i < n;i ++ )printf ( "%lld ", result[i] );return 0;
}

版权申明：
教会我多项式的blog
证明无代码

最后只想说一句，希望各个dalao能完善我的证明，甚至教教这个蒟蒻，部分实在证不来，百度百科长的又丑有问题欢迎评论，受教了