FFT(快速傅里叶变换) 模板

洛谷 P3803 【模板】多项式乘法（FFT）传送门

存个板子，完全弄懂之后找机会再写个详解。

 #include<cstdio>
 #include<cmath>

 struct cpx
 {
     double rl,im;
     friend cpx operator + (cpx q,cpx w)
     {
         return (cpx){q.rl+w.rl,q.im+w.im};
     }
     friend cpx operator - (cpx q,cpx w)
     {
         return (cpx){q.rl-w.rl,q.im-w.im};
     }
     friend cpx operator * (cpx q,cpx w)
     {
         return (cpx){q.rl*w.rl-q.im*w.im,q.rl*w.im+q.im*w.rl};
     }
     friend cpx operator ~ (cpx q)
     {
         return (cpx){q.rl,-q.im};
     }
 }urt[][],af[],bf[];

 void swap(cpx &q,cpx &w)
 {
     cpx t=q;q=w;w=t;
 }

 int n=,cnt,na,nb;
 int r[];
 const double pi=acos(-);

 void prefft()
 {
     for(int i=;i<n;i++)
     {
         urt[i][]=(cpx){cos(*pi*i/n),sin(*pi*i/n)};
         urt[i][]=~urt[i][];
         r[i]=(r[i>>]>>)|((i&)<<(cnt-));
     }
 }

 void fft(cpx *a,int inv)
 {
     for(int i=;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
     for(int l=;l<=n;l<<=)
     {
         int m=l>>;
         for(cpx *p=a;p!=a+n;p+=l)
         {
             for(int i=;i<m;i++)
             {
                 cpx t=urt[n/l*i][inv]*p[i+m];
                 p[i+m]=p[i]-t;
                 p[i]=p[i]+t;
             }
         }
     }
     if(inv)for(int i=;i<n;i++)a[i].rl/=n;
 }

 int main()
 {
     scanf("%d%d",&na,&nb);
     while(n<=na+nb)n<<=,cnt++;
     for(int i=;i<=na;i++)scanf("%lf",&af[i].rl);
     for(int i=;i<=nb;i++)scanf("%lf",&bf[i].rl);
     prefft();
     fft(af,);
     fft(bf,);
     for(int i=;i<n;i++)af[i]=af[i]*bf[i];
     fft(af,);
     for(int i=;i<=na+nb;i++)printf("%d ",(int)(af[i].rl+0.5));
     return ;
 }