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Java Datatypes

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  • CryptCode
    4 years ago+ 0 comments

    This should work without any problem and you don't have to remember range of any Datatypes. Happy Coding!

    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>=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.");
                }
    
            }
        }
    }
    
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