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  1. Prepare
  2. Java
  3. Introduction
  4. Java Datatypes
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Java Datatypes

  • rohitkumar301104
     + 0 comments

    I solved this problem using Java's BigInteger DataType which is ment to handle very large inputs. I believe using BigInteger is appropriate for this problem as this problem is about Java Data Types.

    import java.io.*;
import java.util.*;
import java.math.BigInteger;

public class Solution {
    static BigInteger bn = BigInteger.valueOf(Byte.MIN_VALUE);
    static BigInteger bp = BigInteger.valueOf(Byte.MAX_VALUE);
    static BigInteger sn = BigInteger.valueOf(Short.MIN_VALUE);
    static BigInteger sp = BigInteger.valueOf(Short.MAX_VALUE);
    static BigInteger in = BigInteger.valueOf(Integer.MIN_VALUE);
    static BigInteger ip = BigInteger.valueOf(Integer.MAX_VALUE);
    static BigInteger ln = BigInteger.valueOf(Long.MIN_VALUE);
    static BigInteger lp = BigInteger.valueOf(Long.MAX_VALUE);
    

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
          Scanner sc=new Scanner(System.in);
        int loopcount=sc.nextInt();   
        
        for (int i=1;i<=loopcount;i++){
            
         BigInteger n=sc.nextBigInteger();  
         if (n.compareTo(bn)>=0 && n.compareTo(bp)<=0){
            System.out.println(n+" can be fitted in:\n* byte\n* short\n* int\n* long");} 
             else if (n.compareTo(sn)>=0 && n.compareTo(sp)<=0){
            System.out.println(n+" can be fitted in:\n* short\n* int\n* long");}
            else if (n.compareTo(in)>=0 && n.compareTo(ip)<=0){
            System.out.println(n+" can be fitted in:\n* int\n* long");}
            else if(n.compareTo(ln)>=0 &&n.compareTo(lp)<=0){
            System.out.println(n+" can be fitted in:\n* long");}
            else
            { System.out.println(n+" can't be fitted anywhere.");}
        }
    }
}