blocks|key|4404588|text|我不熟悉你的算法。它是BBP的实现吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4404589|无论如何，您的make_pi都是一个生成器。尝试在for循环中使用它：|offset|length|style|CODE|4404590|for+digit+in+make_pi():
++++print+digit|code-block|syntax|javascript|4404591|请注意，此循环是无限的：make_pi()从不抛出StopIteration|4404592|entityMap^0|0|7|7|0|0|C|9|P|D|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

I'm not familiar with your algorithm. Is it an implementation of BBP?

In any case, your <code>make_pi</code> is a generator. Try using it in a for loop:

<pre><code>for digit in make_pi():
 print digit
</code></pre>

Note that this loop is infinite: <code>make_pi()</code> never throws <code>StopIteration</code>

blocks|key|1431789|text|这是你想要的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1431790|i+=+0;
pi_str+=+""
for+x+in+make_pi():
++++pi_str+%2B=+str(x)
++++i+%2B=+1
++++if+i+==+1001:
++++++++break

print+"pi=+%25s.%25s"+%25+(pi_str[0],pi_str[1:])|code-block|syntax|javascript|1431791|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|$]]

Does this do what you want?

<pre><code>i = 0;
pi_str = ""
for x in make_pi():
 pi_str += str(x)
 i += 1
 if i == 1001:
 break

print "pi= %s.%s" % (pi_str[0],pi_str[1:])
</code></pre>

blocks|key|414389|text|运行此命令|type|unstyled|depth|inlineStyleRanges|entityRanges|data|414390|def+make_pi():
++++q,+r,+t,+k,+m,+x+=+1,+0,+1,+1,+3,+3
++++for+j+in+range(1000):
++++++++if+4+*+q+%2B+r+-+t+<+m+*+t:
++++++++++++yield+m
++++++++++++q,+r,+t,+k,+m,+x+=+10*q,+10*(r-m*t),+t,+k,+(10*(3*q%2Br))//t+-+10*m,+x
++++++++else:
++++++++++++q,+r,+t,+k,+m,+x+=+q*k,+(2*q%2Br)*x,+t*x,+k%2B1,+(q*(7*k%2B2)%2Br*x)//(t*x),+x%2B2


my_array+=+[]

for+i+in+make_pi():
++++my_array.append(str(i))

my_array+=+my_array[:1]+%2B+['.']+%2B+my_array[1:]
big_string+=+"".join(my_array)
print+"here+is+a+big+string:\n+%25s"+%25+big_string+|code-block|syntax|javascript|414391|并从此处阅读有关yield+operator的信息：What+does+the+"yield"+keyword+do?|offset|length|style|CODE|414392|这就是答案：|414393|3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337|414394|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/231767/the-python-yield-keyword-explained^0|0|0|8|5|Q|X|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Run this

<pre><code>def make_pi():
 q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
 for j in range(1000):
 if 4 * q + r - t &lt; m * t:
 yield m
 q, r, t, k, m, x = 10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x
 else:
 q, r, t, k, m, x = q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2


my_array = []

for i in make_pi():
 my_array.append(str(i))

my_array = my_array[:1] + ['.'] + my_array[1:]
big_string = "".join(my_array)
print "here is a big string:\n %s" % big_string 
</code></pre>

And read about <code>yield</code> operator from here:
<a href="https://stackoverflow.com/questions/231767/the-python-yield-keyword-explained">What does the &quot;yield&quot; keyword do?</a>

Here is the answer:

<pre><code>3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337
</code></pre>

blocks|key|365528|text|如果你不想实现自己的算法，你可以使用mpmath。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|365529|try:
++++#+import+version+included+with+old+SymPy
++++from+sympy.mpmath+import+mp
except+ImportError:
++++#+import+newer+version
++++from+mpmath+import+mp

mp.dps+=+1000++#+set+number+of+digits
print(mp.pi)+++#+print+pi+to+a+thousand+places|code-block|syntax|javascript|365530|365531|更新：代码支持SymPy+(请参阅comment).*的旧版和新版安装|style|BOLD|365532|entityMap|0|LINK|mutability|MUTABLE|url|http://mpmath.org/|1|https://sympy.org/|2|https://stackoverflow.com/questions/9004789/1000-digits-of-pi-in-python/13316984#comment68149601_13316984^0|I|6|0|0|0|0|0|3|7|5|1|H|7|2|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|14|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|15|8|@$A|16|B|17|L|M]]|9|@$A|18|B|19|1|1A]|$A|1B|B|1C|1|1D]]|C|$]]|$1|N|3|-4|5|6|7|1E|8|@]|9|@]|C|$]]]|O|$P|$5|Q|R|S|C|$T|U]]|V|$5|Q|R|S|C|$T|W]]|X|$5|Q|R|S|C|$T|Y]]]]

If you don't want to implement your own algorithm, you can use <a href="http://mpmath.org/" rel="noreferrer">mpmath</a>.

