def main():
n = int(argv[1])
if is_prime(n, 25):
print('{} - простое'.format(n))
return
with open('p.txt') as f:
primes_data = f.read()
primes = list(map(int, primes_data.split('\n')))
if '-u' in argv:
p = dixon(n, primes, dixon_usual)
assert n % p == 0
print('Стандартный алгоритм:\n{} * {}'.format(p, n // p)) \
if p != -1 else print('Делитель не найден')
if '-m' in argv:
p = dixon(n, primes, dixon_modified)
assert n % p == 0
print('С добавлением -1 и выбором наименьшего а:\n{} * {}'.format(p, n // p)) \
if p != -1 else print('Делитель не найден')
if '-c' in argv:
p = dixon(n, primes, dixon_chain)
assert n % p == 0
print('С использованием цепных дробей:\n{} * {}'.format(p, n // p)) \
if p != -1 else print('Делитель не найден')
def PollardRho(n):
i = 1
x = random.randint(0, n - 1)
y = x
k = 2
tried = set()
tried.add(x)
while True:
i = i + 1
x = (x ** 2 - 1) % n
if x in tried:
return
tried.add(x)
d = Euclid(y - x, n)
if d != 1 and d != n:
if mutils.is_prime(d):
print d
# print str(d) + " - " + str(is_prime(int(d))) # d is a factor. Print it.
if i == k:
y = x
k = 2 * k
#!/usr/bin/python import sys import mutils # Sieve approach sum = 0 # problem was changed from 1.000.000 to 2.000.000 # primeList = mutils.prime_sieve(1000000) primeList = mutils.prime_sieve(2000000) for i in primeList: sum = sum + i print sum sys.exit(0) # Brute force approach # This is too slow - it takes around 150s (2.5 mins) in an Athlon XP 2600 sum = 2 # 1.000.000 is multiple of ten so we can end the checking at 999.999 for i in range(3, 1000000, 2): if mutils.is_prime(i): sum = sum + i print sum
#!/usr/bin/python import sys import mutils # i is the accumulated number of primes i = 2 # n is the number to be tested n = 3 while i != 10001: n = n + 2 if mutils.is_prime(n): i = i + 1 # print i print "The 10001st prime is: " + str(n)
