多项式基本操作
作者：互联网
多项式的简单四则运算
有加减乘除
加减比较native
多项式乘法
FFT：在\(O(nlogn)\)的时间内实现多项式的系数表示形式与点值表示形式的转换，作用域：实数
NTT：在\(O(nlogn)\)的时间内实现多项式的系数表示形式与点值表示形式的转换，作用域：有原根的模
MTT:
\[考虑一个n次多项式F(x) = \sum_{i = 1}^nx^ia_i\\ 考虑怎么快速求出其n个不重合的点的点值\\ \]
在实数域上FFT：
有一个玩意叫做n次单位根\(W_n^i\)
考虑把这个玩意带入进去,观察一下整个式子
\[不妨考虑n为二次整数次幂\\ F(x) = \sum_{i = 0}^{n - 1} x^ia_i = (x^0a_0 + ...+x^{n - 1}a_{n - 1})\\ = x^0a_0 + x^2_2 + ...+x^{n - 2}a_{n - 2} + x (x^0a_1 + x^2a_3+...+x^{n - 1}a_{n - 1})\\ F_1(x) = a_0 + a_2x^1 + ... +a_{n - 2}x^{\frac{n}{2} - 1}\\ F_2(x) = a_1 + a_3x^1 + ... +a_{n - 1}x^{\frac{n}{2} - 1}\\ F(x) = F_1(x^2) + xF_2(x^2)\\ 将一个单位根w_n^k带进去\\ F(w_n^k) = F_1(w_n^{2k}) + w_n^kF_2(w^{2k}_n)\\ 又因为单位根有一个性质:w_{2n}^{2k} = w^k_n\\ k < \frac{n}{2}\\ 有F(w_n^k) = F_1(w_{\frac{n}{2}}^{k}) + w_n^kF_2(w^{k}_{\frac{n}{2}})\\ F(w_n^{k + \frac{n}{2}}) = F_1(w_{\frac{n}{2}}^{k}) - w_n^kF_2(w^{k}_{\frac{n}{2}})\\ 发现每一次把问题分成了一半，统计是O(n)的\\ 故总的复杂度为O(nlogn)\\ \]考虑如何将点值表达式转换为系数表示
考虑单位根的奇妙性质:
\[w_n^{k + \frac{n}{2}} = -w_n^k \]考虑将n次单位根的倒数带进去得到的点值表达式
\[F(w_n^{0}) , F(w_n^{-1}),F(w_n^{-2})...,F(w_n^{-{n - 1}})\\ G(w_n^{-k}) = \sum_{i = 0}^{n - 1}F(w_n^i)w_n^{-ik}\\ =\sum_{i = 0}^{n - 1} (\sum_{j = 0}^{n - 1}a_jw_n^{ij}) w_n^{-ik}\\ = \sum_{j = 0}^{n - 1}a_j\sum_{i = 0}^{n - 1}w_n^iw_n^{(j - k)}\\ 发现后面这个式子实际上是一个等比数列求和\\ 于是可以推导得 后面这个式子只有\\ j = k : n\\ j != k : 0\\ 有\sum_{j = 0}^{n - 1}a_j\sum_{i = 0}^{n - 1}w_n^iw_n^{(j - k)}\\ =n*a_k\\ 故a_k = \frac{G(w_n^k)}{n} \]回到FFT
上面的递归，每次都要按奇偶划分系数，实际上常数是非常的的
考虑更简便的方法：
不妨考虑整个系数序列递归到叶子的情况
发现叶子的系数分布，是0到\(n - 1\)的二进制倒置后的值
\[000,001,010,011,100,101,110,111\\ ->\\ 000,100,010,110,001,101,011,111\\ \]于是可以\(O(n)\)递推
\[f[i] = f[i >> 1] + (i \& 1) << (log_2n - 1) \]然后在FFT里面以 长度，起点做就好了
代码实现如下
#include<bits/stdc++.h>
#define MAXN 2000005
typedef double ll;
const ll PI = acos(-1.0);
using namespace std;
int n,m;
int lim,len;
int res[MAXN];
struct node{ll dx,dy;}a[MAXN],b[MAXN],c[MAXN];
node operator + (node A , node B){return (node){A.dx + B.dx , A.dy + B.dy};}
node operator - (node A , node B){return (node){A.dx - B.dx , A.dy - B.dy};}
node operator * (node A , node B){return (node){A.dx * B.dx - A.dy * B.dy , A.dx * B.dy + A.dy * B.dx};}
node wn(double sz , int tp){
ll zz = 2.0 * PI / sz;
if(tp == (-1))return (node){cos(zz) , -sin(zz)};
return (node){cos(zz) , sin(zz)};
}
void fft(node f[] , int tp){
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
node w1 = wn(k << 1 , tp) , w , A , B;
for(int i = 0 ; i < len ; i = i + k + k){
w = (node){1 , 0};
for(int j = 0 ; j < k ; j++ , w = w * w1){
A = f[i + j] , B = f[i + j + k] * w;
f[i + j] = A + B;
