1. 程式人生 > >2018.11.14 uoj#34. 多項式乘法()

2018.11.14 uoj#34. 多項式乘法()

傳送門
N O I p NOIp 爆炸不能阻止我搞 o i

oi 的決心
資訊科技課進行一點康復訓練。
f f t fft 板題。
程式碼:

#include<bits/stdc++.h>
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); int n,m,lim=1,tim=0,pos[N]; struct Complex{ double x,y; friend inline Complex operator+(const Complex&a,
const Complex&b){return (Complex){a.x+b.x,a.y+b.y};} friend inline Complex operator-(const Complex&a,const Complex&b){return (Complex){a.x-b.x,a.y-b.y};} friend inline Complex operator*(const Complex&a,const Complex&b){return (Complex){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};} }a[N],b[N]; inline void fft(Complex a[],int type){ for(int i=0;i<lim;++i)if(i<pos[i])swap(a[i],a[pos[i]]); for(int mid=1;mid<lim;mid<<=1){ Complex w_n=(Complex){cos(pi/mid),type*sin(pi/mid)}; for(int j=0,len=mid<<1;j<lim;j+=len){ Complex w=(Complex){1,0}; for(int k=0;k<mid;++k,w=w*w_n){ Complex a0=a[j+k],a1=w*a[j+k+mid]; a[j+k]=a0+a1,a[j+k+mid]=a0-a1; } } } } int main(){ n=read(),m=read(); for(int i=0;i<=n;++i)a[i].x=read(); for(int i=0;i<=m;++i)b[i].x=read(); while(lim<=n+m)lim<<=1,++tim; for(int i=0;i<lim;++i)pos[i]=(pos[i>>1]>>1)|((i&1)<<(tim-1)); fft(a,1),fft(b,1); for(int i=0;i<lim;++i)a[i]=a[i]*b[i]; fft(a,-1); for(int i=0;i<=n+m;++i)printf("%d ",(int)(a[i].x/lim+0.5)); return 0; }