洛谷 P3803 多项式乘法（FFT） —— FFT

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef double db;
int const xn=(1<<21);//*2 n+m
db const Pi=acos(-1.0);
int n,m,rev[xn],lim;
struct com{db x,y;}a[xn],b[xn];
com operator + (com a,com b){return (com){a.x+b.x,a.y+b.y};}
com operator - (com a,com b){return (com){a.x-b.x,a.y-b.y};}
com operator * (com a,com b){return (com){a.x*b.x-a.y*b.y,a.x*b.y+b.x*a.y};}
int rd()
{
  int ret=0,f=1; char ch=getchar();
  while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0; ch=getchar();}
  while(ch>=‘0‘&&ch<=‘9‘)ret=(ret<<3)+(ret<<1)+ch-‘0‘,ch=getchar();
  return f?ret:-ret;
}
void fft(com *a,int tp)
{
  for(int i=0;i<lim;i++)
    if(i<rev[i])swap(a[i],a[rev[i]]);
  for(int mid=1;mid<lim;mid<<=1)//mid<lim
    {
      com wn=(com){cos(Pi/mid),tp*sin(Pi/mid)};
      for(int j=0,len=(mid<<1);j<lim;j+=len)
    {
      com w=(com){1,0};
      for(int k=0;k<mid;k++,w=w*wn)
        {
          com x=a[j+k],y=w*a[j+mid+k];
          a[j+k]=x+y;
          a[j+mid+k]=x-y;
        }
    }
    }
}
int main()
{
  n=rd(); m=rd();
  for(int i=0;i<=n;i++)a[i].x=rd();
  for(int i=0;i<=m;i++)b[i].x=rd();
  lim=1; int l=0;
  while(lim<=n+m)lim<<=1,l++;
  for(int i=0;i<lim;i++)
    rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
  fft(a,1); fft(b,1);
  for(int i=0;i<lim;i++)a[i]=a[i]*b[i];
  fft(a,-1);
  for(int i=0;i<=n+m;i++)
    printf("%d ",(int)(a[i].x/lim+0.5)); puts("");
  return 0;
}