快速傅里叶变换(FFT)递归

#include <cstdio>
#include <cmath>
#define Maxn 5000
#define Pi 3.1415926535898
using namespace std;
int n,m;
struct complex
{complex (double xx=0,double yy=0){x=xx,y=yy;}
double x,y;};
complex operator + (complex a,complex b){
  return complex(a.x+b.x,a.y+b.y);
}complex operator - (complex a,complex b){
  return complex(a.x-b.x,a.y-b.y);
}complex operator * (complex a,complex b){
  return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}complex w[Maxn],b[Maxn],c[Maxn];
void fft(complex *f,int len,short op)
{
  if (!len)return ;
  complex fl[len+1],fr[len+1];
  for (int k=0;k<len;k++)
   {fl[k]=f[k<<1];fr[k]=f[k<<1|1];}
  fft(fl,len>>1,op);
  fft(fr,len>>1,op);
  complex tmp,buf;
  tmp=complex(cos(Pi/len),sin(Pi/len)*op);
  buf.x=1;buf.y=0;
  for (int k=0;k<len;k++){
    complex t=buf*fr[k];
    f[k]=fl[k]+t;
    f[k+len]=fl[k]-t;
    buf=buf*tmp;
  }
}
int main()
{
  scanf("%d%d",&n,&m);
  for (int i=0;i<=n;i++)scanf("%lf",&b[i].x);
  for (int i=0;i<=m;i++)scanf("%lf",&c[i].x);
  for(m+=n,n=1;n<=m;n<<=1);
  fft(b,n>>1,1);
  fft(c,n>>1,1);
  for(int i=0;i<n;++i)b[i]=b[i]*c[i];
  fft(b,n>>1,-1);
  for(int i=0;i<=m;++i)printf("%.0f ",fabs(b[i].x)/n);
  puts("");
  return 0;
}