LeetCode解题报告（494）-- 1276. Number of Burgers with No Waste of Ingredients

Problem

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:

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

Example 3:

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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.

Constraints:

0 <= tomatoSlices, cheeseSlices <= 107

Analysis

这道题目的背景是做汉堡，有两种汉堡，Jumbo Burger要4片tomato和1片cheese，Small Burger要2片tomato和1片cheese，题目给出了tomatoSlices和cheeseSlices的数量，问两种汉堡最多能做多少个，其中要求两种材料都没有剩余，如果做不到则返回空。

这个问题一下子好像很困难，但是两个变量，去组成两种汉堡，这个听起来很小小学的方程题目啊，再仔细一看这不就是鸡兔同笼问题吗。先列方程，假设能做x个Jumbo Burger，能做y个Small Burger，那么有4x + 2y = tomatoSlices，x + y = cheeseSlices。然后我们来看怎么判断能否做到。

先消元y，得到2x = tomatoSlices - 2cheeseSlices，所以这里要保证tomatoSlices >= 2 * chessSlices，因为要求解x，所以tomatoSlices % 2 == 0才能保证x有整数解。求解完x后代入计算y，2y = 4cheeseSlices - tomatoSlices，同理要保证4 * cheesseSlices >= tomatoSlices。因为要求解y，所以tomatoSlices % 2 == 0才能保证y有整数解。

把上面的条件都判断一遍，如果都满足要求就直接代入方程计算出x和y，否则就返回空。

Solution

无。

Code

class Solution {
public:
    vector<int> numOfBurgers(int t, int c) {
        if (t % 2 == 0 && c * 2 <= t && t <= c * 4)
            return {t / 2 - c, c * 2 - t / 2};
        return {};
    }
};

Summary

这道题目实际上就是一道数学题，在解方程的过程中需要注意合法的情况，保证大于0以及保证整数解。这道题目的分享到这里，感谢你的支持！