luogu4328 多项式求逆

题目链接

problem

给出一个多项式f，求一个多项式g使得(f(x)*g(x) equiv 1 (mod x ^ n))

solution

利用倍增。假设现在我们已经求出了(f(x))在(mod x ^ n)的逆元(g(x))，考虑如何求出在(mod x ^{2n})下的逆元(g'(x))。

[f(x)g(x)equiv 1 (mod x ^ n) \ f(x)g(x) - 1 equiv0(mod x ^ n) \ f^2(x)g^2(x)-2f(x)g(x) + 1equiv 0(mod x ^{2n}) \ 2f(x)g(x) - f^2(x)g^2(x) equiv 1 (mod x ^{2n})\ f(x)(2g(x)-f(x)g^2(x))equiv 1(mod x ^ {2n}) ]

所以，f(x)在(mod x^{2n})下的逆元就是(2g(x)-f(x)g^2(x))。递归求解即可。注意NTT的时候每次翻转的范围都要重新计算。

code

/*
* @Author: wxyww
* @Date: 2020-01-28 09:39:19
* @Last Modified time: 2020-01-28 11:46:14
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int N = 400010,mod = 998244353;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int qm(ll x,int y) {
	ll ret = 1;
	for(;y;y >>= 1,x = x * x % mod)
		if(y & 1) ret = ret * x % mod;

	return ret;
}
int rev[N];
void NTT(ll *a,int n,int xs) {
	
		for(int i = 0;i <= n;++i) 
		rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? (n >> 1) : 0);

	for(int i = 0;i <= n;++i)
		if(rev[i] > i) swap(a[i],a[rev[i]]);
	for(int m = 2;m <= n;m <<= 1) {
		ll w1 = qm(3,(mod - 1) / m);
		if(xs == -1) w1 = qm(w1,mod - 2);

		for(int i = 0;i < n;i += m) {
			ll w = 1;
			for(int k = 0;k < (m >> 1);++k) {
				ll u = a[i + k],t = w * a[i + k + (m >> 1)] % mod;
				a[i + k] = (u + t) % mod;a[i + k + (m >> 1)] = (u - t) % mod;
				w = w * w1 % mod;
			}
		}
	}

	if(xs == -1) {
		for(int i = 0,inv = qm(n,mod - 2);i < n;++i) a[i] = a[i] * inv % mod;
	}
}
ll A[N],B[N];
void solve(ll *F,ll *G,ll len) {
	if(len == 1) {
		G[0] = qm(F[0],mod - 2);
		return;
	}
	solve(F,G,len >> 1);
	for(int i = 0;i < len;++i) A[i] = F[i],B[i] = G[i];
	NTT(A,len << 1,1);NTT(B,len << 1,1);
	for(int i = 0;i < (len << 1);++i) 
		A[i] = 1ll * A[i] * B[i] % mod * B[i] % mod;
	NTT(A,len << 1,-1);
	for(int i = 0;i < len;++i) G[i] = (2ll * G[i] % mod - A[i]) % mod;
}


ll f[N],g[N];
int main() {
	int n = read() - 1;

	for(int i = 0;i <= n;++i) f[i] = read();

	int tot = 1;
	while(tot <= n) tot <<= 1;

	solve(f,g,tot);

	for(int i = 0;i <= n;++i) printf("%d ",(g[i] + mod) % mod);
	return 0;
}