P4238 【模板】多项式求逆 ntt

题意：求多项式的逆
题解：多项式最高次项叫度deg，假设我们对于多项式(A(x)*B(x)equiv 1)，已知A，求B
假设度为n-1，(A(x)*B(x)equiv 1(mod x^{lceil frac{n}{2} ceil})),(A(x)*B'(x)equiv 1(mod x^{lceil frac{n}{2} ceil}))
两式相减得(B(x)-B'(x)equiv 0(mod x^{lceil frac{n}{2} ceil}))，平方得(B(x)^2-2*B(x)*B'(x)+B'(x)^2equiv 0(mod x^n))
注意到mod数也平方了，这是因为如果(A(x)equiv 0(modx^n)),就说明A的0-n-1项都是0，对于n到2*n-1项第x项来说有(sum_{i=1}^x A(i)*A(x-i))，一定有一项小于n，则必为0
对于上式，两边同乘A(x)，则有(B(x)=2*B'(x)-A(x)*B'(x))
可以递推解决

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;

ll a[N<<3],b[N<<3],c[N<<3];
int rev[N<<3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void ntt(ll *a,int n,int dft)
{
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        ll wn=qp(3,(mod-1)/(step*2));
        if(dft==-1)wn=qp(wn,mod-2);
        for(int j=0;j<n;j+=step<<1)
        {
            ll wnk=1;
            for(int k=j;k<j+step;k++)
            {
                ll x=a[k];
                ll y=wnk*a[k+step]%mod;
                a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod;
                wnk=wnk*wn%mod;
            }
        }
    }
    if(dft==-1)
    {
        ll inv=qp(n,mod-2);
        for(int i=0;i<n;i++)a[i]=a[i]*inv%mod;
    }
}
void pol_inv(int deg,ll *a,ll *b)
{
    if(deg==1){b[0]=qp(a[0],mod-2);return ;}
    pol_inv((deg+1)>>1,a,b);
    int sz=0;while((1<<sz)<=deg)sz++;sz++;
    getrev(sz);int len=1<<sz;
    for(int i=0;i<deg;i++)c[i]=a[i];
    for(int i=deg;i<len;i++)c[i]=0;
    ntt(c,len,1),ntt(b,len,1);
    for(int i=0;i<len;i++)
        b[i]=(2ll-c[i]*b[i]%mod+mod)%mod*b[i]%mod;
    ntt(b,len,-1);
    for(int i=deg;i<len;i++)b[i]=0;
}
int main()
{
    int n;scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%lld",&a[i]);
    pol_inv(n,a,b);
    for(int i=0;i<n;i++)printf("%lld ",b[i]);puts("");
    return 0;
}
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