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<int> 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 {};
    }
};