def genlistprimeslimit(LIMIT):
l = []
genp = mymaths.prime()
while True:
p = genp.next()
if p < ((LIMIT / 2) + 1):
l.append(p)
else:
break
return l
def genlprimes(n):
""" genera una lista de primos que pueda dividir n """
l = []
p = mymaths.prime()
prime = p.next()
# quiero una lista de primos que dividan a n, por lo tanto con
# buscar solo la mitad me va bien
limit = n + 1
while prime < limit:
l.append(prime)
prime = p.next()
return l
#!/usr/bin/python
# Problem 9
# A Pythagorean triplet is a set of three natural numbers, a b c, for which,
#
# a**2 + b**2 = c**2
# For example, 3**2 + 4**2 = 9 + 16 = 25 = 5**2.
#
# There exists exactly one Pythagorean triplet for which a + b + c = 1000.
# Find the product abc.
import os
import sys
lib_path = os.path.abspath("../../lib")
sys.path.append(lib_path)
import mymaths
p = mymaths.prime()
limite = 2000000
sumprimos = 0
primo = 0
while primo < limite:
sumprimos += primo
primo = p.next()
print("Resultado 0010", sumprimos)
from datetime import datetime
lib_path = os.path.abspath('../../lib')
sys.path.append(lib_path)
import mymaths
MAXIMO = 10000000000
def get_resto(n, p):
return (((p - 1) ** n) + (p + 1) ** n) % (p ** 2)
# controlamos el tiempo de ejecución
start_time = datetime.now()
# generador de primos
primegen = mymaths.prime()
n = 0
while True:
p = primegen.next()
n += 1
r = get_resto(n, p)
if r > MAXIMO:
break
print "Tiempo total: ", datetime.now() - start_time
print "Resultado de 0123 ", n
