## Algorithm

Problem Name: For Loop in C

In this HackerRank Functions in C programming problem solution,

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 i to 0, tests that i is less than 10, and increments i at every iteration. It will execute 10 times.

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


For each integer n n the interval [a,b] (given as input) :

• if 1 <= n <= 9, , then print the English representation of it in lowercase. That is "one" for one for 1,two for 2,and so on.
• Else if n > 9 and it is an even number, then print "even".
• Else if n > 9 and it is an odd number, then print "odd".

Input Format

The first line contains an integer, a.

The seond line contains an integer, b.

Constraints

1 <= a <= b <= 10sqrt6

Output Format

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

Sample Input

8
11


Sample Output

eight
nine
even
odd


## Code Examples

### #1 Code Example with C Programming

Code - C Programming


#include <stdio.h>
int main()
{
int i,j,k;
int a,b;
char *ch[]={"null","one","two","three","four","five",
"six","seven","eight","nine"};
scanf("%d%d",&a,&b);

for(i=a;i<=b;i++)
{
if(i>9 && i%2)
printf("odd\n");
else if(i>9 && i%2==0)
printf("even\n");
else
printf("%s\n",ch[i]);

}
return 0;
}

Copy The Code &

Input

cmd
8 11

Output

cmd
eight nine even odd