复数 一级ADT实现

COMPLEX.h

 1 /*
 2 typedef struct
 3 {
 4     float RE; //实部
 5     float IM; //虚部
 6 }Complex;
 7 */
 8 typedef struct complex * Complex;
 9 
10 Complex COMPLEXinit(float, float);
11 float Re(Complex);
12 float Im(Complex);
13 Complex COMPLEXmult(Complex, Complex);

COMPLEX.c

 1 #include "COMPLEX.h"
 2 
 3 struct complex
 4 {
 5     float RE; //实部
 6     float IM; //虚部
 7 };
 8 
 9 Complex COMPLEXinit(float RE, float IM)
10 {
11     /*
12     Complex t;
13     t.RE=RE;
14     t.IM=IM;
15     return t;
16     */
17     
18     Complex t=malloc(sizeof *t);
19     t->RE=RE;
20     t->IM=IM;
21     return t;
22 }
23 float Re(Complex z)
24 {
25     return z->RE;
26 }
27 float Im(Complex z)
28 {
29     return z->IM;
30 }
31 Complex COMPLEXmult(Complex a, Complex b)
32 {
33     /*
34     Complex t;
35     t.RE=a.RE*b.RE-a.IM*b.IM;
36     t.IM=a.RE*b.IM+a.IM*b.RE;
37     
38     //a实部乘b实部-a虚部乘b虚部
39     //a实部乘b虚部+a虚部乘b实部
40     return t;*/
41     
42     return COMPLEXinit(Re(a)*Re(b)-Im(a)*Im(b),
43                        Re(a)*Im(b)+Im(a)*Re(b));
44 }

main.c

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include "COMPLEX.h"
 4 
 5 #define PI 3.141592625
 6 
 7 int main(void)
 8 {
 9     int N;
10     printf("输入一个参数:");
11     scanf("%d", &N);
12     getchar();
13     
14     Complex t, x;
15     printf("%dth complex roots of unity\n", N);
16     for(int i=0; i<N; i++)
17     {
18         float r=2.0*PI*i/N;
19         //1=e^(2n*pi*i)  ?
20 
21         
22         t=COMPLEXinit(cos(r), sin(r));
23         
24         printf("%2d %6.3f %6.3f ", i, Re(t), Im(t));
25         x=t;
26         for(int j=0; j<N-1; j++)
27             x=COMPLEXmult(t, x);
28             
29         printf("%6.3f %6.3f\n", Re(x), Im(x));
30     }
31     
32     return 0;
33 }