JAVA End Of File

 The challenge here is to read  lines of input until you reach EOF, then number and print all  lines of content.

Hint: Java's Scanner.hasNext() method is helpful for this problem.

Input Format

Read some unknown  lines of input from stdin(System.in) until you reach EOF; each line of input contains a non-empty String.

Output Format

For each line, print the line number, followed by a single space, and then the line content received as input.

Sample Input

Hello world
I am a file
Read me until end-of-file.

Sample Output

1 Hello world
2 I am a file
3 Read me until end-of-file.

Answer:

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
    public static void main(String[] args) {
        int count =0;
        Scanner sc = new Scanner(System.in);
        //String s = sc.readLine();
        
        while(sc.hasNext())
        {
            count ++;
            System.out.println(count+" "+sc.nextLine());
        }
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
    }
}

JAVA Datatypes

 Java has 8 primitive data types; char, boolean, byte, short, int, long, float, and double. For this exercise, we'll work with the primitives used to hold integer values (byte, short, int, and long):

• A byte is an 8-bit signed integer.
• A short is a 16-bit signed integer.
• An int is a 32-bit signed integer.
• A long is a 64-bit signed integer.

Given an input integer, you must determine which primitive data types are capable of properly storing that input.

To get you started, a portion of the solution is provided for you in the editor.

Reference: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

Input Format

The first line contains an integer, , denoting the number of test cases.
Each test case, , is comprised of a single line with an integer, , which can be arbitrarily large or small.

Output Format

For each input variable  and appropriate primitive , you must determine if the given primitives are capable of storing it. If yes, then print:

n can be fitted in:
* dataType

If there is more than one appropriate data type, print each one on its own line and order them by size (i.e.: ).

If the number cannot be stored in one of the four aforementioned primitives, print the line:

n can't be fitted anywhere.

Sample Input

5
-150
150000
1500000000
213333333333333333333333333333333333
-100000000000000

Sample Output

-150 can be fitted in:
* short
* int
* long
150000 can be fitted in:
* int
* long
1500000000 can be fitted in:
* int
* long
213333333333333333333333333333333333 can't be fitted anywhere.
-100000000000000 can be fitted in:
* long

Explanation

 can be stored in a short, an int, or a long.

 is very large and is outside of the allowable range of values for the primitive data types discussed in this problem.

Answer:

import java.util.*;

import java.io.*;

class Solution{

    public static void main(String []argh)

    {

        Scanner sc = new Scanner(System.in);

        int t=sc.nextInt();

        for(int i=0;i<t;i++)

        {

            try

            {

                long x=sc.nextLong();

                System.out.println(x+" can be fitted in:");

                //if(x>=-128 && x<=127)System.out.println("* byte");

                //Complete the code

                if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE)

                  System.out.println("* byte\n* short\n* int\n* long");

                

                else if (x>=Short.MIN_VALUE && x <= Short.MAX_VALUE) 

                  System.out.println("* short\n* int\n* long");

                

                else if (x>=Integer.MIN_VALUE && x <= Integer.MAX_VALUE)

                  System.out.println("* int\n* long");

                

                else

                  System.out.println("* long");

                

            }

            catch(Exception e)

            {

                System.out.println(sc.next()+" can't be fitted anywhere.");

            }

        }

        sc.close();

    }

}