<pre><code>try:
 # import version included with old SymPy
 from sympy.mpmath import mp
except ImportError:
 # import newer version
 from mpmath import mp

mp.dps = 1000 # set number of digits
print(mp.pi) # print pi to a thousand places
</code></pre>

<a href="https://pythonadventures.wordpress.com/2012/04/13/digits-of-pi-part-2/" rel="noreferrer">Reference</a>

Update: Code supports older and newer installations of <a href="https://sympy.org/" rel="noreferrer">SymPy</a> (see <a href="https://stackoverflow.com/questions/9004789/1000-digits-of-pi-in-python/13316984#comment68149601_13316984">comment</a>).*

blocks|key|365590|text|在这里你可以检查你的程序是否输出正确的1000位数字：http://spoj.com/CONSTANT|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|365591|当然，你也可以使用diff或tc，但你必须从某个地方复制这1000个数字，然后你只需提交你的程序，并检查分数是否大于999。|365592|你可以尝试在那里打印更多的数字，从而获得更多的分数。也许你会喜欢它。|365593|entityMap|0|LINK|mutability|MUTABLE|url|http://spoj.com/CONSTANT^0|R|O|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|C|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]]]

Here you can check whether your program outputs correct 1000 digits:
<a href="http://spoj.com/CONSTANT" rel="nofollow">http://spoj.com/CONSTANT</a>

Of course you can use diff or tc as well but you'd have to copy these 1000 digits from somewhere and there you just submit your program and check whether the score is bigger than 999.

You can try to print even more digits there and thus get more points. Perhaps you'd enjoy it.

blocks|key|365645|text|这是我在这里发现的一种不同的方法-->基于Chudnovsky兄弟生成Pi的公式的Python+pi+calculation?近似python，我已经为我的程序做了明显的修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|365646|def+pifunction():
++++numberofdigits+=+int(input("please+enter+the+number+of+digits+of+pi+that+you+want+to+generate"))
++++getcontext().prec+=+numberofdigits

def+calc(n):
++++t+=+Decimal(0)
++++pi+=+Decimal(0)
++++deno+=+Decimal(0)
++++k+=+0
++++for+k+in+range(n):
++++++++t+=+(Decimal(-1)**k)*(math.factorial(Decimal(6)*k))*(13591409%2B545140134*k)
++++++++deno+=+math.factorial(3*k)*(math.factorial(k)**Decimal(3))*(640320**(3*k))
++++++++pi+%2B=+Decimal(t)/Decimal(deno)
++++pi+=+pi+*+Decimal(12)/Decimal(640320**Decimal(1.5))
++++pi+=+1/pi
++++return+str(pi)
print(calc(1))|code-block|syntax|javascript|365647|我希望这对您有帮助，因为您可以生成您希望生成的pi的任意位数。|365648|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/28284996/python-pi-calculation^0|15|M|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Here is a different way I found here --> <a href="https://stackoverflow.com/questions/28284996/python-pi-calculation">Python pi calculation?</a> to approximate python based on the Chudnovsky brothers formula for generating Pi which I have sightly modified for my program.

<pre><code>def pifunction():
 numberofdigits = int(input("please enter the number of digits of pi that you want to generate"))
 getcontext().prec = numberofdigits

def calc(n):
 t = Decimal(0)
 pi = Decimal(0)
 deno = Decimal(0)
 k = 0
 for k in range(n):
 t = (Decimal(-1)**k)*(math.factorial(Decimal(6)*k))*(13591409+545140134*k)
 deno = math.factorial(3*k)*(math.factorial(k)**Decimal(3))*(640320**(3*k))
 pi += Decimal(t)/Decimal(deno)
 pi = pi * Decimal(12)/Decimal(640320**Decimal(1.5))
 pi = 1/pi
 return str(pi)
print(calc(1))
</code></pre>

I hope this helps as you can generate any number of digits of pi that you wish to generate.

blocks|key|365668|text|wallis公式可以达到3.141592661439964，但需要更有效的方法来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365669|https://www.youtube.com/watch?v=EZSiQv_G9HM|offset|length|365670|现在我的代码|365671|x,+y,+summing+=+2,+3,+4

for+count+in+range+(0,100000000):
++++summing+*=+(x/y)
++++x+%2B=+2
++++summing+*=+(x/y)
++++y+%2B=+2

print+(summing)|code-block|syntax|javascript|365672|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|17|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|Z|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

wallis formula can get to 3.141592661439964 but a more efficient way is needed to solve this problem. 

