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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
 
int main(void)
{
	int q,h,s,d,n;
	long cost;
	double i,j,k,l;
	scanf("%d%d%d%d%d\n",&q,&h,&s,&d,&n);
	cost = 9223372036854775806;
	for(i=0;i*0.25<=n;i++){
		for(j=0;j*0.5<=n;j++){
			for(k=0;k*1<=n;k++){
				for(l=0;l*2<=n;l++){
					if (i*0.25+j*0.5+k+l*2 == (double)n) {
						/* code */
						if (cost > (long)(q*i+h*j+s*k+d*l)) {
							/* code */
							cost = (long)(q*i+h*j+s*k+d*l);
						}
					}
				}
			}
		}
	}
	printf("%ld\n",cost);
	return 0;
}