c语言简单运用，直接坐标转极坐标

Question

请编写程序，输入直角坐标 ( x,y )，输出对应的极坐标 ( r,θ )。

输入格式

( x,y )

输出格式

( r,θ )

注：θ 的单位为弧度，且 −π<θ≤π。

要求：r 和 θ 输出10位有效数字，不输出小数末尾多余的零和小数点。

输入样例1

( -1.29903810567665797, 0.75 )

输出样例1

( 1.5, 2.617993878 )

输入样例2

( -2, -3.4641016151377546 )

输出样例2

( 4, -2.094395102 )

作者

李祥

单位

湖北经济学院

代码长度限制

16 KB

时间限制

400 ms

内存限制

Answer 1

#include<stdio.h>
#include<math.h>
struct complex_s{
    double x,y;
};
double real_part(struct complex_s z){
    return z.x;
}
double img_part(struct complex_s z){
    return z.y;
}
double magnitude(struct complex_s z){
    return sqrt(z.x*z.x + z.y*z.y);
}
double angle(struct complex_s z){
    return atan2(z.y, z.x);
}
struct complex_s make_from_real_img(double x, double y){
    struct complex_s z;
    z.x = x;
    z.y = y;
    return z;
}
struct complex_s make_from_mag_ang(double r, double A){
    struct complex_s z;
    z.x = r * cos(A);
    z.y = r * sin(A);
    return z;
}
struct complex_s add_complex(struct complex_s z1,struct complex_s z2){
    return make_from_real_img(real_part(z1)+real_part(z2),
        img_part(z1) + img_part(z2));
}
struct complex_s sub_complex(struct complex_s z1,struct complex_s z2){
    return make_from_real_img(real_part(z1)-real_part(z2),
        img_part(z1) - img_part(z2));
}
struct complex_s mul_complex(struct complex_s z1,struct complex_s z2){
    return make_from_mag_ang(real_part(z1)*real_part(z2),
        img_part(z1) + img_part(z2));
}
struct complex_s div_complex(struct complex_s z1,struct complex_s z2){
    return make_from_mag_ang(real_part(z1)/real_part(z2),
        img_part(z1) + img_part(z2));
}
int main(void){
    struct complex_s z1 = {3.0,4.0};
    struct complex_s z2= {2.0,5.0};
    struct complex_s x;
    x = add_complex(z1,z2);
    printf("x={%f,%f}",x.x,x.y);
    return 0;
}