output_description: Find the number of chocolate pieces prepared at the beginning of the camp.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      D X
      A_1
      A_2
      :
      A_N
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  def sigma(N):
      ans = 0
      for i in range(1, N + 1):
          ans += i
      return ans
  
  
  def comb(n, r):
      if n - r < r:
          r = n - r
      if r == 0:
          return 1
      if r == 1:
          return n
  
      numerator = [n - r + k + 1 for k in range(r)]
      denominator = [k + 1 for k in range(r)]
  
      for p in range(2, r + 1):
          pivot = denominator[p - 1]
          if pivot > 1:
              offset = (n - r) % p
              for k in range(p - 1, r, p):
                  numerator[k - offset] /= pivot
                  denominator[k] /= pivot
  
      result = 1
      for k in range(r):
          if numerator[k] > 1:
              result *= int(numerator[k])
  
      return result
  
  
  def bisearch(L, target):
      low = 0
      high = len(L) - 1
  
      while low <= high:
          mid = (low + high) // 2
          guess = L[mid]
          if guess == target:
              return True
          elif guess < target:
              low = mid + 1
          elif guess > target:
              high = mid - 1
      if guess != target:
          return False
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      N = int(input())
      D, X = li_input()
      A = [int(input()) for _ in range(N)]
  
      e = 0
  
      for a in A:
          e += math.ceil(D / a)
  
      print(e + X)
  
  
  main()
  
problem_description: Statement
  
  Some number of chocolate pieces were prepared for a training camp. The camp
  had N participants and lasted for D days. The i-th participant (1 \leq i \leq
  N) ate one chocolate piece on each of the following days in the camp: the 1-st
  day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result,
  there were X chocolate pieces remaining at the end of the camp. During the
  camp, nobody except the participants ate chocolate pieces.
  
  Find the number of chocolate pieces prepared at the beginning of the camp.
samples: [{"input": "3\n    7 1\n    2\n    5\n    10", "output": "8\n    \n\nThe camp has 3 participants and lasts for 7 days. Each participant eats\nchocolate pieces as follows:\n\n  * The first participant eats one chocolate piece on Day 1, 3, 5 and 7, for a total of four.\n  * The second participant eats one chocolate piece on Day 1 and 6, for a total of two.\n  * The third participant eats one chocolate piece only on Day 1, for a total of one.\n\nSince the number of pieces remaining at the end of the camp is one, the number\nof pieces prepared at the beginning of the camp is 1 + 4 + 2 + 1 = 8.\n\n* * *"}, {"input": "2\n    8 20\n    1\n    10", "output": "29\n    \n\n* * *"}, {"input": "5\n    30 44\n    26\n    18\n    81\n    18\n    6", "output": "56"}]