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首先,我们需要明确每种食材的体积和保质期,然后根据冰箱的容量和每天的消耗量来计算出最多可以存放的食材数量。我们可以先将食材按照保质期从长到短进行排序,然后从保质期最长的食材开始放入冰箱,直到冰箱放不下为止。在这个过程中,我们需要注意的是,每天需要消耗两份食材,所以在计算可以存放的食材数量时,我们需要考虑到这一点。 | |
以下是实现这个算法的Python代码: | |
```python | |
class Food: | |
def __init__(self, name, width, height, depth, shelf_life, quantity): | |
self.name = name | |
self.width = width | |
self.height = height | |
self.depth = depth | |
self.volume = width * height * depth | |
self.shelf_life = shelf_life | |
self.quantity = quantity | |
class Fridge: | |
def __init__(self, width, height, depth, layers): | |
self.width = width | |
self.height = height | |
self.depth = depth | |
self.volume = width * height * depth * layers | |
self.layers = layers | |
def store_food(self, foods): | |
foods.sort(key=lambda x: x.shelf_life, reverse=True) | |
total_volume = 0 | |
for food in foods: | |
while food.quantity > 0 and total_volume + food.volume <= self.volume: | |
total_volume += food.volume | |
food.quantity -= 1 | |
return total_volume | |
# Initialize fridge and foods | |
fridges = [Fridge(40, 40, 30, 2), Fridge(40, 40, 30, 2), Fridge(40, 40, 20, 1)] | |
#fridge.layers.append(Fridge(40, 40, 20, 1)) | |
foods = [ | |
Food('A', 10, 25, 5, 1, 4), | |
Food('B', 20, 25, 2, 2, 5), | |
Food('C', 20, 15, 3, 2, 6) | |
] | |
# Store foods in the fridge | |
total_volume = 0 | |
for fridge in fridges: | |
total_volume += fridge.store_food(foods) | |
print(f'The total volume of food stored in the fridge is {total_volume} cm^3.') #The total volume of food stored in the fridge is 15400 cm^3. | |
``` | |
这段代码首先定义了两个类,分别是食材类和冰箱类。然后,我们初始化了一个冰箱和三种食材,并将食材按照保质期从长到短进行排序。最后,我们调用冰箱的store_food方法,将食材放入冰箱,并打印出存放的食材总体积。 | |
每个食材的体积: | |
A: 10cm * 25cm * 5cm = 1250cm³ | |
B: 20cm * 25cm * 2cm = 1000cm³ | |
C: 20cm * 15cm * 3cm = 900cm³ | |
第一层和第二层的有效容积相同,均为:40cm * 40cm * 30cm = 48000cm³ | |
第三层有效容积为:40cm * 40cm * 20cm = 32000cm³ | |
总容积为:48000cm³ + 48000cm³ + 32000cm³ = 128000cm³ | |
由于实际需求远小于冰箱的实际容纳能力,因此可以将两天所需的全部食材A、B、C都放入冰箱中,且不会超过冰箱的任何一层的容量。 |