擅长:python、mysql、java
<p>我仍然认为<code>old</code>本质上是一个布尔标志,所以让我们将它设为一个(称为<code>previous_1</code>),但这次,我们将根据传递这两个需求的每个<code>i</code>更改它的值,而不是只触发一次True:</p>
<pre><code>count1_1 = 0
previous_1 = False
for i in B:
if B[i] % 10 == 1 and B[i + 2] % 10 == 3:
if some_other_reqirement:
current_1 = i % 10 == 1
if current_1 and previous_1:
count1_1 += 1
previous_1 = current_1
</code></pre>