bzoj2194: 快速傅立叶之二

 1 #include<bits/stdc++.h>
 2 #define pi acos(-1)
 3 using namespace std;
 4 const int inf=4e5+10;
 5 int n;
 6 struct cp{
 7     double x,y;
 8     cp operator * (const cp &o)const{
 9         return (cp){x*o.x-y*o.y,x*o.y+y*o.x};
10     }
11     cp operator + (const cp &o)const{
12         return (cp){x+o.x,y+o.y};
13     }
14     cp operator - (const cp &o)const{
15         return (cp){x-o.x,y-o.y};
16     }
17 }a[inf],b[inf];
18 int r[inf];
19 void fft(cp *a,int l,int type){
20     for(int i=0;i<l;i++)
21         if(i<r[i])swap(a[i],a[r[i]]);
22     for(int i=2;i<=l;i<<=1){
23         cp wn=(cp){cos(2*pi/i),sin(2*pi*type/i)};
24         for(int j=0;j<l;j+=i){
25             cp w=(cp){1,0};
26             for(int k=j;k<j+i/2;k++,w=w*wn){
27                 cp u=a[k],v=a[k+i/2]*w;
28                 a[k]=u+v;a[k+i/2]=u-v;
29             }
30         }
31     }
32 }
33 int main()
34 {
35     scanf("%d",&n);
36     for(int i=0;i<n;i++){
37         scanf("%lf",&a[i].x);
38         scanf("%lf",&b[n-i-1].x);
39     }
40     int l=1,h=0;
41     while(l<n*2-1)l<<=1,h++;
42     for(int i=1;i<l;i++)
43         r[i]=(r[i>>1]>>1)+((i&1)<<(h-1));
44     fft(a,l,1);fft(b,l,1);
45     for(int i=0;i<l;i++)a[i]=a[i]*b[i];
46     fft(a,l,-1);
47     for(int i=n-1;i<2*n-1;i++)printf("%d\n",(int)(a[i].x/l+0.5));
48     return 0;
49 }