In this HackerRank For loop in c programming problem solution, your task is that 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".

HackerRank For Loop in C programming solution


HackerRank For loop in c programming problem solution.

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

int main() 
{
    int a, b;
    scanf("%d\n%d", &a, &b);
    
    char labels[11][6] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "even", "odd"};
    int labels_index;
      for (int i=a; i<=b; i++) {
        labels_index = i <= 9 ? i - 1 : 9 + i % 2;
        printf("%s\n", labels[labels_index]);
    }

    return 0;
}


Second 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 (int i=a; i<b+1; i++) {
        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;
            default:
                if (i % 2)
                    printf("odd\n");
                else
                    printf("even\n");
        }
    }

    return 0;
}