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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll v[4];
ll n;
int main() {
  cin >> v[0] >> v[1] >> v[2] >> v[3];
  cin >> n;
  n *= 4;
  vector<vector<ll>> pp;
  pp.push_back({v[0] * 8, 1, v[0]});
  pp.push_back({v[1] * 4, 2, v[1]});
  pp.push_back({v[2] * 2, 4, v[2]});
  pp.push_back({v[3] * 1, 8, v[3]});

  sort(pp.begin(), pp.end());
  ll ret = 0;
  for (int i = 0; i < 1; ++i) {
    ret += n / (pp[i][1]) * pp[i][2];
    n %= pp[i][1];
  }
  ll r2 = INT_MAX;
  for (int i = 0; i <= 8; ++i)
    for (int j = 0; j <= 4; j++)
      for (int k = 0; k <= 2; ++k)
        for (int m = 0; m <= 1; ++m) {
          if (i + j * 2 + k * 4 + m * 8 == n) {
            if (i * v[0] + j * v[1] + k * v[2] + m * v[3] < r2) {
              r2 = i * v[0] + j * v[1] + k * v[2] + m * v[3];
            }
          }
        }
  cout << ret + r2 << endl;
}