【CodeForces 1207A

【CodeForces 1207A

【CodeForces 1207A --- There Are Two Types Of Burgers】

There are two types of burgers in your restaurant — hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.

You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.

You have to answer t independent queries.

Input

The first line contains one integer t (1≤t≤100) – the number of queries.

The first line of each query contains three integers b, p and f (1≤b, p, f≤100) — the number of buns, beef patties and chicken cutlets in your restaurant.

The second line of each query contains two integers h and c (1≤h, c≤100) — the hamburger and chicken burger prices in your restaurant.

Output

For each query print one integer — the maximum profit you can achieve.

Sample Input

3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100

Sample Output

40
34
0

Note

In first query you have to sell two hamburgers and three chicken burgers. Your income is 2⋅5+3⋅10=40.

In second query you have to ell one hamburgers and two chicken burgers. Your income is 1⋅10+2⋅12=34.

In third query you can not create any type of burgers because because you have only one bun. So your income is zero.

解题思路

水题，直接看代码就行，贪心找最大利益。

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;int main()
{std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t, b, p, f, r = 0, c, h;cin >> t;while (t--) {cin >> b >> p >> f >> h >> c;if (h > c) {if ((b/2)>=p) {r = p*h;b = (b/2)-p;if (b>=f)r=r+f*c;elser=r+b*c;}else r=(b/2)*h;}else {if ((b/2)>=f) {r=f*c;b=(b/2)-f;if (b>=p)r=r+p*h;elser=r+b*h;}else r=(b/2)*c;}cout << r << endl;}return 0;
}