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: Accepted)

/* gcc acG019A.c -o acG019A */
/*
あなたは行きつけの店、インフバリューにアイスティーを買いに来ています。
この店では、様々なサイズのボトル入りアイスティーが様々な価格で売られています。具体的な価格は、0.25 リットルサイズが一つ Q 円、0.5 リットルサイズが一つ H 円、1 リットルサイズが一つ S 円、2 リットルサイズが一つ D 円です。各サイズの在庫は無限にあります。
あなたはちょうど N リットルのアイスティーを買いたいです。これに必要な金額は何円でしょうか？

制約
1≤Q,H,S,D≤108
1≤N≤109
入力値はすべて整数である。
入力は以下の形式で標準入力から与えられる。
Q H S D
N
*/


#include <stdio.h>
//#include <string.h>
//#include <stdlib.h>
//#include <math.h>


int main()
{
int a[4];// 2ℓあたりの値段
/*
int Q, H, S, D;
int N;
int i;
int min1, min2;
int ans;
*/
unsigned long Q, H, S, D;
unsigned long N;
int i;
unsigned long min1, min2;
unsigned long long ans;



  scanf("%d%d%d%d", &Q, &H, &S, &D);
scanf("%d", &N);
a[0] = 8*Q;
a[1] = 4*H;
a[2] = 2*S;
a[3] = D;

min1 = a[0]/2;
for(i = 1; i <= 2; i++){
  if(min1 > a[i]/2){
    min1 = a[i]/2;
  }
}

min2 = a[0];
for(i = 1; i <= 3; i++){
  if(min2 > a[i]){
    min2 = a[i];
  }
}

ans = min2 * (N / 2) + min1 * (N % 2);
//printf("min1 = %d, min2 = %d, ans =%d", min1, min2, ans);
printf("%llu", ans);
return 0;
}