Objective
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++) {
...
}
Task
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
8
11
Sample Output
eight
nine
even
oddCode Using Array:
#include <stdio.h>#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int a=0,b=0,n=0;
char* arr[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
scanf("%d",&a);
scanf("%d",&b);
for(n = a;n<=b;n++)
{
if(n>9)
{
if(n%2 ==0)
printf("even\n");
else {
printf("odd\n");
}
}
else {
puts(arr[n]);
}
}
return 0;
}