For Loop in C - Hackerrank Solution

For Loop in C - Hackerrank Solution

Objective

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

The syntax for this is

for (  ;  ;  )
    

• 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.

A sample loop will be

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 <assert.h> #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 <= 9) { switch(i) { case 1: printf("one\n"); break; case 2: printf("two\n"); break; case 3: printf("three\n"); break; case 4: printf("four\n"); break; case 5: printf("five\n"); break; case 6: printf("six\n"); break; case 7: printf("seven\n"); break; case 8: printf("eight\n"); break; case 9: printf("nine\n"); break; } } else{ if(!(i%2)) printf("even\n"); else printf("odd\n"); } } return 0; }

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