title: Maximum Value Choice Either Dividing Considering
signature: fun maximumValueChoiceEitherDividingConsidering(n: Int): Int {
test_code: package main
  
  import kotlin.text.*
  import kotlin.math.*
  
  
  // SOLUTION CODE
  // ===================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE
  // ===================================
  fun validateSolution(actual:Int, expected:Int):Boolean{
    return actual.equals(expected)
  }
  fun driver(n: Int, expected:Int):String {
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)) {
                return "PASSED"
        }
        return "FAILED"
      } catch (e: Exception){
        return e::class.simpleName as String
      }
  
  }
  
  fun main() {
      var result: String = ""
      result = driver(3, 3)
      println("TEST-0...${result}")
      
      result = driver(19, 24)
      println("TEST-1...${result}")
      
      result = driver(39, 57)
      println("TEST-2...${result}")
      
      result = driver(89, 152)
      println("TEST-3...${result}")
      
      result = driver(96, 192)
      println("TEST-4...${result}")
      
      result = driver(68, 114)
      println("TEST-5...${result}")
      
      result = driver(48, 83)
      println("TEST-6...${result}")
      
      result = driver(5, 5)
      println("TEST-7...${result}")
      
      result = driver(3, 3)
      println("TEST-8...${result}")
      
      result = driver(4, 4)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def maximum_value_choice_either_dividing_considering(n):
      res = list()
      res.append(0)
      res.append(1)
      i = 2
      while (i < (n + 1)):
          res.append(max(i, (((res[int((i / 2))] + res[int((i / 3))]) + res[int((i / 4))]) + res[int((i / 5))])))
          i = (i + 1)
      return res[n]
source_cpp: int maximum_value_choice_either_dividing_considering(int n) {
    int res [ n + 1 ];
    res [ 0 ] = 0;
    res [ 1 ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) res [ i ] = max ( i, ( res [ i / 2 ] + res [ i / 3 ] + res [ i / 4 ] + res [ i / 5 ] ) );
    return res [ n ];
  }