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>
int main(void) {
	long long int q, h, s, d, n;
	int min;
	long long int p;
	scanf("%llu%llu%llu%llu%llu", &q, &h, &s, &d, &n);

	if (q * 8 < h * 4) {
		if (q * 8 < s * 2)
		{
			if (q * 8 < d)	min = 0;
			else	min = 30;
		}
		else
		{
			if (s * 2 < d)	min = 2;
			else	min = 32;
		}
	}
	else
	{
		if (h * 4 < s * 2)
		{
			if (h * 4 < d)	min = 1;
			else	min = 31;
		}
		else
		{
			if (s * 2 < d)	min = 2;
			else	min = 32;
		}
	}

	switch (min)
	{
	case 0:
		p = q*n * 4;
		break;
	case 1:
		p = h*n * 2;
		break;
	case 2:
		p = s*n;
		break;
	case 30:
		p = d*(n / 2) + q*(n % 2) * 4;
		break;
	case 31:
		p = d*(n / 2) + h*(n % 2) * 2;
		break;
	case 32:
		p = d*(n / 2) + s*(n % 2);
		break;
	default:
		break;
	}

	printf("%llu", p);

	return 0;
}