【模板】FFT快速傅里叶变换

 1 struct Complex{
 2     double x, y;
 3     inline Complex(double xx=0, double yy=0){
 4         x=xx; y=yy;
 5     }
 6     inline Complex operator + (Complex a){
 7         return Complex(x+a.x, y+a.y);
 8     }
 9     inline Complex operator - (Complex a){
10         return Complex(x-a.x, y-a.y);
11     }
12     inline Complex operator * (Complex a){
13         return Complex(x*a.x-y*a.y, x*a.y+y*a.x);
14     }
15 }A[N], B[N];
16
17 int n, m, L;
18 int rev[N];
19
20 void FFT(Complex *a, int le, int ty){
21     for (int i=0; i<le; ++i)
22         if (i < rev[i]) swap(a[i], a[rev[i]]);
23     for (int i=1; i<le; i<<=1){
24         Complex wn=Complex(cos(pi/i), ty*sin(pi/i));
25         for (int j=0; j<le; j+=i<<1){
26             Complex w=Complex(1, 0), x, y;
27             for (int k=0; k<i; ++k, w=w*wn){
28                 x=a[j+k]; y=a[j+i+k]*w;
29                 a[j+k]=x+y; a[j+i+k]=x-y;
30             }
31         }
32     }
33 }
34
35 int main(){
36     scanf("%d%d", &n, &m); ++n; ++m;
37     for (L=1; L<n+m-1; L<<=1) continue;
38     for (int i=0; i<L; ++i){
39         rev[i]=rev[i>>1]>>1;
40         if (i&1) rev[i]|=L>>1;
41     }
42     for (int i=0, z; i<n; ++i) scanf("%d", &z), A[i].x=z;
43     for (int i=0, z; i<m; ++i) scanf("%d", &z), B[i].x=z;
44     FFT(A, L, 1);
45     FFT(B, L, 1);
46     for (int i=0; i<L; ++i) A[i]=A[i]*B[i];
47     FFT(A, L, -1);
48     for (int i=0; i<n+m-1; ++i) printf("%d ", (int)(A[i].x/L+0.5));
49
50     return 0;
51 }

原文地址：https://www.cnblogs.com/Dance-Of-Faith/p/8461382.html