【LeetCode第 165 場周賽】不浪費原料的漢堡製作方案
- 2019 年 12 月 3 日
- 筆記
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5276. 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
來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/number-of-burgers-with-no-waste-of-ingredients 著作權歸領扣網路所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。
外國版雞兔同籠？
第一種 4個A，一個B
第二種 2個A，一個B
假設全是第二種就要 2B個A
A-2B就是多出來的，（A-2B）/2就是第一種個數
A-(A-2B)/2就是第二種個數
如果減法出現了負數說明無解
class Solution { public: vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) { vector<int> v; int a=(tomatoSlices-cheeseSlices*2); if(a%2==0&&a>=0&&cheeseSlices>=a/2){ a=a/2; v.push_back(a); v.push_back(cheeseSlices-a); }else { return v; }return v; } };
