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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

struct value{
    long long v;
    double vper,size;
};

bool comp(const value& e1, const value& e2){
    if(e1.vper < e2.vper){
        return true;
    }else{
        return false;
    }
}
int main(void){
    long long Q, H, S, D, N;
    cin >> Q >> H >> S >> D;
    cin >> N;
    value v[4];
    
    v[0].v = Q;
    v[0].vper = Q;
    v[0].size = 0.25;
    
    v[1].v = H;
    v[1].vper = H/2;
    v[1].size = 0.50;

    v[2].v = S;
    v[2].vper = S/4;
    v[2].size = 1.00;

    v[3].v = D;
    v[3].vper = D/8;
    v[3].size = 2.00;

    sort(v, v+4, comp);

    long long tmp =  N / v[0].size;

    long long sum = 0;

    sum += tmp * v[0].v;

    N -= tmp*v[0].size;
    
    if(v[0].size == 2.00){
        tmp = N/v[1].size;
        sum += tmp * v[1].v;
    }

    cout << sum << endl;

    return 0;
}