For Loop in C Hackerrank solution

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;
    scanf("%d\n%d", &a, &b);
    char *str[] = {"one", "two", "three", "four", "five", "six", "seven",  "eight",  "nine", "even", "odd"};

    scanf("%d\n%d", &a, &b);
    for (int i=a; i<=b; i++)// i= 8 
    {
            if (i <= 9) 
        {
                printf ("%s\n", str[i-1]);
        }
            else
        {
                printf ("%s\n", str[9+(i%2)]);
        }
    
    }
    return 0;
}