# Problem

### Objective

In this challenge, you will learn the usage of the for loop, which is a programming language statement that allows code to be repeatedly executed.
The syntax for this is

for ( <expression_1> ; <expression_2> ; <expression_3> )
<statement>
• expression_1 is used for initializing 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

A sample loop will be

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

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

•  If 1<= n , then print the English representation of it in lowercase. That is “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 second line contains an integer, b.

1<=a<=b<=10^6

#### Output Format

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

8
11

eight
nine
even
odd

### Solution :-

// For Loop in C - Hacker Rank Solution
#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 Loop in C - Hacker Rank Solution START

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)]);
}

// For Loop in C - Hacker Rank Solution END

}
return 0;
}

### Another Solution

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
int a, b;
int n=0;
scanf("%d\n%d", &a, &b);
for(n=a;n<=b;n++)
{
if(n==1)
{
printf("one\n");
}
if(n==2)
{
printf("two\n");
}
if(n==3)
{
printf("three\n");
}
if(n==4)
{
printf("four\n");
}
if(n==5)
{
printf("five\n");
}
if(n==6)
{
printf("six\n");
}
if(n==7)
{
printf("seven\n");
}
if(n==8)
{
printf("eight\n");
}
if(n==9)
{
printf("nine\n");
}
if(n>9)
{
if(n%2==0)
{
printf("even\n");
}
else {
printf("odd\n");
}
}

}
return 0;
}