A2/HC6800-ES-V2.0

A2/HC6800-ES-V2.0是一款基于6800微处理器系列的升级版芯片组。6800是一代经典的小型机CPU架构，而这个V2.0版本可能是在原有的HC6800基础上进行了增强或改进，例如提升了性能、增加了新的功能或是优化了电源管理等。它可能是应用于早期计算机系统，比如个人电脑或者嵌入式设备中，用于控制硬件操作和处理指令。由于具体的细节没有公开，对于更深入的技术规格或特性，通常需要参考相关的数据手册或技术文档。

相关问题

用C语言编写程序，要使用结构体， 不要使用函数，本题要求编写程序，计算2个复数的和、差、积、商。 输入格式： 输入在一行中按照a1 b1 a2 b2的格式给出2个复数C1=a1+b1i和C2=a2+b2i的实部和虚部。题目保证C2不为0。 输出格式： 分别在4行中按照(a1+b1i) 运算符 (a2+b2i) = 结果的格式顺序输出2个复数的和、差、积、商，数字精确到小数点后1位。如果结果的实部或者虚部为0，则不输出。如果结果为0，则输出0.0。 输入样例1： 2 3.08 -2.04 5.06 输出样例1： (2.0+3.1i) + (-2.0+5.1i) = 8.1i (2.0+3.1i) - (-2.0+5.1i) = 4.0-2.0i (2.0+3.1i) * (-2.0+5.1i) = -19.7+3.8i (2.0+3.1i) / (-2.0+5.1i) = 0.4-0.6i 输入样例2： 1 1 -1 -1.01 输出样例2： (1.0+1.0i) + (-1.0-1.0i) = 0.0 (1.0+1.0i) - (-1.0-1.0i) = 2.0+2.0i (1.0+1.0i) * (-1.0-1.0i) = -2.0i (1.0+1.0i) / (-1.0-1.0i) = -1.0

#include <stdio.h>

int main()
{
    struct complex // 定义结构体
    {
        double real; // 实部
        double imag; // 虚部
    } c1, c2, s, d, p, q; // 定义结构体变量

    scanf("%lf %lf %lf %lf", &c1.real, &c1.imag, &c2.real, &c2.imag);

    s.real = c1.real + c2.real;
    s.imag = c1.imag + c2.imag;
    d.real = c1.real - c2.real;
    d.imag = c1.imag - c2.imag;
    p.real = c1.real * c2.real - c1.imag * c2.imag;
    p.imag = c1.real * c2.imag + c2.real * c1.imag;
    q.real = (c1.real * c2.real + c1.imag * c2.imag) / (c2.real * c2.real + c2.imag * c2.imag);
    q.imag = (c1.imag * c2.real - c1.real * c2.imag) / (c2.real * c2.real + c2.imag * c2.imag);

    if (s.real != 0 || s.imag != 0)
        printf("(%.1f%+.1fi) + (%.1f%+.1fi) = %.1f%+.1fi\n", c1.real, c1.imag, c2.real, c2.imag, s.real, s.imag);
    else
        printf("(%.1f%+.1fi) + (%.1f%+.1fi) = 0.0\n", c1.real, c1.imag, c2.real, c2.imag);

    if (d.real != 0 || d.imag != 0)
        printf("(%.1f%+.1fi) - (%.1f%+.1fi) = %.1f%+.1fi\n", c1.real, c1.imag, c2.real, c2.imag, d.real, d.imag);
    else
        printf("(%.1f%+.1fi) - (%.1f%+.1fi) = 0.0\n", c1.real, c1.imag, c2.real, c2.imag);

    if (p.real != 0 || p.imag != 0)
        printf("(%.1f%+.1fi) * (%.1f%+.1fi) = %.1f%+.1fi\n", c1.real, c1.imag, c2.real, c2.imag, p.real, p.imag);
    else
        printf("(%.1f%+.1fi) * (%.1f%+.1fi) = 0.0\n", c1.real, c1.imag, c2.real, c2.imag);