f[i + j + k] = A - B;
}
}
}
if(tp == (-1))for(int i = 0 ; i < len ; i++)f[i].dx /= (len * 1.0);
}
int main(){
scanf("%d%d" , &n , &m);
for(int i = 0 ; i <= n ; i++)scanf("%lf" , &a[i].dx);
for(int i = 0 ; i <= m ; i++)scanf("%lf" , &b[i].dx);
len = 1 , lim = 0;
while(len <= (n + m + 2))len = len + len , lim++;
for(int i = 1 ; i < len ; i++){
res[i] = (res[i >> 1] >> 1) | ((i & 1) << (lim - 1));
}
fft(a , 1);
fft(b , 1);
for(int i = 0 ; i < len ; i++)c[i] = a[i] * b[i];
fft(c , -1);
for(int i = 0 ; i < n + m + 1 ; i++)printf("%d " , (int)(c[i].dx + 0.5));
}
考虑在模意义下怎么做
考虑998244353的原根，有一个为3
经过一些数学推导，发现原根在模意义下与单位根的性质相似
有\(w_n = g^{\frac{p - 1}{n}}\ mod p\)
其余的与单位根相似，不明白为什么之前写的常数这么大
代码如下：
#include<bits/stdc++.h>
#define MAXN 5000005
typedef long long ll;
const ll mod = 998244353;
const ll g = 3;
using namespace std;
int n,m,limit,len;
ll a[MAXN],b[MAXN],c[MAXN];
int res[MAXN];
int poww(ll x , int y){
ll zz = 1;
while(y){
if(y & 1)zz = 1ll * x * zz % mod;
x = (1ll * x * x) % mod;
y = y >> 1;
}
return zz;
}
//wn =
void NTT(ll f[] , int tp){
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
ll w1 = poww(3 , (mod - 1) / (k << 1)) , w , A , B;
if(tp == (-1))w1 = poww(w1 , mod - 2);
for(int i = 0 ; i < len ; i = i + k + k){
w = 1;
for(int j = 0 ; j < k ; j++ , w = w * w1 % mod){
A = f[i + j] , B = f[i + j + k] * w % mod;
f[i + j] = ((A + B) % mod + mod) % mod;
f[i + j + k] = ((A - B) % mod + mod) % mod;
}
}
}
if(tp == 1)return;
ll zz = poww(len , mod - 2);
for(int i = 0 ; i < len ; i++)f[i] = (1ll * f[i] * zz) % mod;
}
int main(){
scanf("%d%d" , &n , &m);
for(int i = 0 ; i <= n ; i++)scanf("%lld" , &a[i]);
for(int i = 0 ; i <= m ; i++)scanf("%lld" , &b[i]);
len = 1 , limit = 0;
while(len <= (n + m + 1))len = len + len , limit++;
for(int i = 1 ; i < len ; i++)res[i] = (res[i >> 1] >> 1) | ((i & 1) << (limit - 1));
NTT(a , 1);
NTT(b , 1);
for(int i = 0 ; i < len ; i++)c[i] = 1ll * a[i] * b[i] % mod;
NTT(c , -1);
for(int i = 0 ; i < n + m + 1 ; i++)printf("%lld " , c[i]);
}
P1919 【模板】A*B Problem 升级版（FFT 快速傅里叶变换）
板子题
#include<bits/stdc++.h>
#define MAXN 5000005
typedef long long ll;
const ll mod = 998244353;
const ll g = 3;
using namespace std;
int n,m,limit,len,res[MAXN];
ll a[MAXN],b[MAXN],c[MAXN];
int sz;
char s[MAXN];
ll poww(ll x , int y){
ll zz = 1;
while(y){
if(y & 1)zz = zz * x % mod;
x = x * x % mod;
y = y >> 1;
}
return zz;
}
void NTT(ll f[] , int tp){
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
ll w1 = poww(3 , (mod - 1) / (k << 1)) , w , A , B;
if(tp == (-1))w1 = poww(w1 , mod - 2);
for(int i = 0 ; i < len ; i = i + k + k){
