Objective
In this challenge, you will learn the usage of the for loop, which is a programming language statement which allows code to be repeatedly executed.
The syntax for this is
for ( <expression_1> ; <expression_2> ; <expression_3> )
<statement>
- expression_1 is used for initializing 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
A sample loop will be
for(int i = 0; i < 10; i++) { ... }
Task
For each integer n in the interval [a,b] (given as input) :
- If 1<= n , 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 second 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
For Loop in C – Hacker Rank Solution
Code:
#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 Loop in C - Hacker Rank Solution START
char *str[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "even", "odd"};
scanf("%d\n%d", &a, &b);
for (int i=a; i<=b; i++)// i= 8
{
if (i <= 9)
{
printf ("%s\n", str[i-1]);
}
else
{
printf ("%s\n", str[9+(i%2)]);
}
// For Loop in C - Hacker Rank Solution END
}
return 0;
}
Disclaimer: The above Problem (For Loop in C) is generated by Hackerrank but the Solution is Provided by MultiplexCoder. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.
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