<a href="https://www.youtube.com/watch?v=EZSiQv_G9HM" rel="nofollow noreferrer">https://www.youtube.com/watch?v=EZSiQv_G9HM</a>

and now my code

<pre><code>x, y, summing = 2, 3, 4

for count in range (0,100000000):
 summing *= (x/y)
 x += 2
 summing *= (x/y)
 y += 2

print (summing)
</code></pre>

blocks|key|4404850|text|接受的答案是不正确的，如注释中所述。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4404851|OP的代码似乎是基于从here复制的Spigot's+algorithm实现。|offset|length|4404852|要根据OP的问题修复代码(尽管我重命名了变量和函数以与原始源代码中的变量和函数匹配)，一种解决方案可能是：|4404853|#!/usr/bin/env+python

DIGITS+=+1000

def+pi_digits(x):
++++"""Generate+x+digits+of+Pi."""
++++q,r,t,k,n,l+=+1,0,1,1,3,3
++++while+x+>=+0:
++++++++if+4*q%2Br-t+<+x*t:
++++++++++++yield+n
++++++++++++x+-=+1
++++++++++++q,r,t,k,n,l+=+10*q,+10*(r-n*t),+t,+k,+(10*(3*q+%2B+r))/t-10*n,+l
++++++++else:
++++++++++++q,r,t,k,n,l+=+q*k,+(2*q%2Br)*l,+t*l,+k%2B1,+(q*(7*k%2B2)%2Br*l)/(t*l),+l%2B2

digits+=+[str(n)+for+n+in+list(pi_digits(DIGITS))]
print("%25s.%25s\n"+%25+(digits.pop(0),+"".join(digits)))|code-block|syntax|javascript|4404854|此外，这里有一个更快的*实现，显然也是基于Spigot的算法：|4404855|#!/usr/bin/env+python

DIGITS+=+1000

def+pi_digits(x):
++++"""Generate+x+digits+of+Pi."""
++++k,a,b,a1,b1+=+2,4,1,12,4
++++while+x+>+0:
++++++++p,q,k+=+k+*+k,+2+*+k+%2B+1,+k+%2B+1
++++++++a,b,a1,b1+=+a1,+b1,+p*a+%2B+q*a1,+p*b+%2B+q*b1
++++++++d,d1+=+a/b,+a1/b1
++++++++while+d+==+d1+and+x+>+0:
++++++++++++yield+int(d)
++++++++++++x+-=+1
++++++++++++a,a1+=+10*(a+%25+b),+10*(a1+%25+b1)
++++++++++++d,d1+=+a/b,+a1/b1

digits+=+[str(n)+for+n+in+list(pi_digits(DIGITS))]
print("%25s.%25s\n"+%25+(digits.pop(0),+"".join(digits)))|4404856|我在this在线Pi数字生成器上对这两个工具进行了多次测试。|4404857|所有功劳都归功于deeplook的this+Gist。|4404858|*基于测试10,000个数字，其中我得到了大约7秒，而不是大约1秒。|4404859|entityMap|0|LINK|mutability|MUTABLE|url|https://mail.python.org/pipermail/edu-sig/2006-July/006810.html|1|https://en.wikipedia.org/wiki/Spigot_algorithm|2|https://www.browserling.com/tools/pi-digits|3|https://github.com/deeplook|4|https://gist.github.com/deeplook/4947835^0|0|B|4|0|I|I|1|0|0|0|0|0|2|4|2|0|8|8|3|H|4|4|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@]|9|@$D|1E|E|1F|1|1G]|$D|1H|E|1I|1|1J]]|A|$]]|$1|F|3|G|5|6|7|1K|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1L|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1M|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|1N|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1O|8|@]|9|@$D|1P|E|1Q|1|1R]]|A|$]]|$1|S|3|T|5|6|7|1S|8|@]|9|@$D|1T|E|1U|1|1V]|$D|1W|E|1X|1|1Y]]|A|$]]|$1|U|3|V|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]|16|$5|Z|10|11|A|$12|17]]|18|$5|Z|10|11|A|$12|19]]|1A|$5|Z|10|11|A|$12|1B]]]]

The accepted answer is incorrect, as noted in comments.

The OP's code appears to be based on an implementation of <a href="https://en.wikipedia.org/wiki/Spigot_algorithm" rel="noreferrer">Spigot's algorithm</a> copied from <a href="https://mail.python.org/pipermail/edu-sig/2006-July/006810.html" rel="noreferrer">here</a>.

