擅长:python、mysql、java
<p>根据辛格的回答,我提出了以下解决方案:</p>
<pre><code>import math
# Cumulative distribution function
def CDF(x):
return (1.0 + math.erf(x/math.sqrt(2.0)))/2.0
# Approximation of binomial cdf with continuity correction for large n
# n: trials, p: success prob, m: starting successes
def BCDF(p, n, m):
return 1-CDF((m-0.5-(n*p))/math.sqrt(n*p*(1-p)))
</code></pre>