FFT（模板）

优美

这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>using namespace std;const int N=301000;
const double pi=acos(-1.0);
struct node{double x,y;node (double xx=0,double yy=0){x=xx;y=yy;}
};
node a[N],b[N],omega[N],a_omega[N];
int n,m,fn;node operator +(const node &a,const node &b){return node(a.x+b.x,a.y+b.y);}
node operator -(const node &a,const node &b){return node(a.x-b.x,a.y-b.y);}
node operator *(const node &a,const node &b){return node (a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}void init(int n)
{for (int i=0;i<n;i++){omega[i]=node(cos(2.0*i*pi/n),sin(2.0*i*pi/n));a_omega[i]=node(cos(2.0*i*pi/n),-sin(2.0*i*pi/n));}
} void FFT(int n,node *a,node *w)
{int i,j=0,k;for (i=0;i<n;i++){if (i>j) swap(a[i],a[j]);for (int l=n>>1;(j^=l)<l;l>>=1);   ///}for (i=2;i<=n;i<<=1){int m=i>>1;for (j=0;j<n;j+=i)for (k=0;k<m;k++){node z=a[j+k+m]*w[n/i*k];  ///a[j+m+k]=a[j+k]-z;a[j+k]=a[j+k]+z;}}
}int main()
{scanf("%d%d",&n,&m);for (int i=0;i<=n;i++) scanf("%lf",&a[i].x);for (int i=0;i<=m;i++) scanf("%lf",&b[i].x);fn=1;while (fn<=n+m) fn<<=1;init(fn);  //处理主n次方根 FFT(fn,a,omega);   //转成点值表达 FFT(fn,b,omega);for (int i=0;i<=fn;i++)a[i]=a[i]*b[i];FFT(fn,a,a_omega);  //转成系数表达for (int i=0;i<=n+m;i++) printf("%d ",(int)(a[i].x/fn+0.5)); return 0; 
}