title: Dynamic Programming Set 36 Cut A Rope To Maximize Product 1
signature: function dynamic_programming_set_36_cut_a_rope_to_maximize_product_1(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(62, 6973568802);
      @printf("TEST-0...%s\n",result);
      
      result = driver(53, 258280326);
      @printf("TEST-1...%s\n",result);
      
      result = driver(8, 18);
      @printf("TEST-2...%s\n",result);
      
      result = driver(6, 9);
      @printf("TEST-3...%s\n",result);
      
      result = driver(35, 354294);
      @printf("TEST-4...%s\n",result);
      
      result = driver(35, 354294);
      @printf("TEST-5...%s\n",result);
      
      result = driver(46, 19131876);
      @printf("TEST-6...%s\n",result);
      
      result = driver(74, 564859072962);
      @printf("TEST-7...%s\n",result);
      
      result = driver(69, 94143178827);
      @printf("TEST-8...%s\n",result);
      
      result = driver(3, 2);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def dynamic_programming_set_36_cut_a_rope_to_maximize_product_1(n):
      if ((n == 2) or (n == 3)):
          return (n - 1)
      res = 1
      while (n > 4):
          n -= 3
          res *= 3
      return (n * res)
source_cpp: int dynamic_programming_set_36_cut_a_rope_to_maximize_product_1(int n) {
    if ( n == 2 || n == 3 ) return ( n - 1 );
    int res = 1;
    while ( n > 4 ) {
      n -= 3;
      res *= 3;
    }
    return ( n * res );
  }