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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
import java.util.HashMap;
import java.util.List;
import java.util.Arrays;
import java.util.ArrayList;
import java.lang.Math;

public class Main {

    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        long q = sc.nextLong();
        long h = sc.nextLong();
        long s = sc.nextLong();
        long d = sc.nextLong();
        int n = sc.nextInt();
        long min = 1000000000;
        if (n % 2 == 0) {
          long[] tmp = {q*8, h*4, s*2, d};
          for (int i = 0; i < 4; i++) {
            if (min > tmp[i]) {
              min = tmp[i];
            }
          }
          System.out.println(n/2 * min);
        } else {
          long[] tmp = {q*8, h*4, s*2, d};
          for (int i = 0; i < 4; i++) {
            if (min > tmp[i]) {
              min = tmp[i];
            }
          }
          if (q*4 >= h*2) {
            if (h*2 >= s) {
              System.out.println(n/2 * min + s);
            } else {
              System.out.println(n/2 * min + h*2);
            }
          } else {
            if (q*4 >= s) {
              System.out.println(n/2 * min + s);
            } else {
              System.out.println(n/2 * min + q*4);
            }
          }
        }
    }

}