HackerRank For Loop Solution in 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

If 1 <= n <= 9 then print the English representation of it in lowercase. That is "one" for 1. "two" for 2, and so on.

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

1 <= a <= b <= 10 ^ 6

Output Format

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

Sample Input

8
11

Sample Output

eight
nine
even
odd

HackerRank For Loop Solution in C (Sample-1)

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



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

HackerRank For Loop Solution in C (Sample-2)

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



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

    return 0;
}

HackerRank For Loop Solution in C (Sample-3)

#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++){
    switch(i){
        case 1:
            printf("one");
            break;
        case 2:
            printf("two");
            break;
        case 3:
            printf("three");
            break;
        case 4:
            printf("four");
            break;
        case 5:
            printf("five");
            break;
        case 6:
            printf("six");
            break;
        case 7:
            printf("seven");
            break;
        case 8:
            printf("eight");
            break;
        case 9:
            printf("nine");
            break;
        default:
            printf("%s", (i % 2)?"odd":"even");
            break;
        }
        putchar('\n');
    }

    return 0;
}