blocks|key|2674626|text|请注意，在您的电话中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2674627|fib()+recursively|2674628|You需要一个包装器方法，这样就不会在每次递归调用该方法时都请求输入|unordered-list-item|2674629|
|2674630|，你不需要在列表中发送。仅数字n就足够了。|2674631|2674632|这个方法只会给出序列中的第n个数字。它不打印序列。|2674633|你需要return+fib(n-1)+%2B+fib(n-2)|offset|length|style|CODE|2674634|def+f():
++++n+=+int(input("Please+Enter+a+number:+"))
++++print+fib(n)

def+fib(n):++++
++++if+n+==+0:+
++++++++return+0
++++elif+n+==+1:+
++++++++return+1
++++else:+
++++++++return+fib(n-1)%2Bfib(n-2)|code-block|syntax|javascript|2674635|entityMap^0|0|0|0|0|0|0|0|3|Q|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|13|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|14|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|16|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|17|8|@$P|18|Q|19|R|S]]|9|@]|A|$]]|$1|T|3|U|5|V|7|1A|8|@]|9|@]|A|$W|X]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

Please note, in your call 

<ol>
<li>You are not calling fib() recursively</li>
<li>You need a wrapper method so that the input is not requested every time the method is called recursively </li>
<li>You do not need to send in a list. Just the number n is good enough. </li>
</ol>

This method would only give you the nth number in the sequence. It does not print the sequence. 

You need to <code>return fib(n-1) + fib(n-2)</code>

<pre><code>def f():
 n = int(input("Please Enter a number: "))
 print fib(n)

def fib(n): 
 if n == 0: 
 return 0
 elif n == 1: 
 return 1
 else: 
 return fib(n-1)+fib(n-2)
</code></pre>

blocks|key|6027361|text|def+fib(n):
+++if+n+==+1:
++++++return(1)
+++elif+n+==+0:+++
++++++return(0)++++++++++++
+++else:++++++++++++++++++++++
++++++return+fib(n-1)+%2B+fib(n-2)

my_num+=+int(input("Enter+a+number:"))
print+fib(my_num)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|6027362|我不太确定你的问题是什么.但答案可能是这样的：|unstyled|6027363|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def fib(n):
 if n == 1:
 return(1)
 elif n == 0: 
 return(0) 
 else: 
 return fib(n-1) + fib(n-2)

my_num = int(input("Enter a number:"))
print fib(my_num)
</code></pre>

Im not really sure what your question is... but the answer is probably something like this

blocks|key|1098208|text|def+fibonacci(n):
++if+n+<=+1:
++++return+n
++else:
++++return+fibonacci(n-1)+%2B+fibonacci(n-2)

print(fibonacci(int(input())))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1098209|由于您希望打印到n第th编号：|unstyled|offset|length|style|CODE|1098210|[print(fibonacci(n))+for+n+in+range+(int(input()))]|1098211|对于python2.7，将input更改为raw_input。|1098212|entityMap^0|0|8|1|0|0|D|5|L|9|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|V|8|@$G|W|H|X|I|J]|$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|O|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>def fibonacci(n):
 if n &lt;= 1:
 return n
 else:
 return fibonacci(n-1) + fibonacci(n-2)

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

And since you want to print up to the <code>n</code>th number:

<pre><code>[print(fibonacci(n)) for n in range (int(input()))]
</code></pre>

And for python2.7 change <code>input</code> to <code>raw_input</code>.

blocks|key|2674731|text|非递归解|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2674732|def+fib(n):
++++cur+=+1
++++old+=+1
++++i+=+1
++++while+(i+<+n):
++++++++cur,+old,+i+=+cur%2Bold,+cur,+i%2B1
++++return+cur

for+i+in+range(10):
++++print(fib(i))|code-block|syntax|javascript|2674733|生成器解决方案：|2674734|def+fib(n):
++++old+=+0
++++cur+=+1
++++i+=+1
++++yield+cur
++++while+(i+<+n):
++++++++cur,+old,+i+=+cur%2Bold,+cur,+i%2B1
++++++++yield+cur

for+f+in+fib(10):
++++print(f)|2674735|请注意，生成器解决方案优于非递归解决方案(如果不将记忆化应用于递归解决方案，则非递归解决方案优于递归解决方案)|2674736|再一次，使用记忆化的递归：|2674737|def+memoize(func):
++++memo+=+dict()
++++def+decorated(n):
++++++++if+n+not+in+memo:
++++++++++++memo[n]+=+func(n)
++++++++return+memo[n]

++++return+decorated

@memoize
def+fib(n):
++++#added+for+demonstration+purposes+only+-+not+really+needed
++++global+call_count
++++call_count+=+call_count+%2B+1
++++#end+demonstration+purposes

