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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

typedef long long ll;

int main() {

	ll Q, H, S, D, N;

	scanf("%lld %lld %lld %lld", &Q, &H, &S, &D);
	scanf("%lld", &N);

	ll price[3] = { 4ll * Q ,2ll * H ,S };

	int index;

	int i;

	ll mi = 1000000000ll;

	for (i = 0; i < 3; i++) {
		if (price[i] < mi) {
			mi = price[i];
			index = i;
		}
	}

	ll ans;

	if (mi * 2ll > D) {
		if ((N % 2ll) == 0ll) { ans = N*D / 2ll; }
		else { ans = (N / 2ll)*D + price[index]; }
	}
	else {
		ans = N*mi;
	}

	printf("%lld\n",ans);

	return 0;
}