[模板]FFT

郝神并没有令我明白这个。

但是巨神的题解太强了。

#include <iostream>
#include <complex>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=;
const double pi=acos(-1.0);
struct complexd{
    double x,y;
    complexd (double xx=,double yy=){x=xx,y=yy;}
    complexd operator + (const complexd &rhs) const {return complexd(x+rhs.x,y+rhs.y);}
    complexd operator - (const complexd &rhs) const {return complexd(x-rhs.x,y-rhs.y);}
    complexd operator * (const complexd &rhs) const {return complexd(x*rhs.x-y*rhs.y,x*rhs.y+y*rhs.x);}
};
inline int rd() {
    static int x;x=;static char ch;ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) x=x*+(ch^),ch=getchar();
    return x;
}
int n,m,l,r[N];
void fft(complexd *b,short tag) {
    for(int i=;i<n;i++) if(i<r[i]) swap(b[i],b[r[i]]);
    for(int mid=;mid<n;mid<<=) {
        complexd tp(cos(pi/mid),tag*sin(pi/mid));
        for(int j=;j<n;j+=mid<<) {
            complexd w(,);
            for(int k=;k<mid;k++,w=w*tp) {
                complexd x=b[j+k],y=w*b[j+mid+k];
                b[j+k]=x+y;
                b[j+mid+k]=x-y;
            }
        }
    }
}
complexd a[N],b[N];
int main() {
    n=rd(),m=rd();
    for(int i=;i<=n;i++) a[i].x=rd();
    for(int j=;j<=m;j++) b[j].x=rd();
    m+=n;
    n=;
    while(n<=m) n<<=,l++;
    for(int i=;i<n;i++) r[i]=(r[i>>]>>)|((i&)<<(l-));
    fft(a,);
    fft(b,);
    for(int i=;i<=n;i++) a[i]=a[i]*b[i];
    fft(a,-);
    for(int i=;i<=m;i++) printf("%d ",(int)(a[i].x/n+0.5));
    return ;
}

FFT