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  1. Prepare
  2. Java
  3. Data Structures
  4. Java BitSet
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Java BitSet

  • bogad
     + 1 comment

    Very simple solution with a BitSet array:

    Scanner s = new Scanner(System.in);
int n = s.nextInt();
int m = s.nextInt();
BitSet[] bs = new BitSet[]{new BitSet(n), new BitSet(n)};
for (int i=0; i<m; i++) {
	String c = s.next().trim();
	if ( "AND".equals(c) ) bs[s.nextInt()-1].and(bs[s.nextInt()-1]);
	else if ( "OR".equals(c) ) bs[s.nextInt()-1].or(bs[s.nextInt()-1]);
	else if ( "XOR".equals(c) ) bs[s.nextInt()-1].xor(bs[s.nextInt()-1]);
	else if ( "SET".equals(c) ) bs[s.nextInt()-1].set(s.nextInt());
	else if ( "FLIP".equals(c) ) bs[s.nextInt()-1].flip(s.nextInt());
	System.out.println(bs[0].cardinality() + " " + bs[1].cardinality());
}