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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#include<math.h>
#include<assert.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384
int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int nHr(int a,int b){return nCr(a+b-1,b);}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}
int sankaku(int x){return ((1+x)*x)/2;}
void swap(int *a,int *b){int c;c=(*a);(*a)=(*b);(*b)=c;}
//小さい順
int compareInt(const void* a, const void* b){int aNum = *(int*)a;int bNum = *(int*)b;return aNum - bNum;}

long long _gcd(long long a,long long b){
    long long c;
    while(b!=0){c=a%b;a=b;b=c;}
    return a;
    
}
long long _lcm(long long a,long long b){long long c=_gcd(a,b);a/=c;return a*b;}

int main(){
    int N,D,X;
    int sum=0;
    scanf("%d%d%d",&N,&D,&X);
    int a[N];
    for(int i=0;i<N;i++){
        scanf("%d",&a[i]);
    }
    for(int i=0;i<N;i++){
        for(int j=0;1+j*a[i]<=D;j++){
            sum++;
        }
    }
    printf("%d",sum+X);
}