LeetCode-1276. Number of Burgers with No Waste of Ingredients

LeetCode-1276. Number of Burgers with No Waste of Ingredients

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​​

​​题解：​​

​​解方程，注意判断是否为非负整数。​​

class Solution {public: vector numOfBurgers(int tomatoSlices, int cheeseSlices) { if (tomatoSlices == 0 && cheeseSlices == 0) { return {0, 0}; } float rem = float(tomatoSlices) / float(2); int rems = tomatoSlices / 2; if (rem != rems) { return {}; } int jumbo = rems - cheeseSlices; int small = cheeseSlices - jumbo; if (jumbo >= 0 && small >= 0) { return {jumbo, small}; } return {}; }};

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。