HACKERANK FOR LOOP 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

SOLUTION

#include <stdio.h>

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











    
    return 0;
}