title: Count Number Pairs N B N Gcd B B
signature: function count_number_pairs_n_b_n_gcd_b_b(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return abs(actual - expected) < 1e-09
  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(17, 58.47239288489289);
      @printf("TEST-0...%s\n",result);
      
      result = driver(72, 349.9783310569835);
      @printf("TEST-1...%s\n",result);
      
      result = driver(43, 187.0499406858786);
      @printf("TEST-2...%s\n",result);
      
      result = driver(55, 252.6486716559348);
      @printf("TEST-3...%s\n",result);
      
      result = driver(62, 291.1683590456309);
      @printf("TEST-4...%s\n",result);
      
      result = driver(22, 81.19789150478006);
      @printf("TEST-5...%s\n",result);
      
      result = driver(17, 58.47239288489289);
      @printf("TEST-6...%s\n",result);
      
      result = driver(68, 326.67596370158344);
      @printf("TEST-7...%s\n",result);
      
      result = driver(20, 71.95479314287363);
      @printf("TEST-8...%s\n",result);
      
      result = driver(29, 114.88796013002468);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def count_number_pairs_n_b_n_gcd_b_b(n):
      k = n
      imin = 1
      ans = 0
      while (imin <= n):
          imax = (n / k)
          ans += (k * ((imax - imin) + 1))
          imin = (imax + 1)
          k = (n / imin)
      return ans
source_cpp: int count_number_pairs_n_b_n_gcd_b_b(int n) {
    int k = n;
    int imin = 1;
    int ans = 0;
    while ( imin <= n ) {
      int imax = n / k;
      ans += k * ( imax - imin + 1 );
      imin = imax + 1;
      k = n / imin;
    }
    return ans;
  }