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 <iomanip>
#include <utility>
#include <cmath>
#include <random>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <algorithm>
using namespace std;

#define rep(i,n) for(int i = 0; i<n; ++i)
#define REP(i,n) for(int i = 1; i<=n; ++i)
#define all(x) begin(x),end(x)
#define show(obj) for(auto x:obj)cout<<x<<' ';cout<<endl;
#define line "----------"
typedef long long ll;
typedef pair<int,int> ip;
typedef pair<ll,ll> lp;
const int inf = 1001001000;
const ll INF = 1LL<<60;
const int MOD = (int)1e9 + 7;

int main(){
	double Q,H,S,D;
	int N;
	cin>>Q>>H>>S>>D>>N;
	Q *= 4; H *= 2; D /= 2;
	ll ans = 0LL;
	if(min({Q,H,D,S}) == D){
		ans += N/2 * 2*D;
		N  %= 2;
	}
	ans += min({Q,H,S})*N;
	cout<<ans<<endl;
	return 0;
}