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 v[4];

int main() {
  int a, b, c, d, ab, bb, cb, db, u, p, max, i, x, y;
  float xl, yl, n;
  unsigned long long s;
  scanf( "%d%d%d%d%f", &a, &b, &c , &d, &n );
  cb = b * 2;
  bb = c * 4;
  ab = d * 8;
  v[0] = ab;
  v[1] = bb;
  v[2] = cb;
  v[3] = db;
  for ( u = 3; u > 0; u-- ) { 
    max = v[0];
    p = 0;                        
    for ( i = 1; i <= u; i++ )
      if ( v[i] > max ) {        
        max = v[i];
        p = i;
      }
    v[p] = v[u];
    v[u] = max;
  }
  x = v[3];
  y = v[2];
  if( x == ab ) {
    x = a;
    xl = 0.25;
  }
  else if( x == bb ) {
    x = b;
    xl = 0.50;
  }
  else if( x == cb ) {
    x = c;
    xl = 1;
  }
  else if( x == db ) {
    x = d;
    xl = 2;
  }
  if( y == ab && y != v[3] ) {
    y = a;
    yl = 0.25;
  }
  else if( x == bb && y != v[3] ) {
    y = b;
    yl = 0.50;
  }
  else if( x == cb && y != v[3] ) {
    y = c;
    yl = 1;
  }
  else if( x == db && y != v[3] ) {
    y = d;
    yl = 2;
  }
  s = (n / xl) * x;
  s += (n / yl) * y;
  printf( "%ulld", s ); 
  return 0;
}