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 Ruby (2.3.3) (judge verdict: Accepted)

lines = $stdin.read
array = lines.split("\n")

Q,H,S,D = array[0].split(" ").map(&:to_i)
N = array[1].to_i

litter = {}
litter[:q] = 0.25
litter[:h] = 0.5
litter[:s] = 1
litter[:d] = 2

#
# nリットル
#
def most_cheap_with_four(n)
  prices = {}
  times  = {}

  prices[:q] = (n/0.25) * Q if n/0.25 != 0
  prices[:h] = (n/0.5)  * H if n/0.5  != 0
  prices[:s] = (n/1)    * S if n/1    != 0
  prices[:d] = (n/2)    * D if n/2    != 0

  times[:q]  = n/0.25       if n/0.25 != 0
  times[:h]  = n/0.5        if n/0.5  != 0
  times[:s]  = n/1          if n/1    != 0
  times[:d]  = n/2          if n/2    != 0

  # puts "#{(n/0.25) * Q} yen"
  # puts "#{(n/0.5)  * H} yen"
  # puts "#{(n/1)    * S} yen"
  # puts "#{(n/2)    * D} yen"

  m = prices.sort_by { |key, value| value }.first
  # [:kind, :sum, :times]
  { kind: m[0], sum: m[1].floor, times: times[m[0]].floor }
end

n = N
summary = 0

while n != 0
  ret = most_cheap_with_four(n)
  n = n - litter[ret[:kind]] * ret[:times]
  #puts ret.to_s
  summary = summary + ret[:sum]
  #puts summary.floor
end

puts summary.floor