擅长:python、mysql、java
<p>以下是另一种解决方案:</p>
<pre><code>def boyleslaw(p=None, V=None, k=None):
"""Given two of the values p, V or k this method
returns the third according to Boyle's Law."""
assert (p is None) + (V is None) + (k is None) == 1, \
"exactly 2 of p, V, k must be given"
if p is None:
return 1.0 * k / V
elif V is None:
return 1.0 * k / p
elif k is None:
return 1.0 * p * V
</code></pre>
<p>假设这三个量的第一个函数。然后返回第三个数量。所以你不必明确告诉这个函数要解哪个量,因为从给定的量来看,这是显而易见的。在</p>
^{pr2}$