习题9-2 计算两个复数之积 (15分)
本题要求实现一个计算复数之积的简单函数。
函数接口定义:
struct complex multiply(struct complex x, struct complex y);
其中struct complex
是复数结构体,其定义如下:
struct complex{
int real;
int imag;
};
裁判测试程序样例:
#include <stdio.h>
struct complex{
int real;
int imag;
};
struct complex multiply(struct complex x, struct complex y);
int main()
{
struct complex product, x, y;
scanf("%d%d%d%d", &x.real, &x.imag, &y.real, &y.imag);
product = multiply(x, y);
printf("(%d+%di) * (%d+%di) = %d + %di\n",
x.real, x.imag, y.real, y.imag, product.real, product.imag);
return 0;
}
/* 你的代码将被嵌在这里 */
输入样例:
3 4 5 6
输出样例:
(3+4i) * (5+6i) = -9 + 38i
#include <stdio.h> struct complex{ int real; int imag; }; struct complex multiply(struct complex x, struct complex y); int main() { struct complex product, x, y; scanf("%d%d%d%d", &x.real, &x.imag, &y.real, &y.imag); product = multiply(x, y); printf("(%d+%di) * (%d+%di) = %d + %di\n", x.real, x.imag, y.real, y.imag, product.real, product.imag); return 0; } /* (a+bi)(c+di)=(ac-bd)+(bc+ad)i. */ struct complex multiply(struct complex x, struct complex y){ struct complex z; z.real = x.real * y.real - x.imag * y.imag; z.imag = x.real * y.imag + x.imag * y.real; return z; }