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(void){
  int m[4],n,i,j,pl[4],a=0,x;
  for(i=0;i<4;i++){
    scanf("%d",&m[i]);
  }
  scanf("%d",&n);
  pl[0]=m[0]*4;
  pl[1]=m[1]*2;
  pl[2]=m[2];
  pl[3]=m[3]/2;

  for(i=0;i<4;i++){
    for(j=0;j<4;j++){
      if(pl[i]>pl[j]){
      a = 1;  break;  // aは最小かそうでないかのスイッチ
      }
    }
    if(a != 1){
      a = i; break;  // aに最もコスパの良いを入れた(もう使わないので)
    }else{
      a = 0;
    }
    }
    //for(i=0;i<4;i++) printf("%da",pl[i]);
    if(a != 3){
      printf("%d",n*pl[a]);
    }else{
      if(n%2 == 0){
        printf("%d",n*pl[a]);
      }else{
        if(pl[0]<=pl[1] && pl[0] <= pl[2]){
          printf("%d",(n/2)*m[3]+pl[0]);
        }else if(pl[1]<=pl[0] && pl[1] <= pl[2]){
          printf("%d",(n/2)*m[3]+pl[1]);
        }else{
            printf("%d",(n/2)*m[3]+pl[2]);
        }
        }
      }
      return 0;
    }