w = 1;
for(int j = 0 ; j < k ; j++ , w = w * w1 % mod){
A = f[i + j] , B = f[i + j + k] * w % mod;
f[i + j] = ((A + B) % mod + mod) % mod;
f[i + j + k] = ((A - B) % mod + mod) % mod;
}
}
}
if(tp == 1)return;
ll zz = poww(len , mod - 2);
for(int i = 0 ; i < len ; i++)f[i] = (1ll * f[i] * zz) % mod;
}
int main(){
scanf("%s" , s) , sz = strlen(s) , n = sz - 1;
for(int i = 0 ; i <= n ; i++)a[i] = s[n - i] - '0';
scanf("%s" , s) , sz = strlen(s) , m = sz - 1;
for(int i = 0 ; i <= m ; i++)b[i] = s[m - i] - '0';
len = 1 , limit = 0;
while(len <= (n + m + 4)){
len = len + len , limit++;
}
for(int i = 1 ; i < len ; i++)res[i] = (res[i >> 1] >> 1) | ((i & 1) << (limit - 1));
NTT(a , 1);
NTT(b , 1);
for(int i = 0 ; i < len ; i++)c[i] = (a[i] * b[i] % mod + mod) % mod;
NTT(c , -1);
for(int i = 0 ; i < n + m + 1 ; i++){
c[i + 1] = c[i + 1] + c[i] / 10 , c[i] %= 10;
}
int len = n + m + 1;
while(c[len] >= 10)c[len + 1] = c[len + 1] + c[len] / 10 , c[len] %= 10 , len++;
while(c[len] == 0)len--;
for(int i = len ; i >= 0 ; i--)cout<<c[i];
}
分治FFT
例题：lgP4721 【模板】分治 FFT
\[给定序列g_{1...n - 1},求序列f_{0...n - 1}\\ f_i = \sum_{j = 1}^if_{i - j}g_j , f_0 = 1\\ 对于模数998244353取模 \]考虑cdq分治
递归区间为\([L,R] , mid = (L + R) / 2\)
考虑\([L , mid]\)对区间\([mid + 1 , R]\)的贡献
\[f_i += \sum_{j = l}^{mid}f_j * g_{i - j} \]发现这就是一个很简单的卷积形式
直接类似于cdq分治的形式就好了
复杂度大概是\(O(nlog^2n)\)
代码如下：lgP4721
#include<bits/stdc++.h>
#define MAXN 500005
typedef long long ll;
const ll mod = 998244353;
using namespace std;
int n;
ll g[MAXN],ans[MAXN];
ll poww(ll x , int y){
ll zz = 1;
while(y){
if(y & 1)zz = zz * x % mod;
x = x * x % mod;
y = y >> 1;
}
return zz;
}
int limit,len;
ll a[MAXN],b[MAXN],res[MAXN];
void NTT(ll f[] , int tp){
for(int i = 1 ; i < len ; i++)res[i] = ((res[i >> 1] >> 1) | ((i & 1) << (limit - 1)));
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
ll w1 = poww(3 , (mod - 1) / (k << 1)) , w , A , B;
if(tp == (-1))w1 = poww(w1 , mod - 2);
for(int i = 0 ; i < len ; i = i + k + k){
w = 1;
for(int j = 0 ; j < k ; j++ , w = w * w1 % mod){
A = f[i + j] , B = f[i + j + k] * w % mod;
f[i + j] = ((A + B) % mod + mod) % mod;
f[i + j + k] = ((A - B) % mod + mod) % mod;
}
}
}
if(tp == 1)return;
ll zz = poww(len , mod - 2);
for(int i = 0 ; i < len ; i++)f[i] = (1ll * f[i] * zz) % mod;
}
void cdq(int l , int r){
if(l == r)return;
int mid = (l + r) >> 1;
cdq(l , mid);
for(int i = l ; i <= mid ; i++)a[i - l] = ans[i] , b[i - l] = g[i - l];
for(int i = mid + 1 ; i <= r ; i++)a[i - l] = 0 , b[i - l] = g[i - l];
len = 1 , limit = 0;
while(len <= (r - l + 1))len = len + len , limit++;
