title: Count Number Pairs N B N Gcd B B
signature: function countNumberPairsNBNGcdBB($n) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return abs($actual - $expected) < 1e-09;
  }
  
  function driver($n, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($n), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      printf("TEST-0...%s\n",driver(
        17, 
        58.47239288489289));
      
      printf("TEST-1...%s\n",driver(
        72, 
        349.9783310569835));
      
      printf("TEST-2...%s\n",driver(
        43, 
        187.0499406858786));
      
      printf("TEST-3...%s\n",driver(
        55, 
        252.6486716559348));
      
      printf("TEST-4...%s\n",driver(
        62, 
        291.1683590456309));
      
      printf("TEST-5...%s\n",driver(
        22, 
        81.19789150478006));
      
      printf("TEST-6...%s\n",driver(
        17, 
        58.47239288489289));
      
      printf("TEST-7...%s\n",driver(
        68, 
        326.67596370158344));
      
      printf("TEST-8...%s\n",driver(
        20, 
        71.95479314287363));
      
      printf("TEST-9...%s\n",driver(
        29, 
        114.88796013002468));
      
  }
  
  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;
  }