Problem:
Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	long long int v[4];
	int N;
	long long int ans;
	
	//input
	cin >> v[0] >> v[1] >> v[2] >> v[3];
	cin >> N;
	
	//cost = cost / 1L
	long double cost[4] = {0.25, 0.50, 1.00, 2.00};
	long double mini = 400000000;
	int minS = 0;
	for(int i = 0; i < 4; i++){
		cost[i] = v[i] / cost[i];
		//cout << cost[i] << endl;
		if(mini > cost[i]){
			minS = i;
			mini = cost[i];
		}
	}
	
	
	if(minS == 3){
		if(N%2==0){
			ans = N * mini;
		}else{
			long double miniSec = cost[2];
			if(miniSec >= cost[1]){ miniSec = cost[1]; }
			if(miniSec >= cost[0]){ miniSec = cost[0]; }
			ans = (2*(int)(N / 2)) * cost[3] + miniSec;
		}
		
	}else{
		ans = N * mini;
	}
	
	cout << ans << endl;
	return 0;
}