for(int i = r - l + 1 ; i <= len ; i++)a[i] = b[i] = 0;
NTT(a , 1) , NTT(b , 1);
for(int i = 0 ; i < len ; i++)a[i] = (a[i] * b[i] % mod + mod) % mod;
NTT(a , -1);
for(int i = mid + 1 ; i <= r ; i++)ans[i] = (ans[i] + a[i - l]) % mod;
cdq(mid + 1 , r);
}
int main(){
scanf("%d" , &n) , ans[0] = 1;
for(int i = 1 ; i < n ; i++)scanf("%d" , &g[i]);
len = 1;while(len < n)len = len + len;
cdq(0 , len - 1);
for(int i = 0 ; i < n ; i++)cout<<ans[i]<<" ";
}
也有多项式求逆的做法，之后补
任意模数多项式乘法(MTT)
这个东西一般处理任意模数，可以用于适当的加强题目
1.把答案在三个有原根的模数下求出来，然后对于每一个答案都做一次CRT，9次
2.分解每一个数成\(\sqrt{mod} * p + q = a_i\)的形式 , 然后做多项式乘法暴力乘
3.MTT（论文里面的玩意）
方法2实现
#include<bits/stdc++.h>
#define MAXN 400005
typedef long double ll;
typedef long long LL;
const ll PI = acos(-1.0);
using namespace std;
int n,m,limit,len;
int res[MAXN];
LL ans[MAXN],part = 32768,p;
struct node{ll dx,dy;}a1[MAXN],b1[MAXN],a2[MAXN],b2[MAXN],X[MAXN];
node operator + (node A , node B){return (node){A.dx + B.dx , A.dy + B.dy};}
node operator - (node A , node B){return (node){A.dx - B.dx , A.dy - B.dy};}
node operator * (node A , node B){return (node){A.dx * B.dx - A.dy * B.dy , A.dx * B.dy + A.dy * B.dx};}
node wn(ll sz , int tp){
ll zz = 2.0 * PI / sz;
if(tp == (-1))return (node){cos(zz) , -sin(zz)};
return (node){cos(zz) , sin(zz)};
}
void fft(node f[] , int tp){
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
node w1 = wn(k << 1 , tp) , w , A , B;
for(int i = 0 ; i < len ; i = i + k + k){
w = (node){1 , 0};
for(int j = 0 ; j < k ; j++ , w = w * w1){
A = f[i + j] , B = f[i + j + k] * w;
f[i + j] = A + B;
f[i + j + k] = A - B;
}
}
}
}
void solve(node A[] , node B[] , LL res){
for(int i = 0 ; i < len ; i++)X[i].dx = X[i].dy = 0;
for(int i = 0 ; i < len ; i++)X[i] = A[i] * B[i];
fft(X , -1);
for(int i=0;i<=n+m;++i)(ans[i]+=(LL)(X[i].dx/len+0.5)%p*res%p)%=p;
}
void MTT(node f1[] , node f2[] , node g1[] , node g2[]){
fft(f1 , 1) , fft(f2 , 1) , fft(g1 , 1) , fft(g2 , 1);
solve(f1 , g1 , part * part);
solve(f1 , g2 , part);
solve(f2 , g1 , part);
solve(f2 , g2 , 1);
for(int i = 0 ; i <= n + m ; i++){
cout<<(ans[i] % p + p) % p<<" ";
}
}
int main(){
scanf("%d%d%lld" , &n , &m , &p);LL zz;
for(int i = 0 ; i <= n ; i++){
scanf("%lld" , &zz);
a1[i].dx = zz / part;
a2[i].dx = zz % part;
}
for(int i = 0 ; i <= m ; i++){
scanf("%lld" , &zz);
b1[i].dx = zz / part;
b2[i].dx = zz % part;
}
len = 1 , limit = 0;
while(len <= (n + m + 2))len = len + len , limit++;
for(int i = 1 ; i < len ; i++)res[i] = (res[i >> 1] >> 1) | ((i & 1) << (limit - 1));
MTT(a1 , a2 , b1 , b2);
}
多项式乘法逆
