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[2100005][2],af[2100005],bf[2100005];

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

int n=1,cnt,na,nb;
int r[2100005];
const double pi=acos(-1);

void prefft()
{
    for(int i=0;i<n;i++)
    {
        urt[i][0]=(cpx){cos(2*pi*i/n),sin(2*pi*i/n)};
        urt[i][1]=~urt[i][0];
        r[i]=(r[i>>1]>>1)|((i&1)<<(cnt-1));
    }
}

void fft(cpx *a,int inv)
{
    for(int i=0;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
    for(int l=2;l<=n;l<<=1)
    {
        int m=l>>1;
        for(cpx *p=a;p!=a+n;p+=l)
        {
            for(int i=0;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=0;i<n;i++)a[i].rl/=n;
}

int main()
{
    scanf("%d%d",&na,&nb);
    while(n<=na+nb)n<<=1,cnt++;
    for(int i=0;i<=na;i++)scanf("%lf",&af[i].rl);
    for(int i=0;i<=nb;i++)scanf("%lf",&bf[i].rl);
    prefft();
    fft(af,0);
    fft(bf,0);
    for(int i=0;i<n;i++)af[i]=af[i]*bf[i];
    fft(af,1);
    for(int i=0;i<=na+nb;i++)printf("%d ",(int)(af[i].rl+0.5));
    return 0;
}