title: Count Sum Of Digits In Numbers From 1 To N
signature: countSumOfDigitsInNumbersFrom1ToN :: Integer -> Integer
  countSumOfDigitsInNumbersFrom1ToN n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> String
  driver n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (29) (165) )
  
      printf "TEST-1...%s\n" (driver (97) (865) )
  
      printf "TEST-2...%s\n" (driver (71) (540) )
  
      printf "TEST-3...%s\n" (driver (82) (667) )
  
      printf "TEST-4...%s\n" (driver (69) (525) )
  
      printf "TEST-5...%s\n" (driver (30) (168) )
  
      printf "TEST-6...%s\n" (driver (82) (667) )
  
      printf "TEST-7...%s\n" (driver (32) (177) )
  
      printf "TEST-8...%s\n" (driver (77) (609) )
  
      printf "TEST-9...%s\n" (driver (39) (240) )
  
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 );
  }