7-12 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
复数除法:(a+bi)/(c+di)=(ac+bd)/(c2+d2) +((bc-ad)/(c2+d2))i
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std; 5 6 //小数1位 7 int judge(double x) 8 { 9 return fabs(x)<0.05?0:1; 10 } 11 12 void solve(double e,double f) 13 { 14 if(judge(e)==0&&judge(f)==0){ 15 printf("0.0\n"); 16 } 17 else if(judge(e)==0){ 18 printf("%.1fi\n",f); 19 } 20 else if(judge(f)==0){ 21 printf("%.1f\n",e); 22 } 23 else{ 24 printf("%.1f%c%.1fi\n",e,f<0?'-':'+',fabs(f)); 25 } 26 } 27 28 int main() 29 { 30 double a,b,c,d,e,f; 31 scanf("%lf%lf%lf%lf",&a,&b,&c,&d); 32 printf("(%.1f%c%.1fi) + (%.1f%c%.1fi) = ",a,b<0?'-':'+',fabs(b),c,d<0?'-':'+',fabs(d)); 33 e=a+c; 34 f=b+d; 35 solve(e,f); 36 printf("(%.1f%c%.1fi) - (%.1f%c%.1fi) = ",a,b<0?'-':'+',fabs(b),c,d<0?'-':'+',fabs(d)); 37 e=a-c; 38 f=b-d; 39 solve(e,f); 40 printf("(%.1f%c%.1fi) * (%.1f%c%.1fi) = ",a,b<0?'-':'+',fabs(b),c,d<0?'-':'+',fabs(d)); 41 e=a*c-b*d; 42 f=a*d+b*c; 43 solve(e,f); 44 printf("(%.1f%c%.1fi) / (%.1f%c%.1fi) = ",a,b<0?'-':'+',fabs(b),c,d<0?'-':'+',fabs(d)); 45 e=(a*c+b*d)/(c*c+d*d); 46 f=(b*c-a*d)/(c*c+d*d); 47 solve(e,f); 48 return 0; 49 }