FFT(快速傅里叶变换) 模板

洛谷 P3803 【模板】多项式乘法（FFT）传送门

存个板子，完全弄懂之后找机会再写个详解。

 1 #include
 2 #include
 3 
 4 struct cpx
 5 {
 6     double rl,im;
 7     friend cpx operator + (cpx q,cpx w)
 8     {
 9         return (cpx){q.rl+w.rl,q.im+w.im};
10     }
11     friend cpx operator - (cpx q,cpx w)
12     {
13         return (cpx){q.rl-w.rl,q.im-w.im};
14     }
15     friend cpx operator * (cpx q,cpx w)
16     {
17         return (cpx){q.rl*w.rl-q.im*w.im,q.rl*w.im+q.im*w.rl};
18     }
19     friend cpx operator ~ (cpx q)
20     {
21         return (cpx){q.rl,-q.im};
22     }
23 }urt[2100005][2],af[2100005],bf[2100005];
24 
25 void swap(cpx &q,cpx &w)
26 {
27     cpx t=q;q=w;w=t;
28 }
29 
30 int n=1,cnt,na,nb;
31 int r[2100005];
32 const double pi=acos(-1);
33 
34 void prefft()
35 {
36     for(int i=0;i)
37     {
38         urt[i][0]=(cpx){cos(2*pi*i/n),sin(2*pi*i/n)};
39         urt[i][1]=~urt[i][0];
40         r[i]=(r[i>>1]>>1)|((i&1)<<(cnt-1));
41     }
42 }
43 
44 void fft(cpx *a,int inv)
45 {
46     for(int i=0;iif(i<r[i])swap(a[i],a[r[i]]);
47     for(int l=2;l<=n;l<<=1)
48     {
49         int m=l>>1;
50         for(cpx *p=a;p!=a+n;p+=l)
51         {
52             for(int i=0;i)
53             {
54                 cpx t=urt[n/l*i][inv]*p[i+m];
55                 p[i+m]=p[i]-t;
56                 p[i]=p[i]+t;
57             }
58         }
59     }
60     if(inv)for(int i=0;in;
61 }
62 
63 int main()
64 {
65     scanf("%d%d",&na,&nb);
66     while(n<=na+nb)n<<=1,cnt++;
67     for(int i=0;i<=na;i++)scanf("%lf",&af[i].rl);
68     for(int i=0;i<=nb;i++)scanf("%lf",&bf[i].rl);
69     prefft();
70     fft(af,0);
71     fft(bf,0);
72     for(int i=0;ibf[i];
73     fft(af,1);
74     for(int i=0;i<=na+nb;i++)printf("%d ",(int)(af[i].rl+0.5));
75     return 0;
76 }

 

转载于:https://www.cnblogs.com/eternhope/p/FFT.html