title: Count Sum Of Digits In Numbers From 1 To N
signature: fun countSumOfDigitsInNumbersFrom1ToN(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(29, 165)
      println("TEST-0...${result}")
      
      result = driver(97, 865)
      println("TEST-1...${result}")
      
      result = driver(71, 540)
      println("TEST-2...${result}")
      
      result = driver(82, 667)
      println("TEST-3...${result}")
      
      result = driver(69, 525)
      println("TEST-4...${result}")
      
      result = driver(30, 168)
      println("TEST-5...${result}")
      
      result = driver(82, 667)
      println("TEST-6...${result}")
      
      result = driver(32, 177)
      println("TEST-7...${result}")
      
      result = driver(77, 609)
      println("TEST-8...${result}")
      
      result = driver(39, 240)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def count_sum_of_digits_in_numbers_from_1_to_n(n):
      if (n < 10):
          return ((n * (n + 1)) / 2)
      d = int(math.log10(n))
      a = ([0] * (d + 1))
      a[0] = 0
      a[1] = 45
      for i in range(2, (d + 1)):
          a[i] = ((a[(i - 1)] * 10) + (45 * int(math.ceil(math.pow(10, (i - 1))))))
      p = int(math.ceil(math.pow(10, d)))
      msd = (n // p)
      return int(((((msd * a[d]) + (((msd * (msd - 1)) // 2) * p)) + (msd * (1 + (n % p)))) + count_sum_of_digits_in_numbers_from_1_to_n((n % p))))
source_cpp: int count_sum_of_digits_in_numbers_from_1_to_n(int n) {
    if ( n < 10 ) return n * ( n + 1 ) / 2;
    int d = log10 ( n );
    int * a = new int [ d + 1 ];
    a [ 0 ] = 0, a [ 1 ] = 45;
    for ( int i = 2;
    i <= d;
    i ++ ) a [ i ] = a [ i - 1 ] * 10 + 45 * ceil ( pow ( 10, i - 1 ) );
    int p = ceil ( pow ( 10, d ) );
    int msd = n / p;
    return msd * a [ d ] + ( msd * ( msd - 1 ) / 2 ) * p + msd * ( 1 + n % p ) + count_sum_of_digits_in_numbers_from_1_to_n ( n % p );
  }