Exemplo n.º 1
0
 private ArrayList<String[]> convertSolution(long[] rows, ArrayList<String[]> results) {
   String[] res = new String[rows.length];
   for (int i = 0; i < rows.length; ++i) {
     res[i] = Long.toBinaryString(rows[i]).replace('0', '.').replace('1', 'Q');
     while (res[i].length() < rows.length) res[i] = '.' + res[i];
   }
   results.add(res);
   return results;
 }
Exemplo n.º 2
0
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    String s = "0000000001111111";
    s = s.replaceAll("0000000", "*");
    System.out.println(s);
    StringBuilder bb;
    System.out.printf("%d%d", 1, 1);
    Integer n = 10;
    Map<Integer, Integer> mp = new TreeMap<Integer, Integer>();

    bb = new StringBuilder(input.nextLine());

    Character aaa = new Character('a');

    while (input.hasNext()) {
      long a = input.nextLong();
      long b = input.nextLong();
      String sa = new String(Long.toBinaryString(a));
      String sb = new String(Long.toBinaryString(b));
      int ans = 0;
      int i, j, sa0 = 0, sb0 = 0;
      for (i = 0; i < sa.length(); i++) if (sa.charAt(i) == '0') sa0++;
      for (j = 0; j < sb.length(); j++) if (sb.charAt(j) == '0') sb0++;
      for (i = 0; i < sa.length(); i++) if (sa.charAt(i) == '0') break;
      for (j = 0; j < sb.length(); j++) if (sb.charAt(j) == '0') break;

      if (sa.length() == sb.length()) {
        ans = j - i; // =   sa.length()-i-(sb.length()-j);
        if (sb0 == 1) ans++;
      } else {
        ans += sa.length() - i;
        ans += j - 1; // =   sb.length()-1-(sb.length()-j)
        if (sb0 == 1) ans++;
        if (sb.length() - sa.length() >= 2) {
          ans += (sb.length() - 2 + sa.length()) * (sb.length() - sa.length() - 1) / 2;
        }
      }

      System.out.println(11111);
      System.out.println(ans);
    }
  }
Exemplo n.º 3
0
 static void printBinaryLong(String s, long l) {
   print(s + ", long: " + l + ", binary:\n    " + Long.toBinaryString(l));
 }