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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
bool comp(pair<double, long long> a, pair<double, long long> b) {
    return 1.0 * a.second / a.first < 1.0 * b.second / b.first;
}
 
int main() {
    ios::sync_with_stdio(false);
    pair<double, long long> cost[5];
    cost[1].first = 0.25;
    cost[2].first = 0.5;
    cost[3].first = 1.0;
    cost[4].first = 2.0;
    for (int i = 1; i <= 4; i++) {
        cin >> cost[i].second;
    }
    double n;
    cin >> n;
    sort(cost + 1, cost + 5, comp);
    long long res = 0;
    for (int i = 1; i <= 4; i++) {
        res += floor(1.0 * n / cost[i].first) * cost[i].second;
        n -= floor(1.0 * n / cost[i].first) * cost[i].first;
    }
    cout << res << endl;
    return 0;
}