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For Loop in C HackerRank Solution

Hello Friends in this article i am gone to share Hackerrank C programming Solutions with you. | For Loop in C Hackerrank Solution.


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> )
  • 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++) {


For each integer n 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.


  • 1<a<b<10

Output Format

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

Note: [a,b] = {xez| a<x<b} = {a,a+1,…,b}

Sample Input


Sample Output



Hackerrank  C programming Solutions

For Loop in C Hackerrank 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;

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