Java Stdin and Stdout II


Problem Statement :


In this challenge, you must read an integer, a double, and a String from stdin, then print the values according to the instructions in the Output Format section below. To make the problem a little easier, a portion of the code is provided for you in the editor.

Note: We recommend completing Java Stdin and Stdout I before attempting this challenge.

Input Format

There are three lines of input:

The first line contains an integer.
The second line contains a double.
The third line contains a String.
Output Format

There are three lines of output:

On the first line, print String: followed by the unaltered String read from stdin.
On the second line, print Double: followed by the unaltered double read from stdin.
On the third line, print Int: followed by the unaltered integer read from stdin.
To make the problem easier, a portion of the code is already provided in the editor.

Note: If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

Sample Input

42
3.1415
Welcome to HackerRank's Java tutorials!
Sample Output

String: Welcome to HackerRank's Java tutorials!
Double: 3.1415
Int: 42



Solution :


                            Solution in C :

import java.util.Scanner;

public class Solution {

	public static void main(String[] args) {
			Scanner sc=new Scanner(System.in);
			int x=sc.nextInt();
			double y=sc.nextDouble();
			sc.nextLine();
			String s=sc.nextLine();
			
			
			System.out.println("String: "+s);
			System.out.println("Double: "+y);
			System.out.println("Int: "+x);
	}
}
                        




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 →