++++if+n<=1:
++++++++return+1
++++else:
++++++++return+fib(n-1)+%2B+fib(n-2)

call_count+=+0+#added+for+demonstration+purposes+only+-+not+really+needed
for+i+in+range(100):
++++print(fib(i))
print(call_count)+#prints+100|2674738|这一次，每个fibbonacci数只计算一次，然后存储。因此，此解决方案的性能将优于所有其他解决方案。然而，装饰器的实现只是快速和肮脏，不要让它进入生产。(有关this+amazing+question装饰器的详细信息，请参阅python+on+SO+:)|offset|length|2674739|所以，定义了fib之后，程序应该是这样的(对不起，只是循环很无聊，这里有一些更酷的python东西：list+comprehensions)|style|CODE|2674740|fib_n+=+int(input("Fib+number?"))
fibs+=+[fib(i)+for+i+in+range(fib_n)]
print+"+".join(fibs)+|2674741|这将打印一行中的所有数字，并以空格分隔。如果您希望每一个都在自己的行上-请用"\n"替换"+"|2674742|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/739654/understanding-python-decorators|1|http://en.wikipedia.org/wiki/List_comprehension#Python^0|0|0|0|0|0|0|0|28|L|0|0|6|3|1E|J|1|0|0|12|4|18|3|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1E|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1G|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1H|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@$S|1K|T|1L|1|1M]]|A|$]]|$1|U|3|V|5|6|7|1N|8|@$S|1O|T|1P|W|X]]|9|@$S|1Q|T|1R|1|1S]]|A|$]]|$1|Y|3|Z|5|D|7|1T|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1U|8|@$S|1V|T|1W|W|X]|$S|1X|T|1Y|W|X]]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19]]|1A|$5|15|16|17|A|$18|1B]]]]

Non-recursive solution

<pre><code>def fib(n):
 cur = 1
 old = 1
 i = 1
 while (i &lt; n):
 cur, old, i = cur+old, cur, i+1
 return cur

for i in range(10):
 print(fib(i))
</code></pre>

Generator solution:

<pre><code>def fib(n):
 old = 0
 cur = 1
 i = 1
 yield cur
 while (i &lt; n):
 cur, old, i = cur+old, cur, i+1
 yield cur

for f in fib(10):
 print(f)
</code></pre>

Note that generator solution outperforms the non-recursive (and non-recursive outperforms recursive, if memoization is not applied to recursive solution)

One more time, recursive with memoization:

<pre><code>def memoize(func):
 memo = dict()
 def decorated(n):
 if n not in memo:
 memo[n] = func(n)
 return memo[n]

 return decorated

@memoize
def fib(n):
 #added for demonstration purposes only - not really needed
 global call_count
 call_count = call_count + 1
 #end demonstration purposes

 if n&lt;=1:
 return 1
 else:
 return fib(n-1) + fib(n-2)

call_count = 0 #added for demonstration purposes only - not really needed
for i in range(100):
 print(fib(i))
print(call_count) #prints 100
</code></pre>

This time each fibbonacci number calculated exactly once, and than stored. So, this solution would outperform all the rest. However, the decorator implementation is just quick and dirty, don't let it into production. (see <a href="https://stackoverflow.com/questions/739654/understanding-python-decorators">this amazing question</a> on SO for details about python decorators :)

So, having <code>fib</code> defined, the program would be something like (sorry, just looping is boring, here's some more cool python stuff: <a href="http://en.wikipedia.org/wiki/List_comprehension#Python" rel="noreferrer">list comprehensions</a>)

<pre><code>fib_n = int(input("Fib number?"))
fibs = [fib(i) for i in range(fib_n)]
print " ".join(fibs) 
</code></pre>

this prints all numbers in ONE line, separated by spaces. If you want each on it's own line - replace <code>" "</code> with <code>"\n"</code>

blocks|key|6027475|text|单独的函数是最好的，因为递归函数更容易处理。另一方面，您可以编写一个只接受一个参数的迭代函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6027476|递归：：|6027477|def+fib(n):
++++if+n+==+1:
++++++++return+(1);
++++elif+n+==+0:
++++++++return+(0);
++++else:
++++++++return+fib(n-1)+%2B+fib(n-2);

def+callFib():
++++n+=+int(raw_input('Enter+n::\t'));
++++mylist+=+fib(n);
++++print+mylist;

callFib();|code-block|syntax|javascript|6027478|迭代：：|6027479|def+fib():
++++n+=+int(raw_input('Enter+n::\t'));
++++terms+=+[0,1];
++++i=2;
++++while+i<=n:
++++++++terms.append(terms[i-1]+%2B+terms[i-2]);
++++++++i=i%2B1;
++++print+terms[n];

fib();|6027480|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Separate functions would be best, as the recursive function would be far easier to deal with. On the other hand, you could code an iterative function that would take only one parameter

Recursively::

