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，\(A(x)*B(x)\equiv 1(mod x^{\lceil \frac{n}{2} \rceil})\),\(A(x)*B'(x)\equiv 1(mod x^{\lceil \frac{n}{2} \rceil})\)

两式相减得\(B(x)-B'(x)\equiv 0(mod x^{\lceil \frac{n}{2} \rceil})\)，平方得\(B(x)^2-2*B(x)*B'(x)+B'(x)^2\equiv 0(mod x^n)\)

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
    
     #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
     
      #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
      
       #define pil pair
       
        #define pli pair
        
         #define pii pair
         
          //#define cd complex
          
           #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
           
            inline T const& MAX(T const &a,T const &b){return a>b?a:b;}template
            
             inline T const& MIN(T const &a,T const &b){return 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<
              
               >1]>>1) | ((i&1)<<(bit-1));}void ntt(ll *a,int n,int dft){ for(int i=0;i
               
                >1,a,b); int sz=0;while((1<
                
                 <=deg)sz++;sz++; getrev(sz);int len=1<