PTA C语言结构体 7-5 复数四则运算 (15 分)

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

结尾无空行

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

结尾无空行

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>
#include<math.h>
void result(double a1,double b1,double a2,double b2,int i); 
int main()
{
    double  a1,b1,a2,b2;
    scanf("%lf%lf%lf%lf",&a1,&b1,&a2,&b2);
    
for(int i=0;i&lt;4;i++)
{
    if(i==0)
    {
        if(b1&gt;=0&amp;&amp;b2&gt;=0)
        printf("(%.1lf+%.1lfi) + (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp; b2&gt;=0)
        printf("(%.1lf%.1lfi) + (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp; b2&lt;0)
        printf("(%.1lf%.1lfi) + (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        else
        printf("(%.1lf+%.1lfi) + (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        result(a1,b1,a2,b2,i);
    }
    else if(i==1){
        if(b1&gt;=0&amp;&amp;b2&gt;=0)
        printf("(%.1lf+%.1lfi) - (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp; b2&gt;=0)
        printf("(%.1lf%.1lfi) - (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp;b2&lt;0)
        printf("(%.1lf%.1lfi) - (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        else
        printf("(%.1lf+%.1lfi) - (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        result(a1,b1,a2,b2,i);
    }
    else if(i==2){
        if(b1&gt;=0&amp;&amp;b2&gt;=0)
        printf("(%.1lf+%.1lfi) * (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp; b2&gt;=0)
        printf("(%.1lf%.1lfi) * (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0&amp;&amp;b2&lt;0)
        printf("(%.1lf%.1lfi) * (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        else
        printf("(%.1lf+%.1lfi) * (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        result(a1,b1,a2,b2,i);
    }
    else if(i==3){
        if(b1&gt;=0&amp;&amp;b2&gt;=0)
        printf("(%.1lf+%.1lfi) / (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0 &amp;&amp; b2&gt;=0)
        printf("(%.1lf%.1lfi) / (%.1lf+%.1lfi) = ",a1,b1,a2,b2);
        else if(b1&lt;0&amp;&amp;b2&lt;0)
        printf("(%.1lf%.1lfi) / (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        else
        printf("(%.1lf+%.1lfi) / (%.1lf%.1lfi) = ",a1,b1,a2,b2);
        result(a1,b1,a2,b2,i);
    }
}

}
void result(double a1,double b1,double a2,double b2,int i){
double really,v;
if(i==0){
really=a1+a2;
v=b1+b2;
}
else if(i==1){
really=a1-a2;
v=b1-b2;
}
else if(i==2){
really=a1a2-b1b2;
v=a2b1+a1b2;
}
else if(i==3){
/double fenzi,fenmu;
// MultReally(a1,b1,a2,-b2)+MultV(a1,b1,a2,-b2);
fenmu=a2
a2-b2b2;
really=MultReally(a1,b1,a2,-b2)/fenmu;
v=MultV(a1,b1,a2,-b2)/fenmu;
/

    really=(a1*a2+b1*b2)/(a2*a2+b2*b2);
    v=(a2*b1-a1*b2)/(a2*a2+b2*b2);
    if(a2==0&amp;&amp;b2==0){
        really=0;
        v=0;
        
    }
}

// if(really==0) printf(“yes\n”);
if(fabs(really)<0.1&&fabs(v)>0.1){//实部为0
printf(“%.1fi\n”,v);
}
else if(fabs(really)>0.1&&fabs(v)<0.1)
{//虚部为0
printf(“%.1lf\n”,really);
}
else if(fabs(really)>0.1&&fabs(v)>0.1){
if(v<0)
printf(“%.1lf%.1lfi\n”,really,v);

    else
        printf("%.1lf+%.1lfi\n",really,v);
}
else if(fabs(really)&lt;0.1&amp;&amp;fabs(v)&lt;0.1) printf("0.0\n");

}

 


PTA C语言结构体 7-5 复数四则运算 (15 分)
https://6jackjiang.github.io/2021/11/19/categories/C语言/PTA C语言结构体 7-5 复数四则运算 (15 分)/
作者
米兰的小铁匠
发布于
2021年11月19日
许可协议