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;

int main(){
   long long Q,H,S,D,N; cin>>Q>>H>>S>>D>>N;
  double ans=0;

    double q=4*Q,h=2*H,s=S,d=D*0.5;//1Lあたり
  pair<double,double> dri[4];//コスパ、リットル
  dri[0]=make_pair(q,0.25);
  dri[1]=make_pair(h,0.5);
  dri[2]=make_pair(s,1.0);
  dri[3]=make_pair(d,2.0); 
	sort(dri,dri+4);
  
  ans+=((long long)(N/dri[0].second))*dri[0].second*dri[0].first;
  N-=((long long)(N/dri[0].second))*dri[0].second;
  ans+=((long long)(N/dri[1].second))*dri[1].second*dri[1].first;
  N-=((long long)(N/dri[1].second))*dri[1].second;
  ans+=((long long)(N/dri[2].second))*dri[2].second*dri[2].first;
  N-=((long long)(N/dri[2].second))*dri[2].second;
  ans+=((long long)(N/dri[3].second))*dri[3].second*dri[3].first;
    printf("%lld",(long long)ans);
}