给定你一个n次多项式\(F(x)\),让你求一个多项式\(G(x)\)满足\(F(x) * G(x) = 1\ (mod\ x^n)\)
具体而言，相当于找一个多项式函数关于乘法意义下的逆元
\[F(x)H(x) = 1\ (mod\ x^{\frac{n}{2}})\\ F(x)H(x) = 1\ (mod\ x^{\frac{n}{2}})\\ 有F(x)G(x) = 1 (mod\ x^{\frac{n}{2}})\\ 有F(x)(G(x) - H(x)) = 0 (mod\ x^{\frac{n}{2}})\\ 平方有G^2(x) + H^2(x) = 2H(x)G(x) (mod\ x^n)\\ 两边再同时乘上F(x)\\ G(x) = 2H(x) - H^2(x)F(x) (mod\ x^n)\\ \]于是就可以递归的去构造了
#include<bits/stdc++.h>
#define MAXN 400005
typedef long long ll;
const ll mod = 998244353;
using namespace std;
int n,len,limit,res[MAXN];
ll a[MAXN],c[MAXN];
ll H[MAXN],G[MAXN];
ll poww(ll x , int y){
ll zz = 1;
while(y){
if(y & 1)zz = zz * x % mod;
x = x * x % mod;
y = y >> 1;
}
return zz;
}
void NTT(ll f[] , int tp){
for(int i = 1 ; i < len ; i++)res[i] = (res[i >> 1] >> 1) | ((i & 1) << (limit - 1));
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
ll w1 = poww(3 , (mod - 1) / (k << 1)) , w , A , B;
if(tp == (-1))w1 = poww(w1 , mod - 2);
for(int i = 0 ; i < len ; i = i + k + k){
w = 1;
for(int j = 0 ; j < k ; j++ , w = w * w1 % mod){
A = f[i + j] , B = f[i + j + k] * w % mod;
f[i + j] = ((A + B) % mod + mod) % mod;
f[i + j + k] = ((A - B) % mod + mod) % mod;
}
}
}
if(tp == 1)return;
ll zz = poww(len , mod - 2);
for(int i = 0 ; i < len ; i++)f[i] = (1ll * f[i] * zz) % mod;
}
void solve(int LEN , int L){
if(LEN == 1)return (void)(G[0] = poww(a[0] , mod - 2));
solve((LEN + 1) >> 1 , L - 1);
swap(H , G);
len = 1 , limit = 0;
while(len < (LEN << 1))len = len << 1 , limit++;
for(int i = 0 ; i < LEN ; i++)G[i] = c[i] = 0;
for(int i = 0 ; i < LEN ; i++)c[i] = a[i];
for(int i = LEN ; i < len ; i++)c[i] = 0;
NTT(c , 1) , NTT(H , 1);
for(int i = 0 ; i < len ; i++)G[i] = ((H[i] * ((2ll - H[i] * c[i]) % mod + mod) % mod) % mod + mod) % mod;
NTT(G , -1);
for(int i = LEN ; i < len ; i++)G[i] = 0;
}
int main(){
scanf("%d" , &n);for(int i = 0 ; i < n ; i++)scanf("%lld" , &a[i]);
solve(n , limit);
for(int i = 0 ; i < n ; i++)cout<<G[i]<<" ";
}
多项式ln,exp
引入对多项式的微分求导运算就可以做了
多项式ln
\[G(x) = ln(A(x))\\ G'(x) = \frac{A'(x)}{A(X)}\\ G(x) = \int G'(x) = \int \frac{A'(x)}{A(X)}\\ 直接对右边这个玩意多项式求逆+求导+积分一下就好了 \]#include<bits/stdc++.h>
#define MAXN 400005
typedef long long ll;
const ll mod = 998244353;
using namespace std;
int n,m,len,limit,res[MAXN];
ll a[MAXN],c[MAXN];
ll H[MAXN],G[MAXN];
ll poww(ll x , int y){
ll zz = 1;
while(y){
if(y & 1)zz = zz * x % mod;
x = x * x % mod;
y = y >> 1;
}
return zz;
}
void NTT(ll f[] , int tp){
for(int i = 1 ; i < len ; i++)res[i] = (res[i >> 1] >> 1) | ((i & 1) << (limit - 1));
for(int i = 0 ; i < len ; i++)if(i < res[i])swap(f[i] , f[res[i]]);
for(int k = 1 ; k < len ; k = k + k){
