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 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.
A sample loop will be
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
odd
Now here's the solution in C:-
code snippet:#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>int main() { int a,b,n,i; scanf("%d %d",&a,&b); for(i=a;i<=b;i++) { n=i; if(1<=n && n<=9) { 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 { printf("nine\n"); } } else if(n>9) { if(n%2==0) { printf("even\n"); } else { printf("odd\n"); } } } // Complete the code. return 0;}
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