title: Hardy Ramanujan Theorem
signature: function hardy_ramanujan_theorem(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(99, 2);
      @printf("TEST-0...%s\n",result);
      
      result = driver(33, 2);
      @printf("TEST-1...%s\n",result);
      
      result = driver(50, 2);
      @printf("TEST-2...%s\n",result);
      
      result = driver(17, 1);
      @printf("TEST-3...%s\n",result);
      
      result = driver(18, 2);
      @printf("TEST-4...%s\n",result);
      
      result = driver(69, 2);
      @printf("TEST-5...%s\n",result);
      
      result = driver(23, 1);
      @printf("TEST-6...%s\n",result);
      
      result = driver(18, 2);
      @printf("TEST-7...%s\n",result);
      
      result = driver(94, 2);
      @printf("TEST-8...%s\n",result);
      
      result = driver(16, 1);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def hardy_ramanujan_theorem(n):
      count = 0
      if ((n % 2) == 0):
          count = (count + 1)
          while ((n % 2) == 0):
              n = int((n / 2))
      i = 3
      while (i <= int(math.sqrt(n))):
          if ((n % i) == 0):
              count = (count + 1)
              while ((n % i) == 0):
                  n = int((n / i))
          i = (i + 2)
      if (n > 2):
          count = (count + 1)
      return count
source_cpp: int hardy_ramanujan_theorem(int n) {
    int count = 0;
    if ( n % 2 == 0 ) {
      count ++;
      while ( n % 2 == 0 ) n = n / 2;
    }
    for ( int i = 3;
    i <= sqrt ( n );
    i = i + 2 ) {
      if ( n % i == 0 ) {
        count ++;
        while ( n % i == 0 ) n = n / i;
      }
    }
    if ( n > 2 ) count ++;
    return count;
  }