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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

int main(){

long Q,H,S,D;
long N,min=100000000000000000;
long iceTea[4];
scanf("%ld",&Q);
scanf("%ld",&H);
scanf("%ld",&S);
scanf("%ld",&D);
scanf("%ld",&N);

iceTea[0] = (2*Q+H)*N;
iceTea[1] = 4*Q*N;
iceTea[2] = 2*H*N;
iceTea[3] = S*N;

for (int i = 0; i < 4; i++)
{
  if (min>iceTea[i]){min = iceTea[i];}
}

if (N==1){
  printf("%ld",min);
}else if (N%2==0){
  min > D*N/2 ? printf("%ld",D*N/2) : printf("%ld",min); 
}else{
  min > D*(N-1)/2+min/N ? printf("%ld",D*(N-1)/2+min/N) : printf("%ld",min); 
}

return 0;

}