title: Number Subsequences Ab String Repeated K Times
signature: numberSubsequencesAbStringRepeatedKTimes :: String -> Integer -> Double
  numberSubsequencesAbStringRepeatedKTimes s k = 
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 :: Double -> Double -> Bool
  validateSolution actual expected = do
      (abs (actual-expected)) < 1e-06
  
  driver :: String -> Integer -> Double -> String
  driver s k expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME s k) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver ("abbc") (96) (9312.0) )
  
      printf "TEST-1...%s\n" (driver ("abahk") (7) (49.0) )
  
      printf "TEST-2...%s\n" (driver ("hugbabab") (5) (75.0) )
  
      printf "TEST-3...%s\n" (driver ("abadbc") (60) (7260.0) )
  
      printf "TEST-4...%s\n" (driver ("nkg9") (8) (0.0) )
  
      printf "TEST-5...%s\n" (driver ("jh7dab") (41) (861.0) )
  
      printf "TEST-6...%s\n" (driver ("abd") (87) (3828.0) )
  
      printf "TEST-7...%s\n" (driver ("aabb8yk") (4) (40.0) )
  
      printf "TEST-8...%s\n" (driver ("1111") (18) (0.0) )
  
      printf "TEST-9...%s\n" (driver ("PFXAhr") (8) (0.0) )
  
source_py: def number_subsequences_ab_string_repeated_k_times(s, K):
      n = len(s)
      c1 = 0
      c2 = 0
      C = 0
      for i in range(n):
          if (s[i] == 'a'):
              c1 += 1
          if (s[i] == 'b'):
              c2 += 1
              C += c1
      return ((C * K) + ((((K * (K - 1)) / 2) * c1) * c2))
source_cpp: int number_subsequences_ab_string_repeated_k_times(string s, int K) {
    int n = s . length ( );
    int C, c1 = 0, c2 = 0;
    for ( int i = 0;
    i < n;
    i ++ ) {
      if ( s [ i ] == 'a' ) c1 ++;
      if ( s [ i ] == 'b' ) {
        c2 ++;
        C += c1;
      }
    }
    return C * K + ( K * ( K - 1 ) / 2 ) * c1 * c2;
  }