我知道这很可能是一个非常简单的代码,但我似乎无法找出什么在世界上是错的。每次我得到0作为一个答案,如果我尝试只是直接进行PI =丕它给了我一个“浮动”的错误,并说,这是不可调用的。
def pi(n):
result = 0;
N= int(input("Input desired iterations: "))
M = 1.0
denom = 2.0
Pi = 3.0
for I in range (1, N+1):
Pi += ((4.0/(denom*(denom+1)*(denom+2.0)))*M)
denom += 2.0
M *= -1.0
return n == Pi
m = int(input("M: "))
actual_pi =3.1415926535897932384626433832795028841971693993751058209749445923078164062
for i in range(0, 50000) :
val = pi(i)
if (val - actual_pi) <= (10 ** m):
print(i)
break
这应该是正确的代码,我想,但我不能让其他任何比0 This is the question I'm trying to solve. It also has the answer I should be getting.
编辑:刚才注意到,我并没有包括正确的号码,我应该在图片中可以得到。这是24834。
你的回报表现,n == Pi
总是False
,这是算术0
。只是返回Pi
。
EDIT2:原因输出仍为0是PI(0) - actualPi <10 ** M代表任何非负M.
还有其他问题,这可以通过将测试对于Pi在PI功能在循环内被避免,如下所示。
actual_pi =3.1415926535897932384626433832795028841971693993751058209749445923078164062
def pi(testexp=14):
M = 1.0
denom = 2.0
Pi = 3.0
for i in range(50000):
delta = abs(Pi - actual_pi)
if not i % 1000: print(i, delta)
if delta <= (10 ** -14):
return i - 1 # Last iteration did not change Pi
else:
Pi += ((4.0/(denom*(denom+1)*(denom+2.0)))*M)
denom += 2.0
M *= -1.0
print(pi())
# 24834
一个简单的解决问题的方法是创建一个发电机为“Nilakantha系列”,然后就enumerate()
积累的条款,直到你得到的错误:
import itertools as it
Decimal = float
#from decimal import Decimal # To use arbitrary precision decimal type
def nilakantha_series():
yield Decimal(3)
c = it.cycle([1, -1])
for d in it.count(2, 2)
yield Decimal(next(c)*4)/Decimal(d*(d+1)*(d+2))
def term_count(series, error, exact):
for N, value in enumerate(it.accumulate(series)):
if abs(value - exact) <= error:
return N
In []:
PI = Decimal('3.1415926535897932384626433832795028841971693993751058209749445923078164062')
M = 14
term_count(nilakantha_series(), 10**-M, PI)
Out[]:
24835
注意:如果您取消注释了进口,并删除Decimal = float
然后你:
In []:
term_count(nilakantha_series(), 10**-M, PI)
Out[]:
29240