uoj34 多项式乘法

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
const int N=3e5+10;
const double pi=acos(-1);
struct cp
{
double x,y;
cp(){}
cp(double xx,double yy):x(xx),y(yy){}
inline cp operator +(const cp &other){return cp(x+other.x,y+other.y);}
inline cp operator -(const cp &other){return cp(x-other.x,y-other.y);}
inline cp operator *(const cp &other){return cp(x*other.x-y*other.y,y*other.x+x*other.y);}
inline double real(){return x;}
}a
,b
;
int r
;
int n,m,x,len;
inline void FFT(cp *a,int f)
{
for(int i=0;i<n;i++)
if(i<r[i]) swap(a[i],a[r[i]]);
for(int i=1;i<n;i<<=1)
{
cp wn(cos(pi/i),f*sin(pi/i));
for(int j=0,tmp=i<<1;j<n;j+=tmp)
{
cp w(1,0);
for(int k=0;k<i;k++,w=w*wn)
{
cp x=a[j+k],y=w*a[j+k+i];
a[j+k]=x+y,a[j+k+i]=x-y;
}
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++)
scanf("%d",&x),a[i].x=x;
for(int i=0;i<=m;i++)
scanf("%d",&x),b[i].x=x;
m+=n;
for(n=1;n<=m;n<<=1)
len++;
for(int i=0;i<n;i++)
r[i]=(r[i>>1]>>1)|((i&1)<<(len-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 ",(int)(a[i].real()/n+0.5));
return 0;
}

#include<cmath>
#include<cstdio>
#include<complex>
#include<iostream>
using namespace std;
typedef complex<double> cp;
const int N=3e5+10;
const double pi=acos(-1);
int r
;
int n,m,x,len;
cp a
,b
;
inline void FFT(cp *a,int f)
{
for(int i=0;i<n;i++)
if(i<r[i]) swap(a[i],a[r[i]]);
for(int i=1;i<n;i<<=1)
{
cp wn(cos(pi/i),f*sin(pi/i));
for(int j=0,tmp=i<<1;j<n;j+=tmp)
{
cp w(1,0);
for(int k=0;k<i;k++,w*=wn)
{
cp x=a[j+k],y=w*a[j+k+i];
a[j+k]=x+y,a[j+k+i]=x-y;
}
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++)
scanf("%d",&x),a[i]=x;
for(int i=0;i<=m;i++)
scanf("%d",&x),b[i]=x;
m+=n;
for(n=1;n<=m;n<<=1)
len++;
for(int i=0;i<n;i++)
r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
FFT(a,1),FFT(b,1);
for(int i=0;i<=n;i++)
a[i]*=b[i];
FFT(a,-1);
for(int i=0;i<=m;i++)
printf("%d ",(int)(a[i].real()/n+0.5));
return 0;
}