Problem:
Find the sum of digits in the numerator of the 100th convergent of the
continued fraction for e.
Performance time: ~0.0012s
"""
from fractions import Fraction
from timer import timer
from utils import sum_of_digits
timer.start()
a = [int(n / 3 * 2) if n % 3 == 0 else 1 for n in range(1, 101)]
a[0] = 2
def convergent_e(limit=-1, index=0):
if limit == 0:
return a[0]
elif index == limit - 1:
return a[index] + Fraction(1, a[index+1])
else:
return a[index] + Fraction(1, convergent_e(limit, index+1))
print(sum_of_digits(Fraction(convergent_e(99)).numerator))
timer.stop()
def main():
print "\n\n___ Задание 5: Псевдосумма " + "_"*15
digit = 456
result = sum_of_digits(digit)
print digit, ">>", result
"""
Problem :
Find the sum of the digits in the number 100!
Performance time: ~0.00005s
"""
from utils import sum_of_digits
from math import factorial
from timer import timer
timer.start()
print(sum_of_digits(factorial(100)))
timer.stop()
from math import factorial from utils import sum_of_digits print(sum_of_digits(factorial(100)))
from utils import sum_of_digits print(sum_of_digits(2**1000))
def main():
print "\n\n___ Задание 5: Псевдосумма " + "_" * 15
digit = 456
result = sum_of_digits(digit)
print digit, ">>", result
"""
Problem :
What is the sum of the digits of the number 2^1000?
Performance time: ~0.00010s
"""
from utils import sum_of_digits
from timer import timer
timer.start()
print(sum_of_digits(2**1000))
timer.stop()
Problem:
Find the sum of digits in the numerator of the 100th convergent of the
continued fraction for e.
Performance time: ~0.0012s
"""
from fractions import Fraction
from timer import timer
from utils import sum_of_digits
timer.start()
a = [int(n / 3 * 2) if n % 3 == 0 else 1 for n in range(1, 101)]
a[0] = 2
def convergent_e(limit=-1, index=0):
if limit == 0:
return a[0]
elif index == limit - 1:
return a[index] + Fraction(1, a[index + 1])
else:
return a[index] + Fraction(1, convergent_e(limit, index + 1))
print(sum_of_digits(Fraction(convergent_e(99)).numerator))
timer.stop()
