Check Odd or Even Number Without Using Any Condition and Loops

[CheckMeNot]

Hi friends, in my college We have given one question related to programming and I am not able to solve it. Can anyone help me in this problem? I need just logic of that particular program and that's it. Please, help me in this. The question is Find Out the given number is odd or even, without the use of any condition and loops.

[Praetor]

Hi, I think this will help you. I have created below code in java which Check Odd or Even Number Without Using Any Condition and Loops. Just make use of it.

import java.io.*;
public class EvenOrOdd
{
	public static void main(String arg[])throws IOException
	{
		String arr[]={"Even","Odd"};
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int num;
		System.out.println("Enter number to chcek");
		num=Integer.parseInt(br.readLine());
		num=num%2;
		System.out.println("Entered number is  "+arr[num]);		
	}
}

[Katty]

Hi, I have code for you in C language, which will Check Odd or Even Number Without Using Any Condition and Loops. Try to use it.

void main()
{
       int number;
       char sarr[]={"even","odd"};
       printf("Enter the no.:");
       scanf("%d",&number);
       number=number%2;
       printf("the number you have entered is %s",sarr[number]);
       getch();
}

[kelfro]

Hi, I don't know how to check the number is even or odd without using any condition or loop, but I have a solution given in my book, you can use it for your purpose.

void main()
{
       int n;
       char *s[4]={"even","odd"};
       printf("Enter the no.:");
       scanf("%d",&n);
       n=n%2;
       printf("th no. is %s",s[n]);
       getch();
}

[Zecho]

Hi, I don't know how to check odd or even number without using any condition or loop, I have done this with the help of loop. I am sending this, then just develop your logic and use it.

import java.io.*;

public class EvenOdd
{
public static void main(String[] args) throws IOException
{
try
{
      int n;
      BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      n = Integer.parseInt(in.readLine());
      if (n % 2 == 0)
      {
        System.out.println("Given number is Even.");
      }
      else
      {
        System.out.println("Given number is Odd.");
      }
}
     catch(NumberFormatException e){
      System.out.println(e.getMessage() + " is not a numeric value.");
      System.exit(0);
    }
  }
}

[Santrupti]

#include<stdio.h>
#define bind(k,i,n,d) n##k
int main()
{
int n;
printf("Enter the number:");
scanf("%d",&n);
bind(f,a,i,l)(n%2==0)
{
printf("Even\n");
exit(0);
}
printf("Odd\n");
return 0;
}