funcode.cpp

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;

void optimum_sequence_jobs(vector<int>& V, double P)
{
	int j = 1, N = V.size() - 1;
	double result = 0;

	// Create a min-heap (priority queue)
	priority_queue<int, vector<int>, greater<int> > Queue;

	// Add all goods to the Queue
	for (int i = 1; i <= N; i++)
		Queue.push(V[i]);

	// Pop Goods from Queue as long as it is not empty
	while (!Queue.empty()) {

		// Print the good
		cout << Queue.top() << " ";

		// Add the Queue to the vector
		// so that total volume can be calculated
		V[j++] = Queue.top();
		Queue.pop();
	}

	// Calculating volume of goods left when all
	// are produced. Move from right to left of
	// sequence multiplying each volume by
	// increasing powers of 1 - P starting from 0
	for (int i = N; i >= 1; i--)
		result += pow((1 - P), N - i) * V[i];

	// Print result
	cout << endl << result << endl;
}

// Driver code
int main()
{
	// For implementation simplicity days are numbered
	// from 1 to N. Hence 1 based indexing is used
	vector<int> V{ -1, 3, 5, 4, 1, 2, 7, 6, 8, 9, 10 };

	// 10% loss per day
	double P = 0.10;

	optimum_sequence_jobs(V, P);

	return 0;
}