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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll q,h,s,d,n,m;
struct node {
	ll v,id;
} e[5];
bool cmp(node a,node b) {
	return a.v<b.v;
}
int main() {
	scanf("%lld %lld %lld %lld %lld",&q,&h,&s,&d,&n);
	if (n==1)
		printf("%lld\n",min(s,min(h*2,q*4)));
	else {
		e[1].id=1; e[1].v=q*8;
		e[2].id=2; e[2].v=h*4;
		e[3].id=3; e[3].v=s*2;
		e[4].id=4; e[4].v=d;
		sort(e+1,e+5,cmp);
		m=0;
		for (int i=1;i<=4&&n>0;i++) {
			int t=e[i].id;
			if (t==4) {
				m+=(n/2)*d;
				n%=2;
				continue;
			}
			if (t==3)
				m+=n*s;
			else if (t==2)
				m+=n*2*h;
			else m+=n*4*q;
			break;
		}
		printf("%lld\n",m);
	}
	return 0;
}