FFT(模板)
優美
這里寫代碼片 #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); //處理主n次方根 FFT(fn,a,omega); //轉成點值表達 FFT(fn,b,omega);for (int i=0;i<=fn;i++)a[i]=a[i]*b[i];FFT(fn,a,a_omega); //轉成系數表達for (int i=0;i<=n+m;i++) printf("%d ",(int)(a[i].x/fn+0.5)); return 0; }轉載于:https://www.cnblogs.com/wutongtong3117/p/7673400.html
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