title: Compute Ncr P Set 1 Introduction And Dynamic Programming Solution
signature: function computeNcrPSet1IntroductionAndDynamicProgrammingSolution($n, $r, $p) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return serialize($actual) == serialize($expected);
  }
  
  function driver($n, $r, $p, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($n, $r, $p), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      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));
      
  }
  
  main();
  ?>
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 ];
  }