URI-1022 Solve

TDA Rational

By Neilor Tonin, URI  Brazil

Timelimit: 1

You were invited to do a little job for your Mathematic teacher. The job is to read a Mathematic expression in format of two rational numbers (numerator / denominator) and present the result of the operation. Each operand or operator is separated by a blank space. The input sequence (each line) must respect the following format: number, (‘/’ char), number, operation char (‘/’, ‘*’, ‘+’, ‘-‘), number, (‘/’ char), number. The answer must be presented followed by ‘=’ operator and the simplified answer. If the answer can’t be simplified, it must be repeated after a ‘=’ operator.

Considering N1 and D1 as numerator and denominator of the first fraction, follow the orientation about how to do each one of these 4 operations:

Sum: (N1*D2 + N2*D1) / (D1*D2)
Subtraction: (N1*D2 - N2*D1) / (D1*D2)
Multiplication: (N1*N2) / (D1*D2)
Division: (N1/D1) / (N2/D2), that means (N1*D2)/(N2*D1)

Input

The input contains several cases of test. The first value is an integer N (1 ≤ N ≤ 1*104), indicating the amount of cases of test that must be read. Each case of test contains a rational value X (1 ≤ X ≤ 1000), an operation (-, +, * or /) and another rational value Y (1 ≤ Y ≤ 1000).

Output

The output must be a rational number, followed by a “=“ sign and another rational number, that is the simplification of the first value. In case of the first value can’t be simplified, the same value must be repeated after the ‘=’ sign.

Input Sample	Output Sample
4 1 / 2 + 3 / 4 1 / 2 - 3 / 4 2 / 3 * 6 / 6 1 / 2 / 3 / 4	10/8 = 5/4 -2/8 = -1/4 12/18 = 2/3 4/6 = 2/3

Problem link: https://www.urionlinejudge.com.br/judge/en/problems/view/1022

Solution:

• [tab]
• C++

• #include <cstdio>
  using namespace std;
  
  int euclides(int a, int b)
  {
   int divisor, dividendo, c;
  
   if(a == 0)
    return 1;
  
   if(b > a){
    dividendo = b;
    divisor = a;
   }else{
    dividendo = a;
    divisor = b;
   }
  
   while(dividendo % divisor != 0)
   {
    c = dividendo % divisor;
    dividendo = divisor;
    divisor = c;
   }
   return divisor;
  }
  
  int main()
  {
   char c1, c2, c3;
   int n, N1, N2, D1, D2, num, den, numS, denS, e;
   scanf("%i", &n);
  
   for (int i = 0; i < n; ++i)
   {
    scanf("%i %c %i %c %i %c %i", &N1, &c1, &D1, &c2, &N2, &c3, &D2);
    if(c2 == '+'){
     num = ((N1 * D2) + (N2 * D1));
     den = (D1 * D2);
    }else if(c2 == '-'){
     num = ((N1 * D2) - (N2 * D1));
     den = (D1 * D2);
    }else if(c2 == '*'){
     num = (N1 * N2);
     den = (D1 * D2);
    }else{
     num = (N1 * D2);
     den = (N2 * D1);
    }
  
    e = euclides(num, den);
    numS = num / e; 
    denS = den / e;
  
    if(numS > 0 && denS > 0){
     printf("%i/%i = %i/%i\n", num, den, numS, denS);
    }else{
     if(denS < 0){
      denS = -denS;
      numS = -numS;
     }
     printf("%i/%i = %i/%i\n", num, den, numS, denS);
    }
   }
  
   return 0;
  }