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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<bitset>
#include<vector>
#include<deque>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<iomanip>
#include <functional>
using namespace std;
typedef long long ll;
const int mod = 1000000007;
const double EPS = 1e-10;
//cout << fixed << std::setprecision(9)
//memset(a, 0, sizeof(a));
//--------------------------

int main()
{
	ll q, h, s, d, n;

	cin >> q >> h >> s >> d >> n;

	ll min1 = n * s;
	ll min2 = n * s;

	for(ll i25=0;i25<9;i25++)
		for (ll i50 = 0;i50<9;i50++)
			for (ll i1 = 0;i1<9;i1++)
				for (ll i2 = 0;i2 < 9;i2++) {
					ll temp = i25 / 4 + i50 / 2 + i1 + 2 * i2;
					ll sum = q * i25 + h * i50 + s * i1 + d * i2;

					if (temp == 1) {
						if (min1 > sum) min1 = sum;
					}

					if (temp == 2) {
						if (min2 > sum) min2 = sum;
					}
				}

	if (n % 2) {
		cout << (n / 2) * min2 + min1 << endl;
	}
	else {
		cout << (n / 2) * min2 << endl;
	}

	return 0;
}