擅长:python、mysql、java
<p>这个实现更习惯于python,更容易理解。你知道吗</p>
<pre><code>def approxPI(n):
cal1 = [1.0/float(i) for i in range(1, n, 4)]
cal2 = [1.0/float(i) for i in range(3, n, 4)]
pi = (sum(cal1)-sum(cal2)) * 4
return pi
print(approxPI(400000))
</code></pre>
<p>这是相当快的,计算400000系列在0.05秒的台式电脑上</p>