NumberFormatException while converting from binary string

Question

public class ACM_ICPC_TEAM {
 public static void main(String[] args) {

        Scanner sc=new Scanner(System.in);
        int N=sc.nextInt();
        int M=sc.nextInt();
        sc.nextLine();

        String s[]=new String[N];
        for(int i=0;i<N;i++)
        { 
        for(int j=0;j<M;j++);
        {               
            s[i]=sc.nextLine().toString();              
        }         
        }
      BitSet b1=new BitSet(M);
      BitSet b2=new BitSet(M);
      BitSet b3=new BitSet(M);
      int max=0,count=0,count1=0;
        for(int i=0;i<N;i++)
        {  b1=fromString(s[i]);
         for(int j=i+1;j<N;j++)
         {  
            b3=(BitSet) b1.clone();
            b2=fromString(s[j]);
            b3.or(b2);       
            count=b3.cardinality();    
            if(count>max)
                {
                max=count;
                count1=1;
                }
            else if(count==max)
                 count1++;
         }
        }

        System.out.println(count+"\n"+count1);          
    }

 public static BitSet fromString(String s)
 {  
    return BitSet.valueOf(new long[] { Long.parseLong(s,2) });
 }
 public static String toString(BitSet b)
 {
 return Long.toString(b.toLongArray()[0], 2);   
 }
}

My fromString is giving me NumberFormatException when I pass binary string of lage size(~500), its working absolutely fine for smaller length strings.

Just a guess: You read next int, but the number might be bigger. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Sep 8, 2016 at 16:26
Please edit your question to include the actual text of the exception, including the stack trace. Indicate which line of the code you posted is triggering the exception. — Kenster
– Kenster, Commented Sep 8, 2016 at 16:53
Please have a look at How to create a Minimal, Complete, and Verifiable example, and review your post. — Hopeful Llama
– Hopeful Llama, Commented Sep 8, 2016 at 16:54

Markus Fischer · Accepted Answer · 2016-09-12 08:11:55Z

1

A Java long value is a 64 bit number, i.e. you can store up to 64 bits in it. A String with 500 bits simply cannot be represented by a single long, thus the exception.

A BitSet can represent sets of bits of arbitrary length. Unfortunately, BitSet does not have a method to append one BitSet to another. Therefore, what I would do is go through the String char by char, and set or clear the n'th bit of the BitSet, as needed.

answered Sep 12, 2016 at 8:11

Markus Fischer

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Comments

xenteros · Accepted Answer · 2016-10-13 04:50:11Z

0

There's no way to create an int with 500 bits. The maximum is 64. In Java you can use BigInteger which is a class which is used for huge numbers.

It has a constructor which takes a String and a constructor which takes a String value and int radix.

BigInteger a = new BigInteger("1234567890123456");
BigInteger b = new BigInteger("10000000000000000", 2);

answered Oct 13, 2016 at 4:50

xenteros

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