### Problem Statement :

```HackerLand Enterprise is adopting a new viral advertising strategy. When they launch a new product, they advertise it to exactly 5 people on social media.

On the first day, half of those 5 people (i.e., floor(5/2) = 2 ) like the advertisement and each shares it with 3 of their friends. At the beginning of the second day, floor(5/2)*3 = 2*3 = 6 people receive the advertisement.

Each day, floor(recipients/2) of the recipients like the advertisement and will share it with 3 friends on the following day. Assuming nobody receives the advertisement twice, determine how many people have liked the ad by the end of a given day, beginning with launch day as day 1.

Example
.n = 5

Day Shared Liked Cumulative
1      5     2       2
2      6     3       5
3      9     4       9
4     12     6      15
5     18     9      24
The progression is shown above. The cumulative number of likes on the 5th day is 24.

Function Description

Complete the viralAdvertising function in the editor below.
int n: the day number to report

Returns

int: the cumulative likes at that day

Input Format

A single integer, n, the day number.

Constraints
1 <= n <= 50```

### Solution :

```                            ```Solution in C :

python 3  :

people = 5
total = 0
n = int(input())

for i in range(n):
print(total)

Java  :

import java.io.*;
import java.util.*;

public class Solution {

public static final int INITIAL_AMOUNT_OF_PEOPLE = 5;

public static void main(String[] args) {
try (Scanner scanner = new Scanner(System.in)) {
int n = scanner.nextInt();
int currentAmount = INITIAL_AMOUNT_OF_PEOPLE;
int totalNumber = 0;
for (int i = 0; i < n; i++) {
currentAmount = currentAmount/2;
totalNumber += currentAmount;
currentAmount *= 3;
}
System.out.println(totalNumber);
}
}
}

C++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n;
cin>>n;

int m = 5;
int total;
for(int i=0; i<n; ++i){
m = m/2;
total += m;
m *= 3;
}

cout<<total<<endl;
return 0;
}

C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
int i=0,j,k,l,m,n,t;
scanf("%d",&n);
k=5;
for(i=0;i<n;i++)
{
j=k/2;
k=j*3;
m=m+j;
}
printf("%d",m);
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}```
```

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