中M2019春C入门和进阶练习 7-162 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

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>    //主要是输出格式控制
void f(double a1,double b1,double a2,double b2,double c1,double c2,char c){
    if(c1>-0.05&&c1<0.05){        //c1,c2四舍五入为零时不输出,参考输入样例1
        c1=0.0;
    }
    if(c2>-0.05&&c2<0.05){
        c2=0.0;
    }
    if(b1>=0&&b2>=0){       //先由b1,b2的正负控制等号前面的格式
        if(c1!=0&&c2>0){      //由c1,c2的正负还是0选择右边的格式
            printf("(%.1lf+%.1fi) %c (%.1lf+%.1lfi) = %.1f+%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c1!=0&&c2<0){
            printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = %.1f%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c2==0&&c1){
            printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = %.1f\n",a1,b1,c,a2,b2,c1);
        }else if(c1==0&&c2){
            printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = %.1fi\n",a1,b1,c,a2,b2,c2);
        }else{
            printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = 0.0\n",a1,b1,c,a2,b2);
        }        
    }else if(b1<0&&b2>=0){
        if(c1!=0&&c2>0){
            printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = %.1f+%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c1!=0&&c2<0){
            printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = %.1f%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c2==0&&c1){
            printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = %.1f\n",a1,b1,c,a2,b2,c1);
        }else if(c1==0&&c2){
            printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = %.1fi\n",a1,b1,c,a2,b2,c2);
        }else{
            printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = 0.0\n",a1,b1,c,a2,b2);
        }
    }else if(b1>=0&&b2<0){
        if(c1!=0&&c2>0){
            printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = %.1f+%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c1!=0&&c2<0){
            printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = %.1f%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c2==0&&c1){
            printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = %.1f\n",a1,b1,c,a2,b2,c1);
        }else if(c1==0&&c2){
            printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = %.1fi\n",a1,b1,c,a2,b2,c2);
        }else{
            printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = 0.0\n",a1,b1,c,a2,b2);
        }
    }else if(b1<0&&b2<0){
        if(c1!=0&&c2>0){
            printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = %.1f+%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c1!=0&&c2<0){
            printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = %.1f%.1fi\n",a1,b1,c,a2,b2,c1,c2);
        }else if(c2==0&&c1){
            printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = %.1f\n",a1,b1,c,a2,b2,c1);
        }else if(c1==0&&c2){
            printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = %.1fi\n",a1,b1,c,a2,b2,c2);
        }else{
            printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = 0.0\n",a1,b1,c,a2,b2);
        }
    }
}
int main(){
    double a1,b1,a2,b2,c1,c2;
    scanf("%lf %lf %lf %lf",&a1,&b1,&a2,&b2);   //c1为结果实部,c2为结果虚部,计算公式可百度
    c1=1.0*(a1+a2);
    c2=1.0*(b1+b2);
    f(a1,b1,a2,b2,c1,c2,'+');
    c1=1.0*(a1-a2);
    c2=1.0*(b1-b2);
    f(a1,b1,a2,b2,c1,c2,'-');
    c1=1.0*(a1*a2-b1*b2);
    c2=1.0*(a2*b1+a1*b2);
    f(a1,b1,a2,b2,c1,c2,'*');
    c1=1.0*(a1*a2+b1*b2)/(a2*a2+b2*b2);
    c2=1.0*(b1*a2-a1*b2)/(a2*a2+b2*b2);
    f(a1,b1,a2,b2,c1,c2,'/');
    return 0;
}

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转载自blog.csdn.net/Great_Linlin/article/details/89252955
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