**Viral Advertising**

### 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. viralAdvertising has the following parameter(s): 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):
receive = int(people/2)
total+=receive
people = receive*3
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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