* Structure-06. arithmetic operation with four plural values

1/* 2 * Main. C 3 * F6-structure-06. the number of four arithmetic operations 4 * created on: August 26, 2014 5 * Author: in the boomkeeper 6 ******** section, ************* 7 */8 9 # include <stdio. h> 10 # include <math. h> 11 12 # define epsinon 0.1 13 14/* 15 * plural struct. C1 and C2 correspond to C1, C2 16 */17 struct complexnumber {18 float real; 19 float imaginary; 20} C1, C2; 21 22/* 23 * follow the "(A1 + b1i) operator (A2 + b2i) = The result is formatted in order to output two plural numbers 24 * Ch corresponding to "+" "-" "*" "/" four operators 25 */26 Void printcomplex (struct complexnumber C1, char CH, struct complexnumber C2) {27 28 If (c1.imaginary <0) {29 If (c2.imaginary <0) 30 printf ("(%. 1f %. 1fi) % C (%. 1f %. 1fi) = ", c1.real, c1.imaginary, CH, 31 c2.real, c2.imaginary); 32 If (c2.imaginary> 0) 33 printf (" (%. 1f %. 1fi) % C (%. 1f + %. 1fi) = ", c1.real, c1.imaginary, CH, 34 c2.real, c2.imaginary); 35} else if (c1.imaginary> 0) {36 IF (c2. I Maginary <0) 37 printf ("(%. 1f + %. 1fi) % C (%. 1f %. 1fi) = ", c1.real, c1.imaginary, CH, 38 c2.real, c2.imaginary); 39 if (c2.imaginary> 0) 40 printf (" (%. 1f + %. 1fi) % C (%. 1f + %. 1fi) = ", c1.real, c1.imaginary, CH, 41 c2.real, c2.imaginary ); 42} 43} 44 45/* 46 * output the sum, difference, product, and Quotient of two complex numbers 47 */48 void printresult (float re, float IM) {49 50 if (FABS (re) <epsinon) 51 Re = 0; 52 If (FABS (IM) <epsinon) 53 im = 0; 54 55 if (RE = 0 & im = 0) 56 printf ("0.0 \ n"); 57 else {58 If (re! = 0 & im> 0) 59 printf ("%. 1f + %. 1fi \ n", re, Im); 60 if (re! = 0 & im <0) 61 printf ("%. 1f %. 1fi \ n", re, Im); 62 if (re! = 0 & im = 0) 63 printf ("%. 1f \ n", RE); 64 if (RE = 0 & Im! = 0) 65 printf ("%. 1fi \ n ", Im); 66} 67} 68 69 int main (void) {70 71 float Re = 0, Im = 0; // temporary variable, calculate real and virtual parts 72 73 scanf ("% F", & c1.real, & c1.imaginary, & c2.real, & c2.imaginary ); 74 75 // and 76 printcomplex (C1, '+', C2); 77 Re = c1.real + c2.real; 78 im = c1.imaginary + c2.imaginary; 79 printresult (Re, Im ); 80 81 // difference 82 printcomplex (C1, '-', C2); 83 Re = c1.real-c2.real; 84 im = c1.imaginary-c2.imaginary; 85 printresult (Re, Im ); 86 87 // product 88 printcomplex (C1, '*', C2); 89 Re = c1.real * c2.real-c1.imaginary * c2.imaginary; 90 im = c1.real * c2.imaginary + c1.imaginary * c2.real; 91 printresult (Re, Im); 92 93 // vendor 94 printcomplex (C1, '/', C2); 95 Re = (c1.real * c2.real + c1.imaginary * c2.imaginary) 96/(c2.real * c2.real + c2.imaginary * c2.imaginary); 97 im = (c1.imaginary * c2.real-c1.real * c2.imaginary) 98/(c2.real * c2.real + c2.imaginary * numeric ); 99 printresult (Re, Im); 100 101 return 0; 102}

 

1. The question that makes me grasp the most is the output of the symbol of the imaginary part of the plural number. The arithmetic operation on the four digits of the plural number is still easy to calculate. This is the plus or minus sign, and it will be messy when I am not careful;

2. Compare floating-point numbers with 0. For this question, because the output result only needs to be accurate to 0.1, <= 0.1 &> =-0.1 is regarded as 0. At first I referred to other articles. According to 0.00001, the output results are always irrelevant.

 

Question link:

Http://pat.zju.edu.cn/contests/basic-programming/%E7%BB%93%E6%9E%84-06

Refer:

Http://blog.csdn.net/rabbit8848/article/details/30284651

Http://blog.sina.com.cn/s/blog_64a475370101119h.html

* Structure-06. arithmetic operation with four plural values