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 <fstream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <functional>
#define inf 0X3f3f3f3f
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
const int MAX=20000+10;
const double eps=1e-6;


int main(){
    #ifdef ONLINE_JUDGE
    #else
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    #endif
    double s[10],t,ans=0;
    int n;
    cin>>s[0]>>s[1]>>s[2]>>s[3]>>n;
    map<double,double>val,cost;
   	val[s[0]]=0.25;
   	cost[s[0]]=s[0];

   	t=s[1];
   	s[1]/=2;
   	val[s[1]]=0.5;
   	cost[s[1]]=t;

   	t=s[2];
   	s[2]/=4;
   	val[s[2]]=1;
   	cost[s[2]]=t;

   	t=s[3];
   	s[3]/=8;
   	val[s[3]]=2;
   	cost[s[3]]=t;

   	sort(s,s+4);
   	for(int i=0;i<=3;i++){
   		int temp=n/val[s[i]];
   		//cout<<n<<" "<<val[s[i]]<<endl;
   		n=n-temp*val[s[i]];
   		ans+=temp*cost[s[i]];
   		if(n<=0)	break;
   	}  	
   	printf("%.0f\n",ans);
    return 0;
}