title: Round The Given Number To Nearest Multiple Of 10
signature: function round_the_given_number_to_nearest_multiple_of_10(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(31, 30);
      @printf("TEST-0...%s\n",result);
      
      result = driver(78, 80);
      @printf("TEST-1...%s\n",result);
      
      result = driver(19, 20);
      @printf("TEST-2...%s\n",result);
      
      result = driver(36, 40);
      @printf("TEST-3...%s\n",result);
      
      result = driver(77, 80);
      @printf("TEST-4...%s\n",result);
      
      result = driver(94, 90);
      @printf("TEST-5...%s\n",result);
      
      result = driver(86, 90);
      @printf("TEST-6...%s\n",result);
      
      result = driver(16, 20);
      @printf("TEST-7...%s\n",result);
      
      result = driver(95, 90);
      @printf("TEST-8...%s\n",result);
      
      result = driver(2, 0);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def round_the_given_number_to_nearest_multiple_of_10(n):
      a = ((n // 10) * 10)
      b = (a + 10)
      return (b if ((n - a) > (b - n)) else a)
source_cpp: int round_the_given_number_to_nearest_multiple_of_10(int n) {
    int a = ( n / 10 ) * 10;
    int b = a + 10;
    return ( n - a > b - n ) ? b : a;
  }