To fix the code per the OP's question (although I renamed the variables and functions to match what they were in the original source), one solution might be:

<pre class="lang-py prettyprint-override"><code>#!/usr/bin/env python

DIGITS = 1000

def pi_digits(x):
 """Generate x digits of Pi."""
 q,r,t,k,n,l = 1,0,1,1,3,3
 while x &gt;= 0:
 if 4*q+r-t &lt; x*t:
 yield n
 x -= 1
 q,r,t,k,n,l = 10*q, 10*(r-n*t), t, k, (10*(3*q + r))/t-10*n, l
 else:
 q,r,t,k,n,l = q*k, (2*q+r)*l, t*l, k+1, (q*(7*k+2)+r*l)/(t*l), l+2

digits = [str(n) for n in list(pi_digits(DIGITS))]
print("%s.%s\n" % (digits.pop(0), "".join(digits)))
</code></pre>

Also, here is a much faster* implementation, also apparently based on Spigot's algorithm:

<pre class="lang-py prettyprint-override"><code>#!/usr/bin/env python

DIGITS = 1000

def pi_digits(x):
 """Generate x digits of Pi."""
 k,a,b,a1,b1 = 2,4,1,12,4
 while x &gt; 0:
 p,q,k = k * k, 2 * k + 1, k + 1
 a,b,a1,b1 = a1, b1, p*a + q*a1, p*b + q*b1
 d,d1 = a/b, a1/b1
 while d == d1 and x &gt; 0:
 yield int(d)
 x -= 1
 a,a1 = 10*(a % b), 10*(a1 % b1)
 d,d1 = a/b, a1/b1

digits = [str(n) for n in list(pi_digits(DIGITS))]
print("%s.%s\n" % (digits.pop(0), "".join(digits)))
</code></pre>

I tested both a few times against <a href="https://www.browserling.com/tools/pi-digits" rel="noreferrer">this</a> online Pi digit generator.

All credit to <a href="https://gist.github.com/deeplook/4947835" rel="noreferrer">this</a> Gist by <a href="https://github.com/deeplook" rel="noreferrer">deeplook</a>.

* Based on testing 10,000 digits, where I got about 7 seconds compared to about 1 second.

blocks|key|1431974|text|来自Fabrice网站：Pi+Computation算法。对于如此简单的实现，我很抱歉。10000已经足够快了(对我来说是0.1秒)，但是1000就不是那么快了-71秒:-(|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1431975|import+time
from+decimal+import+Decimal,+getcontext

def+compute(n):
++++getcontext().prec+=+n
++++res+=+Decimal(0)
++++for+i+in+range(n):
++++++++a+=+Decimal(1)/(16**i)
++++++++b+=+Decimal(4)/(8*i%2B1)
++++++++c+=+Decimal(2)/(8*i%2B4)
++++++++d+=+Decimal(1)/(8*i%2B5)
++++++++e+=+Decimal(1)/(8*i%2B6)
++++++++r+=+a*(b-c-d-e)
++++++++res+%2B=+r
++++return+res

if+__name__+==+"__main__":
++++t1+=+time.time()
++++res+=+compute(1000)
++++dt+=+time.time()-t1
++++print(res)
++++print(dt)|code-block|syntax|javascript|1431976|entityMap|0|LINK|mutability|MUTABLE|url|https://bellard.org/pi/^0|C|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

From Fabrice Bellard site: <a href="https://bellard.org/pi/" rel="nofollow noreferrer">Pi Computation</a> algorithm. Sorry for such a straightforward implementation. 1000 is fast enough (0.1s for me), but 10000 isn't such fast - 71s :-(

<pre class="lang-py prettyprint-override"><code>import time
from decimal import Decimal, getcontext

def compute(n):
 getcontext().prec = n
 res = Decimal(0)
 for i in range(n):
 a = Decimal(1)/(16**i)
 b = Decimal(4)/(8*i+1)
 c = Decimal(2)/(8*i+4)
 d = Decimal(1)/(8*i+5)
 e = Decimal(1)/(8*i+6)
 r = a*(b-c-d-e)
 res += r
 return res

if __name__ == "__main__":
 t1 = time.time()
 res = compute(1000)
 dt = time.time()-t1
 print(res)
 print(dt)
</code></pre>

blocks|key|365805|text|我是在5-6年前用bellow公式解决的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365806|类机器公式|offset|length|style|BOLD|365807|维基百科：https://en.wikipedia.org/wiki/Machin-like_formula|365808|​|365809|📷|atomic|365810|365811|对于代码质量，我很抱歉。变量名可能毫无意义。|365812|#-*-+coding:+utf-8+-*-

