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)



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

#define ll long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12




int main(void)
{
	//変数の宣言
	ll int n,q,h,s,d;
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	ll int ans=0;
	int count=0;
	int temp,temp1,temp2;
	int max,min;
	int len;
	int sum=0;
	
	
	//データの読み込み
	scanf("%lld %lld %lld %lld %lld",&q,&h,&s,&d,&n);
//	scanf_s("%lld %lld %lld %lld %lld",&q,&h,&s,&d,&n);



	//	printf("nは%dです\n", n);
	//	printf("データの読み込み終了\n");
	//実際の処理
	
	s=min(q*4,min(h*2,s));

	if(d<2*s){
		ans=n/2*d+n%2*s;
	}else{
		ans=n*s;
	}

	//	printf("計算部分終了\n");
	
	//出力
	
	printf("%lld",ans);
	
	//	printf("結果の出力終了\n");
	
	return 0;
}