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>
#include <vector>
#include <string>
#include <unordered_set>
#include <unordered_map>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <cmath>

using namespace std;

int main()
{
	long long Q = 0, H = 0, S = 0, D = 0;
	long long N = 0;

	cin >> Q >> H >> S >> D;
	cin >> N;

	// 1L の最小値
	long long p1 = min(Q * 4, H * 2);
	p1 = min(p1, S);

	// 2L の最小値
	long long p2 = min(Q * 8, H * 4);
	p2 = min(p2, S * 2);
	p2 = min(p2, D);
		
	long long sum = 0;
	if (N % 2) {
		sum += p1;
	}

	long long num2 = N / 2;
	sum += num2 * p2;
		
	cout << sum << endl;

	return 0;
}