Challenge Code

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In this programming code, you will get to know how to use for loop statement n the code. For loop works when we know a statement will iterate specific number of times, its starting and ending point. When you want to run a specific function or statement a number of times, one should use for loop statements.

The syntax of for loop is mentioned below:

for ( <expression_1> ; <expression_2> ; <expression_3> )
    <statement>

In this syntax,

expression 1 decides the starting point of the loop

expression 2 is the condition which must be satisfied during the process

expression 3 is used for increment and decrement of any variable.

For example:-

for(int i = 0; i < 5; i++) {
    ...
}

This particular loop runs 5 times starting with n=0 and succeeding as n=1, n=2, n=3 and n=4.

Here we solve a particular solution also mentioned in HackerRank. One can solve it to practice implementation of for loop.

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” 2 for , 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

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The first line contains an integer, a.
The second line contains an integer, b.

Constraints

1 ≤ a ≤ b ≤ 10⁶

Output Format

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

Sample Input 1:

8
11

Sample Output 1:

eight
nine
even
odd

Sample Input 2:

3 15

Sample Output 2:

three
four
five
six
seven
eight
nine
even
odd
even
odd
even
odd

Solution:

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

void number(int num1,int num2){
    int n;

    for(n=num1;n<=num2&&n<=9;n++){
        
        if(n==1){
            printf("one\n");
        }
        else if(n==2){
            printf("two\n");
        }
        else if(n==3){
            printf("three\n");
        }
        else if(n==4){
            printf("four\n");
        }
        else if(n==5){
            printf("five\n");
        }
        else if(n==6){
            printf("six\n");
        }
        else if(n==7){
            printf("seven\n");
        }
        else if(n==8){
            printf("eight\n");
        }
        else if(n==9){
            printf("nine\n");
        }
        }
    
}
void check(int num1,int num2){
    int n;

    for(n=10;n<=num2&&n>9;n++){
    if(n%2==0){
        printf("even");
    }
    else{
        printf("odd");
        }
    printf("\n");
    }
    
}
int main() 
{
    int a, b;
    scanf("%d\n%d", &a, &b);
   // Complete the code.
    number(a,b);
    check(a,b);
    
    return 0;
}

Happy Learning! :)