def main(args):
if args.test:
m = 10
else:
m = 1000*1000
prime = PrimeNumberPool()
md = 5
mn = 2
for d in range(2, m):
if d == 7:
continue
n = d * 3 // 7
if (n>mn*d/md and HCF(n, d, prime)):
mn = n
md = d
logging.debug("{}/{}".format(mn, md))
logging.info("answer: {}/{}".format(mn, md))
logging.debug(prime.Factorize(mn))
logging.debug(prime.getPrimeFactor(mn))
logging.debug(prime.Factorize(md))
logging.debug(prime.getPrimeFactor(md))
def main(args):
if args.test:
m = 100 * 1000
maxprime = 1000
else:
m = 10 * 1000 * 1000
maxprime = 100000
t1 = time.time()
prime = PrimeNumberPool(maxprime)
t2 = time.time()
logging.debug("time for build prime pool:{}".format(t2 - t1))
'''
Analysis
By observation, the numbers with high φ(n) are product of two prime numbers
'''
max_ratio = (2, 1)
num_prime = len(prime.numbers)
for i in range(num_prime):
p1 = prime.numbers[i]
if p1 * p1 > m:
break
for j in range(i, num_prime):
p2 = prime.numbers[j]
if p1 * p2 > m:
break
n = p1 * p2
phi = n - p1 - p2 + 1
if (IsPermute(n, phi)):
if (max_ratio[0] * phi > max_ratio[1] * n):
logging.debug("{} factor to {}".format(
n, prime.Factorize(n)))
max_ratio = (n, phi)
logging.debug(max_ratio)
'''
# brute force way
for n in range(3,m,2):
# pre-qualify
pre_qualify = 1
for p in prime.numbers:
if (n % p == 0 and p*max_ratio[1] > (p-1)*max_ratio[0]):
pre_qualify = 0
break
if (p * p > n):
break
if (pre_qualify):
pn = Phi(n,prime)
if (IsPermute(n, pn)):
if (max_ratio[0]*pn > max_ratio[1]*n):
logging.debug("{} factor to {}".format(n, prime.Factorize(n)))
max_ratio = (n, pn)
logging.debug(max_ratio)
'''
logging.info("answer: {}".format(max_ratio[0]))
