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.

# Java Datatypes problem solution | HackerRank

import java.io.*;
import java.util.*;
class Solution {
public static void main(String[] args) {
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");
if (x >= -32768 && x <= 32767) System.out.println("* short");
if (x >= -2147483648 && x <= 2147483647) System.out.println("* int");
if (x >= -9223372036854775808L && x <= 9223372036854775807L) System.out.println("* long");
} catch (Exception e) {
System.out.println(sc.next() + " can't be fitted anywhere.");
}

}
sc.close();
}
}

## Second solution

import java.io.*;
import java.util.*;
class Solution {
public static void main(String[] args) {
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 >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) System.out.println("* byte");
if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) System.out.println("* short");
if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) System.out.println("* int");
if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) System.out.println("* long");
} catch (Exception e) {
System.out.println(sc.next() + " can't be fitted anywhere.");
}

}
sc.close();
}
}