多项式求逆
https://www.luogu.com.cn/problem/P4238
原理
利用倍增来得到答案。
假设现在已经得到 \(H(x)\),使得 \(F(x)H(x)\equiv 1 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)
同时有 \(F(x)G(x)\equiv 1 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)
而且 \(F(x)\not\equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)
因此 \(H(x)-G(X)\equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)
同时我们也有 \(H(x)-G(X)\equiv 0 \pmod{x^{\lfloor\frac{n}{2} \rfloor}}\)
进而有 \((H(x)-G(X))^2 \equiv 0 \pmod{x^n}\)
因为 \(\lceil \frac{n}{2} \rceil + \lfloor\frac{n}{2} \rfloor = n\)
展开得到 \(H^2(X) -2H(X)G(X) + G^2(X) \equiv 0 \pmod{x^n}\)
两边同乘 \(F(X)\),可得 \(H^2(X)F(X) -2H(X) + G(X) \equiv 0 \pmod{x^n}\)
因此 \(2H(X) - H^2(X)F(X) \equiv G(X) \pmod{x^n}\)
实现
// Problem: P4238 【模板】多项式乘法逆
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4238
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define debug(x) cerr << #x << ": " << (x) << endl
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
using pii = pair<int, int>;
using ll = long long;
#define int long long
inline void read(int &x){
int s=0; x=1;
char ch=getchar();
while(ch<'0' || ch>'9') {if(ch=='-')x=-1;ch=getchar();}
while(ch>='0' && ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
x*=s;
}
const int N=3e5+5, rt=3, mod=998244353;
int rev[N], tot=1, bit;
ll fpow(ll x, int p, ll mod){
int res=1;
for(; p; p>>=1, x=x*x%mod) if(p&1) res=res*x%mod;
return res;
}
ll inv(ll x, ll mod){
return fpow(x, mod-2, mod);
}
ll mul(ll x, int p, ll mod){
ll res=0;
for(; p; p>>=1, x=(x+x)%mod) if(p&1) res=(res+x)%mod;
return res;
}
void NTT(ll *a, int type, int mod){
for(int i=0; i<tot; i++){
a[i]%=mod;
if(i<rev[i]) swap(a[i], a[rev[i]]);
}
for(int mid=1; mid<tot; mid<<=1){
ll w1=fpow(rt, (type==1? (mod-1)/(mid<<1): mod-1-(mod-1)/(mid<<1)), mod);
for(int i=0; i<tot; i+=mid*2){
ll wk=1;
for(int j=0; j<mid; j++, wk=wk*w1%mod){
auto x=a[i+j], y=wk*a[i+j+mid]%mod;
a[i+j]=(x+y)%mod, a[i+j+mid]=(x-y+mod)%mod;
}
}
}
if(type==-1){
for(int i=0; i<tot; i++) a[i]=a[i]*inv(tot, mod)%mod;
}
}
int n;
int A[N], B[N], C[N];
void poly_inv(int sz, int *a, int *b){
if(sz==1) return b[0]=inv(a[0], mod), void();
poly_inv(sz+1>>1, a, b);
// init
bit=0, tot=1;
while(tot<=(sz-1<<1)) tot<<=1, bit++;
for(int i=0; i<tot; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
rep(i,0,sz-1) C[i]=a[i];
rep(i,sz,tot-1) C[i]=0;
NTT(C, 1, mod), NTT(b, 1, mod);
rep(i,0,tot-1) b[i]=(2-C[i]*b[i]%mod+mod)%mod*b[i]%mod;
NTT(b, -1, mod);
rep(i,sz,tot-1) b[i]=0;
}
signed main(){
cin>>n;
rep(i,0,n-1) read(A[i]);
poly_inv(n, A, B);
rep(i,0,n-1) cout<<B[i]<<' ';
cout<<endl;
return 0;
}