**For Loops in c**

### Problem Statement :

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> 1. expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop. 2. expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated. 3. 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 in the interval [a,b] (given as input) : 1. If 1<=n<=9, then print the English representation of it in lowercase. That is "one" for 1, "two" for 2, and so on. 2. Else if n>9 and it is an even number, then print "even". 3. 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<=10^6 Output Format: Print the appropriate English representation,even, or odd, based on the conditions described in the 'task' section.

### Solution :

` ````
Solution in C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int a, b,t,s;
scanf("%d\n%d", &a, &b);
t=a>b?a:b;
s=a<b?a:b;
for(s;s<=t;s++)
{
if(s<10)
{
if(s==1)
printf("one\n");
if(s==2)
printf("two\n");
if(s==3)
printf("three\n");
if(s==4)
printf("four\n");
if(s==5)
printf("five\n");
if(s==6)
printf("six\n");
if(s==7)
printf("seven\n");
if(s==8)
printf("eight\n");
if(s==9)
printf("nine\n");
}
if(s>=10)
{
if(s%2==0)
printf("even\n");
else
printf("odd\n");
}
}
return 0;
}
```

## View More Similar Problems

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →