[英]Finding the last digits sum with recursion
I'm attempting to create a function that will sum all the digits我正在尝试创建一个对所有数字求和的函数
and will return the sum of the summarized number.并将返回汇总数字的总和。
Example:例子:
For Input getNumValue(1589)对于输入getNumValue(1589)
The Output will be: 5输出将是:5
Becuase: 1 + 5 + 8 + 9 = 23因为:1 + 5 + 8 + 9 = 23
And 2 + 3 = 5 2 + 3 = 5
So the output will be 5所以输出将是 5
Because we can't split it into more digits.因为我们不能把它分成更多的数字。
I did managed to create a recursion function that summarize the digits:我确实设法创建了一个总结数字的递归函数:
def getNumValue(number: int):
if number == 0:
return 0
return (number % 10 + getNumValue(int(number / 10)))
But I can't seem to utilize it to my cause.但我似乎无法将它用于我的事业。
Btw顺便提一句
I don't want to use any strings我不想使用任何字符串
And I'm trying to use recursion so far no luck.到目前为止,我正在尝试使用递归,但没有运气。
I bet this is a known mathematic problem I'm just unfamiliar with.我敢打赌,这是一个我不熟悉的已知数学问题。
Any advice?有什么建议吗?
Even shorter:更短:
def getNumValue(number: int): return ((number-1) % 9) + 1
The digit sum is always in the same remainder class mod 9 as the original decimal number, this applies recursively, and thus reducing it to one digit just is the remainder under division by 9.数字和总是与原始十进制数处于相同的余数类 mod 9 中,这递归适用,因此将其减少到一位就是除以 9 的余数。
The shift by 1
just serves the purpose that the remainder class 0
is represented by 9
.移位1
只是为了将余数类0
用9
表示。
you can make a final check before returning the answer.您可以在返回答案之前进行最后检查。
def getNumValue(number: int):
if number == 0:
return 0
answer = (number % 10 + getNumValue(int(number // 10)))
if answer < 10:
return answer
return getNumValue(answer)
print(getNumValue(15899999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999))
OUTPUT :输出 :
9
You can check if number is greater than 9. If yes, then call the function again:您可以检查数字是否大于 9。如果是,则再次调用该函数:
def getNumValue(number: int):
if number == 0:
return 0
j=(number % 10 + getNumValue(int(number // 10)))
if j>9:
return getNumValue(j)
return j
print(getNumValue(getNumValue(1589)))
Without recursion:没有递归:
def getNumValue(number: int) -> int:
while True:
total = 0
while number:
total += number % 10
number //= 10
number = total
if total <= 9:
return total
>>> getNumValue(number)
5
pythonic recursion :) pythonic递归:)
def getNumValue(number=1589):
ans = number % 10 + getNumValue(int(number / 10)) if number else 0
return getNumValue(ans) if ans > 10 else ans
output输出
5
With two lines of code用两行代码
getNumValue = lambda x: 0 if x == 0 else x % 10 + getNumValue(int(x / 10))
output = getNumValue(getNumValue(1589))
print(output)
//5
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.