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#!/usr/bin/env python3
# Challenge 006
#
# Challenge #2
# Create a script to calculate Ramanujan's constant with at least 32 digits of
# precision. Find out more about it here.
#
# The standard IEEE 754 double-precision binary floating-point format: binary64
# gives only 15 to 17 significant decimal digits
# Therefore must use the bignum library
from decimal import *
getcontext().prec = 128
def pi():
getcontext().prec += 2 # extra digits for intermediate steps
three = Decimal(3) # substitute "three=3.0" for regular floats
lasts, t, s, n, na, d, da = 0, three, 3, 1, 0, 0, 24
while s != lasts:
lasts = s
n, na = n+na, na+8
d, da = d+da, da+32
t = (t * n) / d
s += t
getcontext().prec -= 2
return +s # unary plus applies the new precision
def exp(x):
getcontext().prec += 2
i, lasts, s, fact, num = 0, 0, 1, 1, 1
while s != lasts:
lasts = s
i += 1
fact *= i
num *= x
s += num / fact
getcontext().prec -= 2
return +s
e = pi() * Decimal(163).sqrt()
k = exp(e)
print(str(k)[:31])
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