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

源码

struct complex multiply(struct complex x, struct complex y)
{
	struct complex product;
	product.real = x.real*y.real - x.imag*y.imag;
	product.imag = x.real*y.imag + x.imag*y.real;
	return product;
}

***谢谢！！！

标签：real,product,15,struct,int,imag,之积,complex,习题
来源： https://blog.csdn.net/weixin_45784564/article/details/113069234