title: Compute Ncr P Set 1 Introduction And Dynamic Programming Solution
signature: function compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(n, r, p)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, r, p, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n, r, p), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(82, 5, 94, 50);
      @printf("TEST-0...%s\n",result);
      
      result = driver(45, 24, 95, 80);
      @printf("TEST-1...%s\n",result);
      
      result = driver(44, 68, 61, 0);
      @printf("TEST-2...%s\n",result);
      
      result = driver(88, 24, 43, 0);
      @printf("TEST-3...%s\n",result);
      
      result = driver(90, 75, 57, 0);
      @printf("TEST-4...%s\n",result);
      
      result = driver(98, 55, 43, 2);
      @printf("TEST-5...%s\n",result);
      
      result = driver(80, 54, 88, 0);
      @printf("TEST-6...%s\n",result);
      
      result = driver(60, 75, 65, 0);
      @printf("TEST-7...%s\n",result);
      
      result = driver(52, 73, 86, 0);
      @printf("TEST-8...%s\n",result);
      
      result = driver(71, 26, 45, 12);
      @printf("TEST-9...%s\n",result);
      
  end
  
  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 ];
  }