FOR loops solution for hackerrank problem in C

 Objective

In this challenge, you will learn the usage of the for loop, which is a programming language statement which allows code to be executed until a terminal condition is met. They can even repeat forever if the terminal condition is never met.

The syntax for the for loop is:

for ( <expression_1> ; <expression_2> ; <expression_3> )
    <statement>

• expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop.
• expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
• expression_3 is generally used to update the flags/variables.

The following loop initializes  to 0, tests that  is less than 10, and increments  at every iteration. It will execute 10 times.

for(int i = 0; i < 10; i++) {
    ...
}

Task

For each integer  in the interval  (given as input) :

• If , then print the English representation of it in lowercase. That is "one" for , "two" for , and so on.
• Else if  and it is an even number, then print "even".
• Else if  and it is an odd number, then print "odd".

Input Format

The first line contains an integer, .
The seond line contains an integer, .

Constraints

Output Format

Print the appropriate English representation,even, or odd, based on the conditions described in the 'task' section.

Note: 

Sample Input

8
11

Sample Output

eight
nine
even
odd

SOLUTION:

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

int main() 

{

    int a, b,n;

    scanf("%d\n%d", &a, &b);

    for(int i=a;i<=b;i++)

    {

    

            switch(i)

            {

                case 1: printf("one\n");

                        break;

                case 2: printf("two\n");

                        break;

                case 3: printf("three\n");

                        break;

                case 4: printf("four\n");

                        break;

                case 5: printf("five\n");

                        break;

                case 6: printf("six\n");

                        break;

                case 7: printf("seven\n");

                        break;

                case 8: printf("eight\n");

                        break;

                case 9: printf("nine\n");

                        break;

            }   

             

            if(i>9)

            {

                if(i%2==0)

                    printf("even\n");

                else

                    printf("odd\n");

            }

        

    }

    

    return 0;

}