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
对这类题目还是很犯浑等天梯赛回来,在改进代码
1 #include <iostream> 2 #include <algorithm> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <cstring> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cmath> 10 #include <stack> 11 #include <vector> 12 #include <set> 13 using namespace std; 14 const int minn=1e-6; 15 double trans(double n){ 16 double zhi; 17 if(n-0>=minn) 18 zhi=(int)(n*10+0.5)/10.0; 19 else zhi=(int)(n*10-0.5)/10.0; 20 return zhi; 21 } 22 23 void printff(double a,double b,double c,double d,char s){ 24 printf("(%.1f",a); 25 if(b>=0) printf("+"); 26 printf("%.1fi)",b); 27 printf(" %c ",s); 28 printf("(%.1f",c); 29 if(d>=0) printf("+"); 30 printf("%.1fi)",d); 31 printf(" = "); 32 } 33 void shuchu(double shi,double xu){ 34 if(fabs(shi)<1e-1&&fabs(xu)<1e-1) printf("0.0 "); 35 else if(fabs(shi)<1e-1&&fabs(xu)>1e-1) printf("%.1fi ",xu); 36 else if(fabs(shi)>1e-1&&fabs(xu)<1e-1) printf("%.1f ",shi); 37 else 38 { 39 printf("%.1f",shi); 40 if(xu>0) printf("+"); 41 printf("%.1fi ",xu); 42 } 43 } 44 int main() 45 { 46 double a1,b1,a2,b2,a3,b3,a4,b4; 47 double shi,xu; 48 scanf("%lf%lf%lf%lf",&a1,&b1,&a2,&b2); 49 a3=trans(a1),b3=trans(b1),a4=trans(a2),b4=trans(b2); 50 printff(a3,b3,a4,b4,'+'); 51 shi=a1+a2,xu=b1+b2; 52 shuchu(shi,xu); 53 54 printff(a3,b3,a4,b4,'-'); 55 shi=a1-a2,xu=b1-b2; 56 shuchu(shi,xu); 57 58 printff(a3,b3,a4,b4,'*'); 59 shi=a1*a2+(b1*b2*(-1.0)),xu=a1*b2+b1*a2; 60 shuchu(shi,xu); 61 62 printff(a3,b3,a4,b4,'/'); 63 shi=(a1*a2+b1*b2)*1.0/(a2*a2+b2*b2),xu=(a2*b1-a1*b2)*1.0/(a2*a2+b2*b2); 64 shuchu(shi,xu); 65 return 0; 66 }