ll w1 = poww(3 , (mod - 1) / (k << 1)) , w , A , B;
if(tp == (-1))w1 = poww(w1 , mod - 2);
for(int i = 0 ; i < len ; i = i + k + k){
w = 1;
for(int j = 0 ; j < k ; j++ , w = w * w1 % mod){
A = f[i + j] , B = f[i + j + k] * w % mod;
f[i + j] = ((A + B) % mod + mod) % mod;
f[i + j + k] = ((A - B) % mod + mod) % mod;
}
}
}
if(tp == 1)return;
ll zz = poww(len , mod - 2);
for(int i = 0 ; i < len ; i++)f[i] = (1ll * f[i] * zz) % mod;
}
void solve(int LEN){
if(LEN == 1)return (void)(G[0] = poww(a[0] , mod - 2));
solve((LEN + 1) >> 1);
swap(H , G);
len = 1 , limit = 0;
while(len < (LEN << 1))len = len << 1 , limit++;
for(int i = 0 ; i < LEN ; i++)G[i] = c[i] = 0;
for(int i = 0 ; i < LEN ; i++)c[i] = a[i];
for(int i = LEN ; i < len ; i++)c[i] = 0;
NTT(c , 1) , NTT(H , 1);
for(int i = 0 ; i < len ; i++)G[i] = ((H[i] * ((2ll - H[i] * c[i]) % mod + mod) % mod) % mod + mod) % mod;
NTT(G , -1);
for(int i = LEN ; i < len ; i++)G[i] = 0;
}
int main(){
scanf("%d" , &n) , m = n - 1;for(int i = 0 ; i < n ; i++)scanf("%lld" , &a[i]);
solve(n);memset(H , 0 , sizeof(H));
for(int i = 1 ; i < n ; i++)H[i - 1] = ((a[i] * (1ll * i)) % mod + mod) % mod;
len = 1 , limit = 0;
while(len <= (n + m + 1))len = len + len , limit++;
NTT(H , 1) , NTT(G , 1);
for(int i = 0 ; i < len ; i++)H[i] = (H[i] * G[i] % mod + mod) % mod;
NTT(H , -1);memset(G , 0 , sizeof(G));
for(int i = 0 ; i < len ; i++){
G[i + 1] = (H[i] * poww(i + 1 , mod - 2) % mod + mod) % mod;
}
for(int i = 0 ; i < n ; i++)cout<<G[i]<<" ";
}
牛顿迭代
实数域上的牛顿迭代
常用来逼近某一些函数的零点
\[求函数F(x)的零点\\ 先大致故一下零点在哪里\\ 在零点附近取一个点(x_0 , F(x_0))\\ 在该点做一条切线与x轴相切 k = F'(x_0)\\ 交点为x_1\\ 即y - F(x_0) = F'(x_0)(x - x_0)\\ \frac{0 - F(x_0)}{F'(x_0)} + x_0 = x_1 \]按上述过程反复迭代即可
函数域上的牛顿迭代，虽然上下两个部分关系并不大，但思想是类似的
\[给定一个F(x),让你求一个多项式f\ .st F(f) = 0\\ 考虑倍增构造\\ F(f) = 0\ (mod\ x^n) , f_0 = f\ (mod\ x^n)\\ 考虑F(f)在f = f_0除的泰勒展开\\ F(f) = \sum_{k = 0}\frac{F^{k}(f - f_0)^k}{k!}\\ k >= 2时，(f - f_0) ^ k = 0\ (mod\ x^{2*n})显然\\ 则有 F(f) = 0 = F(f_0) + F'(f_0)(f - f_0) \ (mod\ x^{2n})\\ 则f = f_0 - \frac{F(f_0)}{F'(f_0)} \ (mod\ x^{2n})\\ 如此下去就可以得到f \]注意，这里的f虽然是函数，但我们把它当成常数来处理
多项式求逆就是一个很好的例子
多项式exp
\[B(x) = e^{A(x)}\\ 两边取对数 ln(B(x)) = A(x) \\ ln(B(x)) - A(x) = 0\\ 构造函数F(f) = ln(f) - A(x) \\ 采用牛顿迭代得到这个f就好了\\ 即f = (1 + A(x) - ln(f_0))*f_0 \]多项式开方
\[B(x) = \sqrt{A(x)}\\ B^2(x) = A(x)\\ B^2(x) - A(x) = 0\\ 构造函数F(f) = f^2 - A(x) = 0\\ 即f = \frac{f_0^2 + A(f_0)}{2f_0} 直接套用牛顿迭代就好了 \]下降幂多项式乘法
\[A(x) = \sum_{i = 0}a_ix^{\underline{i}}这个个下降幂多项式的例子 \]咕~~~~~
标签：node,int,多项式,ll,MAXN,zz,基本操作,mod 来源： https://www.cnblogs.com/Yeyuqing0913/p/15837221.html