<pre><code>def fib(n):
 if n == 1:
 return (1);
 elif n == 0:
 return (0);
 else:
 return fib(n-1) + fib(n-2);

def callFib():
 n = int(raw_input('Enter n::\t'));
 mylist = fib(n);
 print mylist;

callFib();
</code></pre>

Iteratively::

<pre><code>def fib():
 n = int(raw_input('Enter n::\t'));
 terms = [0,1];
 i=2;
 while i&lt;=n:
 terms.append(terms[i-1] + terms[i-2]);
 i=i+1;
 print terms[n];

fib();
</code></pre>

blocks|key|550613|text|对于递归解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550614|def+fib(n):
++++if+n+==+0:
++++++++return+0
++++elif+n+==+1:
++++++++return+1
++++else:
++++++++return+fib(n-1)+%2B+fib(n-2)
x=input('which+fibonnaci+number+do+you+want?')
print+fib(x)|code-block|syntax|javascript|550615|说明:如果n是0，那么“0”项当然是0，而第一项是1。从这里，您知道下一个数字将是前两个数字的总和，这就是else之后的代码行所推断的结果。|550616|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

for a recursive solution:

<pre><code>def fib(n):
 if n == 0:
 return 0
 elif n == 1:
 return 1
 else:
 return fib(n-1) + fib(n-2)
x=input('which fibonnaci number do you want?')
print fib(x)
</code></pre>

explanation:
if n is 0, then ofcourse the '0th' term is 0, and the 1st term is one. From here, you know that the next numbers will be the sum of the previous 2. That is what's inferred by the line after the else.

blocks|key|2674841|text|这在长列表的情况下可能会更快|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2674842|#+Get+nth+Fibonacci+number+
def+nfib(nth):
++sq5+=+5**.5
++phi1+=+(1%2Bsq5)/2
++phi2+=+-1+*+(phi1+-1)
++resp+=+(phi1**(nth%2B1)+-+phi2**(nth%2B1))/sq5
++return+long(resp)

for+i+in+range(10):
++print+i%2B1,+":+",++nfib(i)|code-block|syntax|javascript|2674843|输出|2674844|1+:++1
2+:++1
3+:++2
4+:++3
5+:++5
6+:++8
7+:++13
8+:++21
9+:++34
10+:++55|2674845|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|$]]

This might be faster incase of long list

<pre><code># Get nth Fibonacci number 
def nfib(nth):
 sq5 = 5**.5
 phi1 = (1+sq5)/2
 phi2 = -1 * (phi1 -1)
 resp = (phi1**(nth+1) - phi2**(nth+1))/sq5
 return long(resp)

for i in range(10):
 print i+1, ": ", nfib(i)
</code></pre>

<h3>OUTPUT</h3>

<pre><code>1 : 1
2 : 1
3 : 2
4 : 3
5 : 5
6 : 8
7 : 13
8 : 21
9 : 34
10 : 55
</code></pre>

blocks|key|1098277|text|看起来你可能正在尝试解决和我一样的家庭作业问题，你实际上不需要用户输入。只需在调用函数时传入参数即可。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1098278|def+compute_nth_fib(num):
+++
++++if+num+==+0:
++++++++return+0
++++elif+num+==+1:
++++++++return+1
++++else:
++++++++return+compute_nth_fib(num-1)+%2B+compute_nth_fib(num-2)

#test+with+different+parameters++++
print(compute_nth_fib(3))|code-block|syntax|javascript|1098279|希望这对某些人有帮助！|1098280|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

It looks like you might be trying to solve the same homework problem I was, where you don't actually need user input. You just pass in the parameter when you call the function.
<pre><code>
def compute_nth_fib(num):
 
 if num == 0:
 return 0
 elif num == 1:
 return 1
 else:
 return compute_nth_fib(num-1) + compute_nth_fib(num-2)

#test with different parameters 
print(compute_nth_fib(3))
</code></pre>
Hope that's helpful to someone!

I have a homework assignment that I'm stumped on. I'm trying to write a program that outputs the fibonacci sequence up the nth number. Here's what I have so far: 

<pre><code>def fib():
 n = int(input("Please Enter a number: "))

 if n == 1:
 return(1)
 elif n == 0: 
 return(0) 
 else: 
 return (n-1) + (n-2)


mylist = range[0:n]
print(mylist)
</code></pre>

I'm thinking I could use separate functions but I can't figure out how to pass the argument that calculates the fibonacci sequence. Then the next step would be to to print out the sequence of numbers up to that number.

How do I print a fibonacci sequence to the nth number in Python?

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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我有一项家庭作业，我被难倒了。我正在试着写一个程序，输出斐波那契数列的第n个数字。这是我到目前为止所知道的：def fib():   n = int(input("Please Enter a number: "))   if n == 1:      return(1)   elif n == 0:         ...

问如何在Python中将fibonacci序列打印到第n个数字？
EN

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问如何在Python中将fibonacci序列打印到第n个数字？EN