# For Loop in C | hackerrank practice problem solution

Problem:-

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++) {    ...}``

For each integer  in the interval  (given as input) :

• If , then print the English representation of it in lowercase. That is “one” for , “two” for , and so on.
• Else if  and it is an even number, then print “even”.
• Else if  and it is an odd number, then print “odd”.

Input Format

The first line contains an integer, .
The seond line contains an integer, .

Constraints

Output Format

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

Note:

Sample Input

``811``

Sample Output

``eightnineevenodd``

Solution:-

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

int main()
{
int a,n,b;
scanf(“%dn%d”, &a, &b);
n=a;
while(n<=b)
{
if(n==1)
printf(“onen”);
if(n==2)
printf(“twon”);
if(n==3)
printf(“threen”);
if(n==4)
printf(“fourn”);
if(n==5)
printf(“fiven”);
if(n==6)
printf(“sixn”);
if(n==7)
printf(“sevenn”);
if(n==8)
printf(“eightn”);
if(n==9)
printf(“ninen”);
if(n>9)
{
if(n%2==0)
printf(“evenn”);
else {
printf(“oddn”);
}
}
n++;
}
return 0;
}

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