<p>如果你还没有改正,这些都是需要改正的错误</p>
<pre class="lang-py prettyprint-override"><code>kara = 10**((n2+1)//2)*c*number1 + d*number1 #in base case 2
</code></pre>
<pre class="lang-py prettyprint-override"><code>kara = 10**((n1+1)//2)*a*number2 + b*number2 #in base case 3. your code has n2+1
</code></pre>
<p>传统的Karatsuba有3个递归。但我明白你为什么要进行4次递归。但不能说哪个更快</p>
<p>上面注释中给出的示例的工作代码</p>
<pre class="lang-py prettyprint-override"><code>def karatsuba(number1, number2):
n1 = len(str(number1)) # number of digits in the first number
n2 = len(str(number2)) # number of digits in the second number
if n1 == 1 and n2 == 1: # base case number 1 - both numbers are single-digit
kara = number1*number2
return kara
elif n1 == 1: # base case number 2 - only one number is single-digit
c = int(str(number2)[:(n2//2)])
d = int(str(number2)[(n2//2):])
kara = 10**((n2+1)//2)*c*number1 + d*number1 #a mistake here
return kara
elif n2 == 1: # base case number 3 - only one number is single digit
a = int(str(number1)[:(n1//2)])
b = int(str(number1)[(n1//2):])
kara = 10**((n1+1)//2)*a*number2 + b*number2 #a mistake here
return kara
elif n1 != 1 and n2 != 1: # loop
a = int(str(number1)[:(n1 // 2)])
b = int(str(number1)[(n1 // 2):])
c = int(str(number2)[:(n2 // 2)])
d = int(str(number2)[(n2 // 2):])
z1 = karatsuba(a, c)
z2 = karatsuba(a, d)
z3 = karatsuba(b, c)
z4 = karatsuba(b, d)
kara = 10**((n1+1)//2+(n2+1)//2)*z1 + 10**((n1+1)//2)*z2 + 10**((n2+1)//2)*z3 + z4
return kara
num1 = 3141592653589793238462643383279502884197169399375105820974944592
num2 = 2718281828459045235360287471352662497757247093699959574966967627
k_res = karatsuba(num1,num2)
ac_res = num1*num2
print(k_res)
print(ac_res)
assert k_res==ac_res
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