title: Compute Ncr P Set 1 Introduction And Dynamic Programming Solution
signature: computeNcrPSet1IntroductionAndDynamicProgrammingSolution :: Integer -> Integer -> Integer -> Integer
  computeNcrPSet1IntroductionAndDynamicProgrammingSolution n r p = 
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 -> Integer -> Integer -> String
  driver n r p expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n r p) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (82) (5) (94) (50) )
  
      printf "TEST-1...%s\n" (driver (45) (24) (95) (80) )
  
      printf "TEST-2...%s\n" (driver (44) (68) (61) (0) )
  
      printf "TEST-3...%s\n" (driver (88) (24) (43) (0) )
  
      printf "TEST-4...%s\n" (driver (90) (75) (57) (0) )
  
      printf "TEST-5...%s\n" (driver (98) (55) (43) (2) )
  
      printf "TEST-6...%s\n" (driver (80) (54) (88) (0) )
  
      printf "TEST-7...%s\n" (driver (60) (75) (65) (0) )
  
      printf "TEST-8...%s\n" (driver (52) (73) (86) (0) )
  
      printf "TEST-9...%s\n" (driver (71) (26) (45) (12) )
  
source_py: def compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(n, r, p):
      C = [0 for i in range((r + 1))]
      C[0] = 1
      for i in range(1, (n + 1)):
          for j in range(min(i, r), 0, (- 1)):
              C[j] = ((C[j] + C[(j - 1)]) % p)
      return C[r]
source_cpp: int compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(int n, int r, int p) {
    int C [ r + 1 ];
    memset ( C, 0, sizeof ( C ) );
    C [ 0 ] = 1;
    for ( int i = 1;
    i <= n;
    i ++ ) {
      for ( int j = min ( i, r );
      j > 0;
      j -- ) C [ j ] = ( C [ j ] + C [ j - 1 ] ) % p;
    }
    return C [ r ];
  }