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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 这个太麻烦了，我实在不想写它了……下面代码来自CSDN用户：#include#include //打印相同的部分void Print_same(double a1, double b1, double a2, double b2, char c);//打印最后的结果void Print_result(double res1, double res2);//对结果进行四舍五入操作double rounding(double num);int main(){ double a1, b1, a2, b2; scanf("%lf%lf%lf%lf", &a1, &b1, &a2, &b2); //加法 Print_same(a1, b1, a2, b2, '+'); Print_result(rounding(a1 + a2), rounding(b1 + b2)); //减法 Print_same(a1, b1, a2, b2, '-'); Print_result(rounding(a1 - a2), rounding(b1 - b2)); //乘法 Print_same(a1, b1, a2, b2, '*'); Print_result(rounding(a1 * a2 - b1 * b2), rounding(a1 * b2 + a2 * b1)); //除法 Print_same(a1, b1, a2, b2, '/'); Print_result(rounding((a1 * a2 + b1 * b2) / (a2 * a2 + b2 * b2)), rounding((-a1 * b2 + a2 * b1) / (a2 * a2 + b2 * b2))); return 0;}//打印相同的部分 void Print_same(double a1, double b1, double a2, double b2, char c) { if (b1 < 0 && b2 < 0) printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = ", a1, b1, c, a2, b2); else if (b1 < 0) printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = ", a1, b1, c, a2, b2); else if (b2 < 0) printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = ", a1, b1, c, a2, b2); else printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = ", a1, b1, c, a2, b2);}//打印结果 void Print_result(double res1, double res2) { if (res1 == 0 && res2 == 0) printf("0.0\n"); else if (res1 == 0) printf("%.1lfi\n", res2); else if (res2 == 0) printf("%.1lf\n", res1); else if (res2 < 0) printf("%.1lf%.1lfi\n", res1, res2); else printf("%.1lf+%.1lfi\n", res1, res2);}//四舍五入 double rounding(double num){ if (num > 0) num = (int)(num * 10 + 0.5) / 10.0; else num = (int)(num * 10 - 0.5) / 10.0; return num;}
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