2018.11.18 bzoj2194: 快速傅立叶之二（fft）

传送门

模板题。

将bbb序列反过来然后上fftfftfft搞定。

代码：

#include<bits/stdc++.h>
#define ri register int
using namespace std;
inline int read(){
	int ans=0;
	char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans;
}
const int N=4e5+5;
const double pi=acos(-1.0);
struct Complex{
	double x,y;
	inline Complex operator+(const Complex&b){return (Complex){x+b.x,y+b.y};}
	inline Complex operator-(const Complex&b){return (Complex){x-b.x,y-b.y};}
	inline Complex operator*(const Complex&b){return (Complex){x*b.x-y*b.y,x*b.y+y*b.x};}
	inline Complex operator/(const double&b){return (Complex){x/b,y/b};}
}a[N],b[N];
int n,pos[N],lim,tim;
inline void init(){
	lim=1,tim=0;
	while(lim<=n*2)lim<<=1,++tim;
	for(ri i=0;i<lim;++i)pos[i]=(pos[i>>1]>>1)|((i&1)<<(tim-1));
}
inline void fft(Complex *a,int type){
	for(ri i=0;i<lim;++i)if(i<pos[i])swap(a[i],a[pos[i]]);
	for(ri mid=1;mid<lim;mid<<=1){
		Complex wn=(Complex){cos(pi/mid),type*sin(pi/mid)};
		for(ri j=0,len=mid<<1;j<lim;j+=len){
			Complex w=(Complex){1,0};
			for(ri k=0;k<mid;++k,w=w*wn){
				Complex a0=a[j+k],a1=w*a[j+k+mid];
				a[j+k]=a0+a1,a[j+k+mid]=a0-a1;
			}
		}
	}
	if(type==-1)for(ri i=0;i<lim;++i)a[i]=a[i]/lim;
}
int main(){
	freopen("lx.in","r",stdin);
	n=read()-1,init();
	for(ri i=0;i<=n;++i)a[i].x=read(),b[n-i].x=read();
	fft(a,1),fft(b,1);
	for(ri i=0;i<lim;++i)a[i]=a[i]*b[i];
	fft(a,-1);
	for(ri i=n;i<=n*2;++i)printf("%d\n",(int)(a[i].x+0.5));
	return 0;
}