If Else Statement
Problem - 03:
In this challenge, we test your knowledge of using if-else conditional statements to automate decision-making processes. An if-else statement has the following logical flow:
Task
Given an integer, , perform the following conditional actions:
Given an integer, , perform the following conditional actions:
- If is odd, print
Weird - If is even and in the inclusive range of to , print
Not Weird - If is even and in the inclusive range of to , print
Weird - If is even and greater than , print
Not Weird.
Complete the stub code provided in your editor to print whether or not is weird.
Input Format
A single line containing a positive integer, .
Constraints
Output Format
Print Weird if the number is weird; otherwise, print Not Weird.
Sample Input 0
3
Sample Output 0
Weird
Sample Input 1
24
Sample Output 1
Not Weird
Explanation
Sample Case 0:
is odd and odd numbers are weird, so we print Weird.
Sample Case 1:
and is even, so it isn't weird. Thus, we print Not Weird.
code in java:
// Import the java library in place of "Scanner" you can use "*" which mean
you are calling all library....
import java.util.Scanner;
class Solution //class is keyword with user-define first letter should be
cap.
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
//use the if else statement if(n%2==1) //if n is odd
System.out.print("Weird\n");
else
{
if(n>=2 && n<=5){//range 2-5
System.out.print("Not Werid\n");
}
else if(n>6 && n<=20){//range 6-20
System.out.print("Weird\n");
}
else {
System.out.print("Not Weird\n");
}
}
}
}
output:
6
Werid
2
Not Weird

Comments
Post a Comment