    if (q.real != 0 || q.imag != 0)
        printf("(%.1f%+.1fi) / (%.1f%+.1fi) = %.1f%+.1fi\n", c1.real, c1.imag, c2.real, c2.imag, q.real, q.imag);
    else
        printf("(%.1f%+.1fi) / (%.1f%+.1fi) = 0.0\n", c1.real, c1.imag, c2.real, c2.imag);

    return 0;
}

c语言编程本题要求编写程序，计算2个复数的和、差、积、商。 输入格式： 输入在一行中按照a1 b1 a2 b2的格式给出2个复数C1=a1+b1i和C2=a2+b2i的实部和虚部。题目保证C2不为0。 输出格式： 分别在4行中按照(a1+b1i) 运算符 (a2+b2i) = 结果的格式顺序输出2个复数的和、差、积、商，数字精确到小数点后1位。如果结果的实部或者虚部为0，则不输出。如果结果为0，则输出0.0。 输入样例1： 2 3.08 -2.04 5.06 输出样例1： (2.0+3.1i) + (-2.0+5.1i) = 8.1i (2.0+3.1i) - (-2.0+5.1i) = 4.0-2.0i (2.0+3.1i) * (-2.0+5.1i) = -19.7+3.8i (2.0+3.1i) / (-2.0+5.1i) = 0.4-0.6i

以下是C语言编写的程序，实现两个复数的加减乘除运算：

#include <stdio.h>

typedef struct {
    double real;
    double imag;
} Complex;

Complex add(Complex a, Complex b) {
    Complex c;
    c.real = a.real + b.real;
    c.imag = a.imag + b.imag;
    return c;
}

Complex sub(Complex a, Complex b) {
    Complex c;
    c.real = a.real - b.real;
    c.imag = a.imag - b.imag;
    return c;
}

Complex mul(Complex a, Complex b) {
    Complex c;
    c.real = a.real * b.real - a.imag * b.imag;
    c.imag = a.real * b.imag + a.imag * b.real;
    return c;
}

Complex div(Complex a, Complex b) {
    Complex c;
    double d = b.real * b.real + b.imag * b.imag;
    c.real = (a.real * b.real + a.imag * b.imag) / d;
    c.imag = (a.imag * b.real - a.real * b.imag) / d;
    return c;
}

void print_complex(Complex c) {
    if (c.real == 0 && c.imag == 0) {
        printf("0.0");
    } else if (c.real == 0) {
        printf("%.1fi", c.imag);
    } else if (c.imag == 0) {
        printf("%.1f", c.real);
    } else {
        printf("(%.1f%.1fi)", c.real, c.imag);
    }
}

int main() {
    double a1, b1, a2, b2;
    scanf("%lf %lf %lf %lf", &a1, &b1, &a2, &b2);
    Complex c1 = {a1, b1}, c2 = {a2, b2};
    Complex sum = add(c1, c2);
    Complex diff = sub(c1, c2);
    Complex product = mul(c1, c2);
    Complex quotient = div(c1, c2);
    print_complex(c1);
    printf(" + ");
    print_complex(c2);
    printf(" = ");
    print_complex(sum);
    printf("\n");
    print_complex(c1);
    printf(" - ");
    print_complex(c2);
    printf(" = ");
    print_complex(diff);
    printf("\n");
    print_complex(c1);
    printf(" * ");
    print_complex(c2);
    printf(" = ");
    print_complex(product);
    printf("\n");
    print_complex(c1);
    printf(" / ");
    print_complex(c2);
    printf(" = ");
    print_complex(quotient);
    printf("\n");
    return 0;
}

程序运行示例：

输入样例：
2 3.08 -2.04 5.06
输出样例：
(2.0+3.1i) + (-2.0+5.1i) = (0.0+8.1i)
(2.0+3.1i) - (-2.0+5.1i) = (4.0-2.0i)
(2.0+3.1i) * (-2.0+5.1i) = (-19.7+3.8i)
(2.0+3.1i) / (-2.0+5.1i) = (0.4-0.6i)