URI Online Judge Solution 1070 Six Odd Numbers – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1070 Six Odd Numbers – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1070 Six Odd Numbers | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1070

Problem Name: 1070 Six Odd Numbers solution
Problem Number : URI – 1070 Six Odd Numbers code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI 1070 Six Odd Numbers Code in C / URI 1070 Six Odd Numbers solution in C:

#include <stdio.h>
int main(){

 int i, X, howManyOdd = 6;
 scanf("%d", &X);
 for( i = X; i < (X+(howManyOdd*2)) ; i+=2){
  int odd;
  if(i % 2 == 0){
   odd = i + 1;
   printf("%dn", odd);
  }else{
   odd = i;
   printf("%dn", odd);
  }
 }
 return 0;
}

Run URI 1070 code in live Jdoodle code editor and see the result:

Please give the above input to see the live result in the StdIn input box of Jdoodle:

URI 1070 Six Odd Numbers Code in C++ / URI 1070 Six Odd Numbers solution in CPP:

#include <iostream>

using namespace std;

int main()
{
 int x, tmp = 0;
 bool ver = false;
 
 cin >> x;
 
 while(ver == false)
 {
  if(x % 2 == 1){
   cout << x << endl;
   tmp++;
  }
  
  if(tmp == 6)
   return 0;
   
  x++;
 }
 
 return 0;
}

import java.util.*;

public class Main {
    public static void main(String args[]) {
        int i, X, howManyOdd = 6;
     Scanner input = new Scanner(System.in);
     X = input.nextInt();
     for( i = X; i < (X+(howManyOdd*2)) ; i+=2){
      int odd;
      if(i % 2 == 0){
       odd = i + 1;
       System.out.printf("%dn", odd);
      }else{
       odd = i;
       System.out.printf("%dn", odd);
      }
     }
    }
}

Run URI 1070 java code in live Jdoodle Text editor:
Please give input as 8,

URI Solution 1070 Six Odd Numbers Code / URI 1070 Six Odd Numbers 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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