模板题。

将$b$序列反过来然后上$fft$搞定。

代码：

#include
    
     #define ri register intusing 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
     
      >1]>>1)|((i&1)<<(tim-1));}inline void fft(Complex *a,int type){
    	for(ri i=0;i