本文整理匯總了Python中Euler.previous_prime方法的典型用法代碼示例。如果您正苦於以下問題：Python Euler.previous_prime方法的具體用法？Python Euler.previous_prime怎麽用？Python Euler.previous_prime使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類Euler的用法示例。
在下文中一共展示了Euler.previous_prime方法的2個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: evaluate_prime
# 需要導入模塊: import Euler [as 別名]
# 或者: from Euler import previous_prime [as 別名]
def evaluate_prime(prime):
# Evaluates a prime number and adds a key: value pair to prime_table
# with a list containing each set of pairs for that prime organized by
# number of pairs
candidate = Euler.previous_prime(prime)
answer = [] # This will be the value we return
finished = False
pairs = [prime]
while finished == False:
while candidate > 2:
if is_pair(prime, candidate) == True:
pairs.append(candidate)
candidate -= 2
if len(pairs) > 1:
answer.append((len(pairs), pairs))
candidate = Euler.previous_prime(pairs[-1])
pairs = pairs[:-1]
if len(pairs) < 2:
finished = True
if len(answer) == 0:
prime_table[prime] = False
else:
prime_table[prime] = answer示例2: int
# 需要導入模塊: import Euler [as 別名]
# 或者: from Euler import previous_prime [as 別名]
# Problem 70: Totient permutation
# candidate: 7026037
import time, Euler, math
t1 = time.clock()
limit = 10000000
root = Euler.previous_prime(int(math.sqrt(limit)))
# 3137 is the largest prime the square of which is < 10,000,000,
# Equivalent to Euler.previous_prime(int(math.sqrt(10000000)))
primes = {0: root}
large_prime_counter = 0
small_prime_counter = 0
composites = {}
large = root
small = root
while small > 2:
composite = large*small
# Deal with the case where composite is less than limit
if composite < limit:
phi = composite - (large + small - 1)
# Put the composite in the dictionary if it's a permutation
if (int("".join(sorted([x for x in str(composite)]))) ==
int("".join(sorted([x for x in str(phi)])))):
composites[composite] = [large, small, phi]
# Increase large and add it to primes if necessary