Algorithm
Problem Name:
In this HackerRank Functions in C programming problem 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> )
<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 i to 0, tests that i is less than 10, and increments i at every iteration. It will execute 10 times.
for(int i = 0; i < 10; i++) {
...
}
Task
For each integer n n the interval [a,b] (given as input) :
- if 1 <= n <= 9, , then print the English representation of it in lowercase. That is "one" for
onefor 1,twofor 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.
Constraints
1 <= a <= b <= 10sqrt6
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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main()
{
int i,j,k;
int a,b;
char *ch[]={"null","one","two","three","four","five",
"six","seven","eight","nine"};
scanf("%d%d",&a,&b);
for(i=a;i < =b;i++)
{
if(i>9 && i%2)
printf("odd\n");
else if(i>9 && i%2==0)
printf("even\n");
else
printf("%s\n",ch[i]);
}
return 0;
}
Input
Output