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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 41 + 26 = 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
鸡兔同笼问题
Jumbo Burger: 4 tomato slices and 1 cheese slice.
Small Burger: 2 Tomato slices and 1 cheese slice.
当tomatoSlices为奇数,或者tomatoSlices大于4cheeseSlices,或者tomatoSlices小于2cheeseSlices时,无解;
否则,令small burger=cheeseSlices,即所有的cheese都用来做small burger,这时会消耗tomatoSlices=2* cheeseSlices,剩余tomatoSlices-=2* cheese(2* small也一样)。
对于剩余的tomatoSlices,只能全部用来做 jumbo burger。所以对于每2片剩余的 tomatoSlices,都另外需要small burger的2片tomatoSlices 和 1片 cheeseSlices原料来制作jumbo burger。
class Solution {
public:
vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
vector<int>res;
if(tomatoSlices%2!=0||tomatoSlices>4*cheeseSlices||tomatoSlices<2*cheeseSlices) return res;
int small=cheeseSlices,jumbo=0;
tomatoSlices-=2*small;
if(tomatoSlices!=0) small-=tomatoSlices/2,jumbo+=tomatoSlices/2;
res.push_back(jumbo); res.push_back(small);
return res;
}
};