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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-burgers-with-no-waste-of-ingredients
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外国版鸡兔同笼？

第一种 4个A，一个B

第二种 2个A，一个B

假设全是第二种就要 2B个A

A-2B就是多出来的，（A-2B）/2就是第一种个数

A-(A-2B)/2就是第二种个数

如果减法出现了负数说明无解

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;
        
    }
};