在python中获取动物数量的函数[关闭]

Question

你好，这些函数根据这两个动物的头和腿的输入返回猪和鸡的数量。我对python真的很陌生。我无法理解它是如何工作的。你能详细解释一下吗？

def solve(numLegs, numHeads):
    for numChick in range(0, numHeads + 1):
        numPigs = numHeads - numChicks
        totLegs = 4 * numPigs + 2* numChicks
        if totLegs == numLegs:
           return [numPigs, numChicks]
    return[None, None]

def barnYard():
    heads = int(raw_input('Enter number of heads:'))
    legs = int(raw_input('Enter number of legs:'))
    pigs, chickens = solve(legs, heads)
    if pigs = None:
       print 'there is no solution'
    else:
        print 'number of pigs:' , pigs
        pirnt 'number of chickes:', chickens

所以当我运行 barnYard 函数时，它会询问头数，所以我放了 20 个头，然后放了 56 条腿。它打印猪的数量 8 和鸡的数量 12。但我真的不明白它在这个阶段是如何达到的。特别是solve() 函数。帮助将不胜感激。

Answer 1

基本上：

def solve(numLegs, numHeads):
    for numChick in range(0, numHeads + 1): #for every number in the range 0 - the number of heads + 1, numChick = that number
        numPigs = numHeads - numChicks #the number of Pigs equals the number of heads entered minus the current number
        totLegs = 4 * numPigs + 2* numChicks #the amount of legs is 4(amount of legs) * number of heads + 2(chicken legs) * the current number
        if totLegs == numLegs: if the pigs legs + the chicken legs = the total number of legs, 
           return [numPigs, numChicks] #return the number of pigs and chickens
    return[None, None] #else, return none, triggering "no solution"

def barnYard():
    heads = int(raw_input('Enter number of heads:')) #
    legs = int(raw_input('Enter number of legs:'))
    pigs, chickens = solve(legs, heads)
    if pigs = None:
       print 'there is no solution'
    else:
        print 'number of pigs:' , pigs
        pirnt 'number of chickes:', chickens

所以：

基本上，它会一遍又一遍地运行该函数，直到计算出的边数等于输入的边数。如果它永远不等于，它只会返回 [none, none]。 （return [numPigs, numChicks] 打破 for 循环）

编辑： 我尝试在第 2 行中取出 +1，它仍然可以正常工作。