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: Wrong Answer)

#include<stdio.h>

int main(){
  long long int sum=0, Q, H, S, D, N, min=1000000000000;
  int index, j, i;
  scanf("%lld%lld%lld%lld", &Q, &H, &S, &D);
  long long int a[4] = {Q*8, 4*H, S*2, D};
  scanf("%lld", &N);
  
  for(i=0;i<4;i++){
    if(a[i]<min){
      min = a[i];
      index = i;}
  }
  min=1000000000;
  if(index==3){
    sum += (N/2)*D;
    N = N%2;
    for(j=0;j<3;j++){
      if(a[j]<min)
        min = a[j];
    } 
    sum += min * N;
  }
  else
    sum += a[index] * N / 2;
  
  printf("%lld", sum);
  return 0;  
}