URI Online Judge Solution 1022 TDA Rational - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1002 Area of a Circle - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1022 TDA Rational  | Beginner
URI Problem Link - https://www.urionlinejudge.com.br/judge/en/problems/view/1022

Problem Name: 1022 TDA Rational solution
Problem Number : URI - 1022 TDA Rational code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1022 TDA Rational  problem simple input output:

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

URI Solution 1022 TDA Rational  Code in C:

#include <stdio.h>

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);

    int i;
    for (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;
}

URI Solution 1022 TDA Rational  Code in c++/ URI 1022 solution in CPP:

#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;
}

URI Solution 1022 TDA Rational  Code / URI 1022 solution in Java:

URI Solution 1022 TDA Rational  Code / URI 1022 solution in  Python:

URI Solution 1022 TDA Rational  Code / URI 1022 solution in  C# (C Sharp):

Demonstration:

Just implement this in coding. Since having any problem just put a comment below. Thanks



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