For Loop in C

Task

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

• If 1≤ n ≤9, 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 seond line contains an integer, b.

Constraints

a ≤ b≤ 10⁶

 

Output Format

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

 

Note: [a,b] = { x ∈ Z | a ≤ x ≤b } = { a, a+1, ..., b }

 

Sample Input

• 8
• 11

Sample Output

• eight
• nine
• even
• odd

---

#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 0:
          printf("zero\n");
          break;
        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 == 0)
          printf("even\n");
        else
          printf("odd\n");
      }
    }

    return 0;
}