这里用的拆系数\(FFT\)（咱实在是不会三模……）

就是说把\(F_i\)和\(G_i\)中的每一个数字都拆成\(k\times M+b\)的形式，这里\(M\)可以取\(2^{15}=32768\)

于是把\(F\)拆出来的两个数列和\(G\)拆出来的两个数列分别相乘，最终对应的系数要乘上\(1,2^{15},2^{15},2^{30}\)

//minamoto#include
    
     #define R register#define ll long long#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int K=-1,Z=0;inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}void print(R int x){    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++K]=z[Z],--Z);sr[++K]=' ';}const int N=5e5+5;const double Pi=acos(-1.0);struct cp{    double x,y;    cp(double xx=0,double yy=0){x=xx,y=yy;}    inline cp operator +(cp b)const{return cp(x+b.x,y+b.y);}    inline cp operator -(cp b)const{return cp(x-b.x,y-b.y);}    inline cp operator *(cp b)const{return cp(x*b.x-y*b.y,x*b.y+y*b.x);}    inline cp operator *(const double &b)const{return cp(x*b,y*b);}}A[N],B[N],C[N],D[N],H[N],F[N],G[N],w[N];int r[N],lim=1,l,n,m,P,x;void FFT(cp *A,int ty){    fp(i,0,lim-1)if(i
      
       >1]>>1)|((i&1)<<(l-1));    for(R int i=1;i
       
        <<=1)fp(k,0,i-1)w[i+k]=cp(cos(Pi*k/i),sin(Pi*k/i));    fp(i,0,n)x=read(),A[i].x=x>>15,B[i].x=x&32767;    fp(i,0,m)x=read(),C[i].x=x>>15,D[i].x=x&32767;    FFT(A,1),FFT(B,1),FFT(C,1),FFT(D,1);    fp(i,0,lim-1)        F[i]=A[i]*C[i],G[i]=A[i]*D[i]+B[i]*C[i],H[i]=B[i]*D[i];    FFT(F,-1),FFT(G,-1),FFT(H,-1);    fp(i,0,n+m)print((((ll)(F[i].x+0.5)%P<<30)+((ll)(G[i].x+0.5)<<15)+((ll)(H[i].x+0.5)))%P);    return Ot(),0;}