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法律研究用 C 语言编写程序以相加两个复数
已知有两个复数 a1+ ib1 和 a2 + ib2，任务是相加这两个复数。
复数是可以表达为“a+ib”形式的数，其中“a”和“b”是实数，i 是虚数，即关于表达式 𝑥 2 = −1 的解，因为没有实数满足该方程，所以称其为虚数。
输入
a1 = 3, b1 = 8 a2 = 5, b2 = 2
输出
Complex number 1: 3 + i8 Complex number 2: 5 + i2 Sum of the complex numbers: 8 + i10
解释
(3+i8) + (5+i2) = (3+5) + i(8+2) = 8 + i10
输入
a1 = 5, b1 = 3 a2 = 2, b2 = 2
输出
Complex number 1: 5 + i3 Complex number 2: 2 + i2 Sum of the complex numbers: 7 + i5
解释
(5+i3) + (2+i2) = (5+2) + i(3+2) = 7 + i5
用于解决问题的如下方法
首先声明一个用于存储实数和虚数的结构。
接收输入，并相加所有复数的实数和虚数。
算法
Start
Decalre a struct complexnum with following elements
1. real
2. img
In function complexnum sumcomplex(complexnum a, complexnum b)
Step 1→ Declare a signature struct complexnum c
Step 2→ Set c.real as a.real + b.real
Step 3→ Set c.img as a.img + b.img
Step 4→ Return c
In function int main()
Step 1→ Declare and initialize complexnum a = {1, 2} and b = {4, 5}
Step 2→ Declare and set complexnum c as sumcomplex(a, b)
Step 3→ Print the first complex number
Step 4→ Print the second complex number
Step 5→ Print the sum of both in c.real, c.img
Stop示例
#include <stdio.h>
//structure for storing the real and imaginery
//values of complex number
struct complexnum{
int real, img;
};
complexnum sumcomplex(complexnum a, complexnum b){
struct complexnum c;
//Adding up two complex numbers
c.real = a.real + b.real;
c.img = a.img + b.img;
return c;
}
int main(){
struct complexnum a = {1, 2};
struct complexnum b = {4, 5};
struct complexnum c = sumcomplex(a, b);
printf("Complex number 1: %d + i%d
", a.real, a.img);
printf("Complex number 2: %d + i%d
", b.real, b.img);
printf("Sum of the complex numbers: %d + i%d
", c.real, c.img);
return 0;
}输出
如果运行以上代码，会生成以下输出 −
Complex number 1: 1 + i2 Complex number 2: 4 + i5 Sum of the complex numbers: 5 + i7
更新日期： 13-Aug-2020
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