【Luogu3808】多项式乘法FFT

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<complex>
#include<algorithm>
using namespace std;
#define MAX 2700000
inline int read()
{
register int x=0,t=1;
register char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-'){t=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
return x*t;
}
const double Pi=acos(-1);
int N,M,r[MAX],l;
complex<double> a[MAX],b[MAX];
void FFT(complex<double> *P,int opt)
{
for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
for(int i=1;i<N;i<<=1)
{
complex<double> W(cos(Pi/i),opt*sin(Pi/i));
for(int p=i<<1,j=0;j<N;j+=p)
{
complex<double> w(1,0);
for(int k=0;k<i;k++,w*=W)
{
complex<double> X=P[j+k],Y=w*P[j+k+i];
P[j+k]=X+Y;P[j+k+i]=X-Y;
}
}
}
}

int main()
{
N=read();M=read();
for(int i=0;i<=N;++i)a[i]=read();
for(int i=0;i<=M;++i)b[i]=read();
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 ",(int)(a[i].real()/N+0.5));
return 0;
}