LG3803 「模板」快速傅里叶变换 FFT

Posted 2020-11-25 liubainian

问题描述

LG3803

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题解

点我

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(mathrm{Code})

#include<bits/stdc++.h>
using namespace std;

const int maxn=1350000;
const double Pi=acos(-1);
int n,m;

struct CP{
    CP(double X=0,double Y=0) {x=X,y=Y;}
    double x,y;
    CP operator + (CP const &a) const
    {return CP(x+a.x,y+a.y);}
    CP operator - (CP const &a) const
    {return CP(x-a.x,y-a.y);}
    CP operator * (CP const &a) const
    {return CP(x*a.x-y*a.y,x*a.y+y*a.x);}
}f[maxn<<1],p[maxn<<1];

int tr[maxn<<1];

void fft(CP *f,bool type){
    for(int i=0;i<n;i++) if(i<tr[i]) swap(f[i],f[tr[i]]);
    for(int p=2;p<=n;p<<=1){
        int len=p>>1;
        CP tG(cos(2*Pi/p),sin(2*Pi/p));
        if(!type) tG.y*=-1;
        for(int k=0;k<n;k+=p){
            CP buf(1,0);
            for(int l=k;l<k+len;l++){
                CP tt=buf*f[l+len];
                f[l+len]=f[l]-tt;
                f[l]=f[l]+tt;
                buf=buf*tG;
            }
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;i++) scanf("%lf",&f[i].x);
    for(int i=0;i<=m;i++) scanf("%lf",&p[i].x);
    for(m+=n,n=1;n<=m;n<<=1);
    for(int i=0;i<n;i++) tr[i]=(tr[i>>1]>>1)|((i&1)?n>>1:0);
    fft(f,1);fft(p,1);
    for(int i=0;i<n;i++) f[i]=f[i]*p[i];
    fft(f,0);
    for(int i=0;i<=m;i++){
        printf("%d ",(int)(f[i].x/n+0.49));
    }
}