擅长:python、mysql、java
<p>怎么样:</p>
<pre><code>n = 1000
for exp, name in zip(range(9, -13, -3), ('GMk1munp')):
if exp == 0:
continue
if isinstance(n, int):
if n % 10**exp == 0:
n = '{0:d}{1}'.format(n / 10**exp, name)
break
elif isinstance(n, basestring):
if n[-exp:] == '0' * exp:
n = '{0}{1}'.format(n[:-exp], name)
break
elif n[-1] == name:
n = n[:-1] + '0' * exp
break
</code></pre>
<p>通过向后运行指数,可以确保找到正确的匹配项。你知道吗</p>