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c++ code run the output Q3 A company ABC requires to use an online server for research experimentation for a certain number of days. The

c++ code run the output

Q3

A company ABC requires to use an online server for research experimentation for a certain number of days. The server will be used on certain days but not every day.

To use an online server, the company is required to subscribe to a cloud provider. A cloud provider offers 3 subscriptions options: 1-day subscription, a 1-week subscription, or a 30-day subscription, each one of them has a price that can be found in a given array of prices, where prices[0] represents the price of a 1-day subscription, prices[1] represents the price of a 1-week subscription, and prices[2] represents the price of a 30-days subscription.

Write a function MinSubscription to find the minimum amount of money that company ABC needs to have in order to use a server during all the experimentation days.

Your function MinSubscription takes as parameters 3 values:

E which represents the number of experimentation days.

Days which has E integer values that represent the experimentation days Di where 1< Di <=365

Prices which has 3 values: 1-day Subscription, 1-Week Subscription, and 30-day Subscription

Output:

Your function should return the minimum cost of subscription(s) required to cover the given experimentation days

Example 1:

prices = [4, 10, 25]

Experimentation Days = [1, 3, 8, 9, 22, 23, 28, 31]

output:

28

Explanation:

30-day subscription is not enough to cover 31 days.

28 = 10 + 10 + 4 + 4

10 = 1-week subscription to cover days 3 to 9

10 = 1-week subscription to cover days 22 to 28

4 = 1-day subscription to cover day 1

4 = 1-day subscription to cover day 31

#include

using namespace std;

#include "Q3.cpp"

int main(){

int Prices[]={4 ,10 ,25};

int D=8;

int Days[]={1,3 ,8 , 9 , 22, 23, 28 , 31};

cout << MinSubscription(D, Days, Prices)<

// answer should be 28

return 0;

}

#include

#include

int MinSubscription(int D, int * Days, int * Prices){

return dp[D][D];

}

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