blocks|key|137973|text|以下是您可以做的几件事：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|137974|Num+=+int(input())
i=0+++++++++++++++++++++#+your+for+loop+will+initialize+i,+you+don't+need+to+do+this+here

for+i+in+range(0,Num):++
++i+=+i+%2B+1+++++++++++++#+your+for+loop+will+increment+i,+no+need+to+do+this+either
++x+=+Num/i+++++++++++++#+you+don't+need+the+extra+variable+'x'+here
++Num+=+x
++if+(x==1):
++++++print(i)|code-block|syntax|javascript|137975|您可以重写此代码，如下所示：|137976|for+index+in+range(1,+number):+++++++#+start+range+at+1

++++++++number+/=+index++++++++++++++#+this+means;+number+=+number+/+index

++++++++if+number==1:
++++++++++++return+index|137977|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

A few things you can do:

<pre><code>Num = int(input())
i=0 # your for loop will initialize i, you don't need to do this here

for i in range(0,Num): 
 i = i + 1 # your for loop will increment i, no need to do this either
 x = Num/i # you don't need the extra variable 'x' here
 Num = x
 if (x==1):
 print(i)
</code></pre>

You can rewrite this to look something like:

<pre><code>for index in range(1, number): # start range at 1

 number /= index # this means; number = number / index

 if number==1:
 return index
</code></pre>

blocks|key|2178573|text|按升序计算阶乘，直到达到(或超过)您正在寻找的阶乘，使用前一个阶乘有效地计算下一个阶乘。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2178574|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Compute the factorials in ascending order until you reach (or exceed) the factorial you are looking for, using the previous factorial to efficiently compute the next.

blocks|key|137979|text|def+reverse_factorial(num):
++++i+=+1
++++while+num+>+1:
++++++++i+%2B=+1
++++++++num+/=+i
++++return+i

print(reverse_factorial(int(input())))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|137980|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def reverse_factorial(num):
 i = 1
 while num &gt; 1:
 i += 1
 num /= i
 return i

print(reverse_factorial(int(input())))
</code></pre>

blocks|key|2178569|text|乘法比除法快得多。你应该尝试用阶乘来达到这个数字，而不是迭代地除以它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2178570|def+unfactorial(n):
++++f,i+=+1,1
++++while+f+<+n:
++++++++i+%2B=+1
++++++++f+*=+i
++++return+i+if+f+==+n+else+None


unfactorial(120)+#+5|code-block|syntax|javascript|2178571|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Multiplications are much faster than divisions. You should try to reach the number with a factorial instead of dividing it iteratively:

<pre><code>def unfactorial(n):
 f,i = 1,1
 while f &lt; n:
 i += 1
 f *= i
 return i if f == n else None


unfactorial(120) # 5
</code></pre>

I am just starting to learn python and made a program where it calculates the factorial number based on the factorial.

For example if I give the program the number 120 it will tell me it's factorial is 5

anyways my question is how can I make this code more efficient and faster.

<pre><code>Num = int(input())
i=0

for i in range(0,Num):
 i = i + 1
 x = Num/i
 Num = x
 if (x==1):
 print(i)






</code></pre>

Is there a way to make this reverse factorial code run more efficiently

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我刚刚开始学习python，并编写了一个程序，它根据阶乘计算阶乘数。例如，如果我给程序一个数字120，它会告诉我它的阶乘是5无论如何，我的问题是如何让这段代码更高效、更快。Num = int(input())i=0for i in range(0,Num):  i = i + 1  x = Num/i  Num = x...

问有没有一种方法可以让这种反向阶乘代码更高效地运行？
EN

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问有没有一种方法可以让这种反向阶乘代码更高效地运行？EN