1276-不浪费原料的汉堡制作方案(Number of Burgers with No Waste of Ingredients)

原文链接: https://leetcode-cn.com/problems/number-of-burgers-with-no-waste-of-ingredients

英文原文

Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:

Jumbo Burger: 4 tomato slices and 1 cheese slice.
Small Burger: 2 Tomato slices and 1 cheese slice.

Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].

Example 1:

Input: tomatoSlices = 16, cheeseSlices = 7
Output: [1,6]
Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese. There will be no remaining ingredients.

Example 2:

Input: tomatoSlices = 17, cheeseSlices = 4
Output: []
Explantion: There will be no way to use all ingredients to make small and jumbo burgers.

Example 3:

Input: tomatoSlices = 4, cheeseSlices = 17
Output: []
Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.

Example 4:

Input: tomatoSlices = 0, cheeseSlices = 0
Output: [0,0]

Example 5:

Input: tomatoSlices = 2, cheeseSlices = 1
Output: [0,1]

Constraints:

0 <= tomatoSlices <= 10^7
0 <= cheeseSlices <= 10^7

中文题目

圣诞活动预热开始啦，汉堡店推出了全新的汉堡套餐。为了避免浪费原料，请你帮他们制定合适的制作计划。

给你两个整数 tomatoSlices 和 cheeseSlices，分别表示番茄片和奶酪片的数目。不同汉堡的原料搭配如下：

巨无霸汉堡：4 片番茄和 1 片奶酪
小皇堡：2 片番茄和 1 片奶酪

请你以 [total_jumbo, total_small]（[巨无霸汉堡总数，小皇堡总数]）的格式返回恰当的制作方案，使得剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量都是 0。

如果无法使剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量为 0，就请返回 []。

示例 1：

输入：tomatoSlices = 16, cheeseSlices = 7
输出：[1,6]
解释：制作 1 个巨无霸汉堡和 6 个小皇堡需要 4*1 + 2*6 = 16 片番茄和 1 + 6 = 7 片奶酪。不会剩下原料。

示例 2：

输入：tomatoSlices = 17, cheeseSlices = 4
输出：[]
解释：只制作小皇堡和巨无霸汉堡无法用光全部原料。

示例 3：

输入：tomatoSlices = 4, cheeseSlices = 17
输出：[]
解释：制作 1 个巨无霸汉堡会剩下 16 片奶酪，制作 2 个小皇堡会剩下 15 片奶酪。

示例 4：

输入：tomatoSlices = 0, cheeseSlices = 0
输出：[0,0]

示例 5：

输入：tomatoSlices = 2, cheeseSlices = 1
输出：[0,1]

提示：

0 <= tomatoSlices <= 10^7
0 <= cheeseSlices <= 10^7

通过代码

高赞题解

这一题实在对不起中等这个标签，就是简单的二元一次方程求非负整数解。
$$
\begin{cases}
4x + 2y = tomato \
x + y = cheese
\end{cases}
==>
\begin{cases}
x = \frac{tomato - 2 * cheese}{2} \
y = cheese - x
\end{cases}
$$
如果解非整数，或者出现负数，那么则无解

class Solution {
public:
    vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
        vector<int> re;
        if (0 == (tomatoSlices - 2 * cheeseSlices) % 2) {
            int j = (tomatoSlices - 2 * cheeseSlices) / 2;
            int s = cheeseSlices- j;
            if (j >= 0 && s >= 0) {
                re.push_back(j);
                re.push_back(s);
            }
        }
        return re;
    }
};