Question

我今天创建了这行代码，用于确定数字是否均匀划分

if (numerator / denominator * denominator) == numerator:
    print "Divides evenly!"
else:
    print "Doesn't divide evenly."

然而，当我除以0时遇到了一个问题，因为我仍然希望它告诉我，即使分母是0，通过查看分子并查看它是偶数还是奇数，它是否会分裂。我想出了这个：

if denominator != 0 and (numerator / denominator * denominator) == 
numerator:
    print "Divides evenly!"
elif denominator == 0 and numerator % 2 == 0:
    print "Divides evenly!"
else:
    print "Doesn't divide evenly."

有没有办法缩短这个？另外，有没有办法不必添加新的'elif'语句？

Answer 1

简而言之，如果denominator为0，您希望它为2：

if denominator == 0:
    denominator = 2

if numerator % denominator == 0:
    print "Divides evenly!"
else:
    print "Doesn't divide evenly."

甚至更短：

if numerator % (denominator or 2) == 0:
    print "Divides evenly!"
else:
    print "Doesn't divide evenly."

Answer 2

def fun(n,d):
    return (d==0 and n%2==0) or (d!=0 and n%d==0)

有什么方法可以缩短这段代码吗？

2 个答案: