蟒蛇函數參數

Question

我試圖寫一個簡單的計算器，採用四面體的給定數量（用戶輸入他們的邊緣）和水給定的量（升），並計算如何四面體的許多可充滿水。我已經定義了計算單個四面體體積的函數，但是當我嘗試寫第二個函數（用水填充它們）時，我失敗了。

water = int(raw_input("water:")) #Enter the amount of water in litres 
n = raw_input("Enter the number of tetrahedrons:") 
tetrahedrons = [] 
for i in range(int(n)): 
    edge = raw_input("Enter the edge of a tetrahedron:") 
    tetrahedrons.append(int(edge)) 
    tetrahedrons.sort() 

#full_tetrahedrons = 0 # counter for the full tetrahedrons 

def fill_tetrahedron(num): 
    volume = (int(num) ** 3) // (6.0 * (2 ** 0.5)) 
    volume_in_l = volume/1000 

def tetrahedron_filled(tetrahedrons, water): 
    if water >= 0: 
     for number in tetrahedrons: 
      value = fill_tetrahedron(number) 
      water = water - value 
      full_tetrahedrons = 0 
      full_tetrahedrons += 1 
     return full_tetrahedrons 

print tetrahedron_filled(tetrahedrons, water) 

編輯：

OK，我已經解決了我的錯誤與爭論，但現在我進入什麼價值，我總是得到1作爲一個答案。

Answer 1

你只有通過一個函數參數tetrahedron_filled()。

這個函數有兩個參數 - tetrahedrons和water。