[UOJ#34]多项式乘法（FFT）

#include <cstdio>
#include <iostream>
#include <cmath>
#define N 300005
using namespace std;
const double pi=acos(-1.0);
int n,m,L,r
;
struct complex
{
double x,y;
complex(double X=0,double Y=0)
{x=X,y=Y;}

}a
,b
;
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);}
void FFT(complex a
,int id)
{
for (int i=0;i<n;i++)
if (i<r[i]) swap(a[i],a[r[i]]);
for (int k=1;k<n;k<<=1)
{
complex wn=complex(cos(pi/k),id*sin(pi/k));
for (int i=0;i<n;i+=(k<<1))
{
complex w=complex(1,0);
for (int j=0;j<k;j++,w=w*wn)
{
complex x=a[i+j],y=w*a[i+j+k];
a[i+j]=x+y,a[i+j+k]=x-y;
}
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for (int i=0;i<=n;i++) scanf("%lf",&a[i].x);
for (int i=0;i<=m;i++) scanf("%lf",&b[i].x);
m+=n;
for (n=1;n<=m;n<<=1) ++L;
for (int i=0;i<n;i++)
r[i]=(r[i>>1]>>1) | ((i&1)<<(L-1));
FFT(a,1); FFT(b,1);
for (int i=0;i<=n;i++) a[i]=a[i]*b[i];
FFT(a,-1);
for (int i=0;i<=m;i++)
printf("%d%c",(int)(a[i].x/n+0.5)," \n"[i==m]);
}