Question 7 || Hacker Rank
For Loop in C || Hacker Rank
Objective
In this test, you will gain proficiency with the use of the for circle, which is a programming language explanation which permits code to be executed until a terminal condition is met. They might in fact rehash always assuming that the terminal condition is rarely met.
The punctuation for the for circle is:
for ( <expression_1> ; <expression_2> ; <expression_3> )
<statement>
expression_1 is utilized for intializing factors which are for the most part utilized for controlling the ending banner for the circle.
expression_2 is utilized to check for the ending condition. On the off chance that this assesses to misleading, the circle is ended.
expression_3 is by and large used to refresh the banners/factors.
The accompanying circle introduces to 0, tests that is under 10, and additions at each cycle. It will execute multiple times.
for(int I = 0; I < 10; i++) {
...
}
Task
For every whole number in the span (given as information) :
In the event that ,, print the English portrayal of it in lowercase. That is "one" for , "two" for ,, etc.
Else on the off chance that and it is a considerably number, print "even".
Else in the event that and it is an odd number, print "odd".
Input Arrangement
The primary line contains a whole number, .
The seond line contains a number, .
Imperatives
Yield Arrangement
Print the proper English representation,even, or odd, in view of the circumstances depicted in the 'task' area.
Note:
Test Information
8
11
Test Result
eight
nine
indeed
odd
#include <stdio.h>
int main()
{
int a,b,n;
scanf("%d",&a);
scanf("%d",&b);
for(n = a;n<=b;n++)
{
if(n == 1)
printf("one\n");
else if(n == 2)
printf("two\n");
else if(n == 3)
printf("three\n");
else if(n == 4)
printf("four\n");
else if(n == 5)
printf("five\n");
else if(n == 6)
printf("six\n");
else if(n == 7)
printf("seven\n");
else if(n == 8)
printf("eight\n");
else if(n == 9)
printf("nine\n");
else {
if(n %2 ==0)
printf("even\n");
else
printf("odd\n");
}}
return 0;
}
Output
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