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: Wrong Answer)

#include <stdio.h>
static double q, h, s, d, n;
double f(double cost, double size) {
    return cost / size;
}
int main(void) {
    long sum = 0;
    scanf("%lf%lf%lf%lf%lf", &q, &h, &s, &d, &n);
    if (f(s, 1) <= f(d, 2))
        d = 2 * (long)s;
    if (f(h, 0.5) <= f(s, 1))
        s = 2 * (long)h;
    if (f(h, 0.5) <= f(d, 2))
        d = 4 * (long)h;
    if (f(q, 0.25) <= f(h, 0.5))
        h = 2 * (long)q;
    if (f(q, 0.25) <= f(s, 1))
        s = 4 * (long)q;
    if (f(q, 0.25) <= f(d, 2))
        d = 8 * (long)q;
    while (n >= 2) {
        sum += d;
        n -= 2;
    }
    while (n >= 1) {
        sum += s;
        n -= 1;
    }
    while (n >= 0.5) {
        sum += h;
        n -= 0.5;
    }
    while (n >= 0.25) {
        sum += q;
        n -= 0.25;
    }
    printf("%ld\n", sum);
    return 0;
}