Problem:
Score : 200 points

Problem 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.

Constraints

1 \leq N \leq 100

1 \leq D \leq 100

1 \leq X \leq 100

1 \leq A_i \leq 100 (1 \leq i \leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D X
A_1
A_2
:
A_N

Output

Find the number of chocolate pieces prepared at the beginning of the camp.

Sample Input 1

3
7 1
2
5
10

Sample Output 1

8

The camp has 3 participants and lasts for 7 days.
Each participant eats chocolate pieces as follows:

The first participant eats one chocolate piece on Day 1, 3, 5 and 7, for a total of four.

The second participant eats one chocolate piece on Day 1 and 6, for a total of two.

The third participant eats one chocolate piece only on Day 1, for a total of one.

Since the number of pieces remaining at the end of the camp is one, the number of pieces prepared at the beginning of the camp is 1 + 4 + 2 + 1 = 8.

Sample Input 2

2
8 20
1
10

Sample Output 2

29

Sample Input 3

5
30 44
26
18
81
18
6

Sample Output 3

56

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <sstream>
#include <cmath>
#include <algorithm>
#include <utility>
#include <functional>
#include <iomanip>
#include <typeinfo>
#include <iterator>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <tuple>
#include <map>
#include <set>
#include <numeric>
using namespace std;

#define repr(itr, bgn, end) for (ll itr = (ll)bgn; itr < (ll)end; ++itr)
#define rep(itr, end) repr(itr, 0, end)
#define reprrev(itr, bgn, end) for (ll itr = (ll)end - 1; itr >= (ll)bgn; --i)
#define reprev(itr, end) reprrev(itr, 0, end)
#define all(container) container.begin(), container.end()
#define sortall(container) sort(all(container))
#define yn(cond) cond ? (cout << "Yes" << endl) : (cout << "No" << endl)
#define YN(cond) cond ? (cout << "YES" << endl) : (cout << "NO" << endl)
#define LLONG_MAX numeric_limits<ll>::max()
#define INT_MAX numeric_limits<int>::max()

typedef long long       ll;
typedef struct edge_ { ll to; ll cost; } edge;
typedef vector<int>     vi;
typedef vector<ll>      vll;
typedef vector<double>  vd;
typedef vector<bool>    vb;
typedef vector<string>  vs;
typedef vector<char>    vc;
typedef vector<edge>    ve;
typedef pair<int, int>  pii;
typedef pair<ll, ll>    pll;
typedef vector<pii>     vpii;
typedef vector<pll>     vpll;
typedef vector<vll>		vvll;
typedef vector<vi>		vvi;
typedef vector<ve>      vve;
typedef vector<vc>		vvc;

constexpr double	PI = 3.14159265358979323846;
constexpr ll		INF = 1 << 15;
constexpr ll		MOD = (ll)1e9 + 7;
constexpr double	EPS = 1e-7;
constexpr int		dy4[4] = { 0, 1, 0, -1 };
constexpr int		dx4[4] = { 1, 0, -1, 0 };
constexpr int		dy8[8] = { 0, 1, 1, 1, 0, -1, -1, -1 };
constexpr int		dx8[8] = { 1, 1, 0, -1, -1, -1, 0, 1 };

template <int mod>
class modint {
private:
	long long num;
	modint set(int _num) { num = _num % mod; return *this; }

public:
	modint() { num = 0; }
	modint(int num) : num(num % mod) {}

	modint operator =  (const int& x) { return set(x % mod); }
	modint operator +  (const int& x) { return modint<mod>(num + (x % mod)); }
	modint operator -  (const int& x) { return modint<mod>(num - (x % mod)); }
	modint operator *  (const int& x) { return modint<mod>(num * (x % mod)); }
	modint operator /  (const int& x) { return modint<mod>(num / (x % mod)); }
	modint operator += (const int& x) { return set(num + (x % mod)); }
	modint operator -= (const int& x) { return set(num - (x % mod)); }
	modint operator *= (const int& x) { return set(num * (x % mod)); }

	friend ostream& operator << (ostream& os, const modint<mod>& x) {
		os << x.num;
		return os;
	}
};

template<typename T>
istream& operator >> (istream& is, vector<T>& vec) {
	for (T& x : vec) is >> x;
	return is;
}

template<typename T>
ostream& operator << (ostream& os, const vector<T>& vec) {
	for (const T& x : vec) os << x << " ";
	return os;
}

template<typename S, typename T>
istream& operator >> (istream& is, pair<S, T>& p) {
	is >> p.first >> p.second;
	return is;
}

template<typename S, typename T>
ostream& operator << (ostream& os, const pair<S, T>& p) {
	os << p.first << " " << p.second;
	return os;
}

/******************************************************************************/
/**********************         Don't make WA !!         **********************/
/******************************************************************************/


int main()
{
	int N, D, X;
	cin >> N >> D >> X;
	vi A(N);
	cin >> A;

	int ans = X;
	for (int i = 1; i <= D; i++) {
		for (int n = 0; n < N; n++) {
			if (i % A[n] == 1) ans++;
			else if (A[n] == 1) ans++;
		}
	}

	cout << ans << endl;

	return 0;
}