#+Author:++++Fatih+Mert+Doğancan
#+Date:++++++02.12.2014

def+arccot(x,+u):
++++sum+=+ussu+=+u+//+x
++++n+=+3
++++sign+=+-1
++++while+1:
++++++++ussu+=+ussu+//+(x*x)
++++++++term+=+ussu+//+n
++++++++if+not+term:
++++++++++++break
++++++++sum+%2B=+sign+*+term
++++++++sign+=+-sign
++++++++n+%2B=+2
++++return+sum

def+pi(basamak):
++++u+=+10**(basamak%2B10)
++++pi+=+4+*+(4*arccot(5,u)+-+arccot(239,u))
++++return+pi+//+10**10

if+__name__+==+"__main__":
++++print+pi(1000)+#+1000+|code-block|syntax|javascript|365813|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Machin-like_formula|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/7b9f81e31f2805b35e19e64d366c42b3.png|imageAlt^0|0|0|5|0|5|1D|0|0|0|0|1|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@$D|1F|E|1G|1|1H]]|A|$]]|$1|J|3|K|5|6|7|1I|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1J|8|@]|9|@$D|1K|E|1L|1|1M]]|A|$]]|$1|O|3|K|5|6|7|1N|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1O|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|1P|8|@]|9|@]|A|$U|V]]|$1|W|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|15|10|16|A|$17|18|19|-4]]]]

I was solved with bellow formula 5-6 years ago. 

Machin-like formula

Wikipedia: <a href="https://en.wikipedia.org/wiki/Machin-like_formula" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Machin-like_formula</a>

<a href="https://i.stack.imgur.com/iWxjH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iWxjH.png" alt="Math formula"></a>

Sorry for the code quality. Variable names can be meaningless.

<pre><code>#-*- coding: utf-8 -*-

# Author: Fatih Mert Doğancan
# Date: 02.12.2014

def arccot(x, u):
 sum = ussu = u // x
 n = 3
 sign = -1
 while 1:
 ussu = ussu // (x*x)
 term = ussu // n
 if not term:
 break
 sum += sign * term
 sign = -sign
 n += 2
 return sum

def pi(basamak):
 u = 10**(basamak+10)
 pi = 4 * (4*arccot(5,u) - arccot(239,u))
 return pi // 10**10

if __name__ == "__main__":
 print pi(1000) # 1000 
</code></pre>

blocks|key|416047|text|对于最多一百万个pi数字，请使用math_pi+(注意:我是该模块的作者)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|416048|使用pip安装：|416049|pip+install+math-pi|code-block|syntax|javascript|416050|在Python中：|416051|>>>+import+math_pi
>>>+print(math_pi.pi(b=1000))
3.1415926535...|416052|entityMap|0|LINK|mutability|MUTABLE|url|https://pypi.org/project/math-pi/^0|G|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

For up to 1 million digits of pi use <a href="https://pypi.org/project/math-pi/" rel="nofollow noreferrer">math_pi</a> (note: I am the author of the module)
Install with pip:
<pre><code>pip install math-pi
</code></pre>
In Python:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; import math_pi
&gt;&gt;&gt; print(math_pi.pi(b=1000))
3.1415926535...
</code></pre>

I have been thinking about this issue and I can't figure it out. Perhaps you can assist me. The problem is my code isn't working to output 1000 digits of pi in the Python coding language. 

Here's my code:

<pre><code>def make_pi():
 q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
 while True:
 if 4 * q + r - t &lt; m * t:
 yield m
 q, r, t, k, m, x = (10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x)
 else:
 q, r, t, k, m, x = (q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2)

digits = make_pi()
pi_list = []
my_array = []
for i in range(1000):
 my_array.append(str("hello, I'm an element in an array \n" ))
big_string = "".join(my_array)

print "here is a big string:\n %s" % big_string 
</code></pre>

I know this code can be fixed to work, but I'm not sure what to fix... The <code>print</code> statement saying here is a big string and the <code>my_array.append(str("hello, im an element in an array \n))</code> is just a filler for now. I know how all the code is used to work, but like I said before, I can't get it to shoot out that code.

1000 digits of pi 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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我一直在思考这个问题，但我想不出来。也许你能帮我。问题是我的代码不能用Python编码语言输出1000位pi。下面是我的代码：def make_pi(): q, r, t, k, m, x = 1, 0, 1, 1, 3, 3 while True: if 4 * q + r - t < m ...

问Python中的1000位pi
EN

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