多項式求逆

https://www.luogu.com.cn/problem/P4238

原理

利用倍增來得到答案。

假設現在已經得到 \(H(x)\)，使得 \(F(x)H(x)\equiv 1 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)​

同時有 \(F(x)G(x)\equiv 1 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)

而且 \(F(x)\not\equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)

因此 \(H(x)-G(X)\equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}\)

同時我們也有 \(H(x)-G(X)\equiv 0 \pmod{x^{\lfloor\frac{n}{2} \rfloor}}\)​

進而有 \((H(x)-G(X))^2 \equiv 0 \pmod{x^n}\)

因為 \(\lceil \frac{n}{2} \rceil + \lfloor\frac{n}{2} \rfloor = n\)​

展開得到 \(H^2(X) -2H(X)G(X) + G^2(X) \equiv 0 \pmod{x^n}\)

兩邊同乘 \(F(X)\)，可得 \(H^2(X)F(X) -2H(X) + G(X) \equiv 0 \pmod{x^n}\)

因此 \(2H(X) - H^2(X)F(X) \equiv G(X) \pmod{x^n}\)

實現

// Problem: P4238 【模板】多項式乘法逆
// 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;
}