URI 1070 solution in C, Java language || Six odd numbers

URI Online Judge | 1070

Six Odd Numbers

URI 1070 solution in java language || Six odd numbers:

import java.io.IOException;
import java.util.Scanner;

public class Main{

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

 }

URI 1070 Solution in C language | Six odd Number's solution:

#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("%d\n", odd);
  }else{
   odd = i;
   printf("%d\n", odd);
  }
 }
 return 0;
}

Tags: URI 1070 solution in C, URI online judge solution, URI solution 1070 in java


[Update : May 25 2017 -> After getting a comment from Gk Polash]