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60 lines
1.7 KiB
Markdown
60 lines
1.7 KiB
Markdown
# 6-1 使用函数计算两个复数之和与之积
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若两个复数分别为：`complex1 = real1 + imag1 i` 和 `complex2 = real2 + imag2 i`，则
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它们的和为：`complex1 + complex2 = (real1 + real2) + (imag1 + imag2) i`
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它们的乘积为：`complex1 * complex2 = (real1 * real2 - imag1 * imag2) + (real1 * imag2 + real2 * imag1) i`
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本题要求实现两个函数分别计算两个复数之和、两个复数之积。
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### 函数接口定义：
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```c++
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void complex_add (double real1, double imag1, double real2, double imag2);
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void complex_prod (double real1, double imag1, double real2, double imag2);
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```
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其中用户传入的参数为两个复数`real1 + imag1 i` 和`real2 + imag2 i`；函数`complex_add`和函数`complex_prod`应将计算结果的实部存放在全局变量result_real中、虚部存放在全局变量result_imag中。
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### 裁判测试程序样例：
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```c++
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#include <stdio.h>
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double result_real, result_imag;
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void complex_add(double real1, double imag1, double real2, double imag2);
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void complex_prod(double x1, double y1, double x2, double y2);
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int main(void)
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{
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double imag1, imag2, real1, real2;
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scanf("%lf %lf", &real1, &imag1);
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scanf("%lf %lf", &real2, &imag2);
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complex_add(real1, imag1, real2, imag2);
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printf("addition of complex is (%f)+(%f)i\n", result_real, result_imag);
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complex_prod(real1, imag1, real2, imag2);
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printf("product of complex is (%f)+(%f)i\n", result_real, result_imag);
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return 0;
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}
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/* 请在这里填写答案 */
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```
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### 输入样例：
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```in
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1 1
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-2 3
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```
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### 输出样例：
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```out
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addition of complex is (-1.000000)+(4.000000)i
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product of complex is (-5.000000)+(1.000000)i
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```
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