Luogu4238 【模板】多项式乘法逆

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

\(NTT\)

递归求解

\[假设已知F(n)H(n) \equiv 1 (\mod x^{\lceil \frac{n}{2} \rceil} )\\ F(n)G(n)\equiv 1 (\mod x^{\lceil \frac{n}{2} \rceil})\\ \therefore F(n)(G(n)-H(n)) \equiv 0 (\mod x^{\lceil \frac{n}{2} \rceil} )\\ G(n)-H(n) \equiv 0 (\mod x^{\lceil \frac{n}{2} \rceil })\\ 令P(n)=G(n)-H(n),C(n)=P(n)^2\\ C_{i}=\sum_{j=0}^i P_j *P_{i-j}\\ j和i-j中，必然有一项值小于等于\lceil \frac{n}{2} \rceil\\ 那么P_i和P_{i-j}中必有一项为0(\because P(n) \equiv 0 (\mod x^{\lceil \frac{n}{2} \rceil}))\\ \therefore C(n) \equiv 0 (\mod x^n)\\ (G(n)-H(n))^2 \equiv 0 (\mod x^n)\\ G(n)^2-2G(n)H(n)+H(n)^2 \equiv 0 (\mod x^n)\\ F(n)G(n)^2-2F(n)G(n)H(n)+F(n)H(n)^2 \equiv 0 (\mod x^n)\\ G(n)\equiv 2H(n)-F(n)H(n)^2 (\mod x^n) \]

\(C++ Code:\)

#include<iostream>
#include<cstdio>
#include<algorithm>
#define p 998244353
#define Gi 3
#define N 400005
#define ll long long
using namespace std;
int n,l,lg2[N << 2],rev[N << 2];
ll G[2][24],s,f[N],g[N],a[N],b[N];
inline ll ksm(ll x,ll y)
{
    ll ans=1;
    while (y)
    {
        if (y & 1)
            ans=ans*x%p;
        x=x*x%p;
        y >>=1;
    }
    return ans;
}
#define inv(x) (ksm(x,p-2))
inline void NTT(ll *a,int t)
{
    for (int i=0;i<s;i++)
        if (i<rev[i])
            swap(a[i],a[rev[i]]);
    for (int mid=1,o=1;mid<s;mid <<=1,o++)
    {
        ll gn=G[t][o];
        for (int j=0;j<s;j+=(mid << 1))
        {
            ll g=1;
            for (int k=0;k<mid;k++,g=g*gn%p)
            {
                ll x=a[j+k],y=g*a[j+k+mid]%p;
                a[j+k]=(x+y)%p;
                a[j+k+mid]=(x-y)%p;
            }
        }
    }
}
void dfs(int l,int r)
{
    if (l==r)
    {
        g[l]=inv(f[l]);
        return;
    }
    int mid=(l+r) >> 1;
    dfs(l,mid);
    s=1 << (lg2[r-l+1]+1);
    for (int i=0;i<s;i++)
        rev[i]=(rev[i >> 1] >> 1) | ( (i & 1) << lg2[r-l+1] );
    for (int i=l;i<=mid;i++)
        a[i]=g[i];
    for (int i=mid+1;i<s;i++)
        a[i]=0;
    for (int i=l;i<=r;i++)
        b[i]=f[i];
    for (int i=r+1;i<s;i++)
        b[i]=0;
    NTT(a,0);
    NTT(b,0);
    for (int i=0;i<s;i++)
        a[i]=a[i]*a[i]%p*b[i]%p;
    NTT(a,1);
    ll t=inv(s);
    for (int i=0;i<s;i++)
        a[i]=a[i]*t%p;
    for (int i=0;i<r-l+1;i++)
        g[i]=(g[i]*2-a[i])%p;
}
void Pre()
{
     G[0][23]=ksm(3,(p-1)/(1 << 23));
     G[1][23]=inv(G[0][23]);
     for (int i=22;i>=1;i--)
     {
         G[0][i]=G[0][i+1]*G[0][i+1]%p;
         G[1][i]=G[1][i+1]*G[1][i+1]%p;
     }
}
int main()
{
    Pre();
    scanf("%d",&n);
    for (int i=0;i<n;i++)
        scanf("%lld",&f[i]);
    lg2[0]=0;
    l=0;
    for (int i=1;i<=n;i++)
    {
        if ((1 << l)<i)
            l++;
        lg2[i]=l;
    }
    for (int i=n+1;i<=(1 << lg2[n]);i++)
        lg2[i]=lg2[n];
    dfs(0,(1 << lg2[n])-1);
    for (int i=0;i<n;i++)
        g[i]=(g[i]%p+p)%p;
    for (int i=0;i<n;i++)
        printf("%lld ",g[i]);
    putchar('\n');
    return 0;
}