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

URI Online Judge Solution URI Solution - Data Structure

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

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/%in", num, den, numS, denS);
}
else
{
if(denS < 0)
{
denS = -denS;
numS = -numS;
}
printf("%i/%i = %i/%in", 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/%in", num, den, numS, denS);
}else{
if(denS < 0){
denS = -denS;
numS = -numS;
}
printf("%i/%i = %i/%in", 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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