多项式乘法逆元 - NTT

递归求解即可

#include 
using namespace std;
#define int long long
namespace NTT {
    #define pw(n) (1<>1]>>1)|((i&1)<<(l-1));
        }
    }
    int fastpow(int a,int b){
        int ans=1;
        for(;b;b>>=1,a=1LL*a*a%mod){
            if(b&1)ans=1LL*ans*a%mod;
        }
        return ans;
    }
    void NTT(int *s,int op){
        for(int i=0;i A,vector  B,vector  &C) {
        n=A.size()-1;
        m=B.size()-1;
        for(int i=0;i<=n;i++) a[i]=A[i];
        for(int i=0;i<=m;i++) b[i]=B[i];
        m+=n;
        bitnum=0;
        for(bit=1;bit<=m;bit<<=1)bitnum++;
        getrev(bitnum);
        NTT(a,1);
        NTT(b,1);
        for(int i=0;i a;
    void cut(int n) {
        while(a.size()>n) a.pop_back();
    }
    poly operator *(int b) {
        poly c=*this;
        for(int i=0;i>=1) if(q&1) r*=p, r%=mod;
    return r;
}

int inv(int p) {
    return qpow(p, mod-2);
}

poly solve(poly A, int n) {
    A.cut(n);
    poly B;
    if(n==1) {
        B.a.push_back(inv(A.a[0]));
    }
    else {

        poly Bi = solve(A,(n-1)/2+1);
        B = Bi*2 - A*Bi*Bi;
        B.cut(n);
    }
    return B;
}

signed main() {
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=0;i>a[i];
    poly A;
    for(int i=0;i