### Problem Statement :

```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>
1. expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop.
2. expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
3. 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 in the interval [a,b] (given as input) :

1. If 1<=n<=9, then print the English representation of it in lowercase. That is "one" for 1, "two" for 2, and so on.
2. Else if n>9 and it is an even number, then print "even".
3. 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<=10^6

Output Format:

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

### Solution :

```                            ```Solution in C :

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

int main()
{
int a, b,t,s;
scanf("%d\n%d", &a, &b);
t=a>b?a:b;
s=a<b?a:b;
for(s;s<=t;s++)
{
if(s<10)
{
if(s==1)
printf("one\n");
if(s==2)
printf("two\n");
if(s==3)
printf("three\n");
if(s==4)
printf("four\n");
if(s==5)
printf("five\n");
if(s==6)
printf("six\n");
if(s==7)
printf("seven\n");
if(s==8)
printf("eight\n");
if(s==9)
printf("nine\n");
}
if(s>=10)
{
if(s%2==0)
printf("even\n");
else
printf("odd\n");
}

}

return 0;
}```
```

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