HackerRank C Solution - For Loop 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

#include <stdio.h>
	#include <string.h>
	#include <math.h>
	#include <stdlib.h>
	int main()
	{
	int a, b;
	scanf("%d\n%d", &a, &b);
	// Complete the code.
	for (int i = a; i <= b; i ++) {
	if(i >= 1 && i <= 9) {
	if(i == 1) {
	printf("one\n");
	} else if(i == 2) {
	printf("two\n");
	} else if(i == 3) {
	printf("three\n");
	} else if(i == 4) {
	printf("four\n");
	} else if(i == 5) {
	printf("five\n");
	} else if(i == 6) {
	printf("six\n");
	} else if(i == 7) {
	printf("seven\n");
	} else if(i == 8) {
	printf("eight\n");
	} else if(i == 9) {
	printf("nine\n");
	}
	} else {
	if(i % 2 == 0) {
	printf("even\n");
	} else {
	printf("odd\n");
	}
	}
	}
	return 0;
	}