本文整理汇总了Python中Euler.nextPrime方法的典型用法代码示例。如果您正苦于以下问题：Python Euler.nextPrime方法的具体用法？Python Euler.nextPrime怎么用？Python Euler.nextPrime使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Euler的用法示例。
在下文中一共展示了Euler.nextPrime方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: main
# 需要导入模块: import Euler [as 别名]
# 或者: from Euler import nextPrime [as 别名]
def main(n):
# Returns the nth prime number.
counter = 0
testnum = 1
while counter < n:
testnum = Euler.nextPrime(testnum)
counter += 1
return testnum示例2: permutePrimes
# 需要导入模块: import Euler [as 别名]
# 或者: from Euler import nextPrime [as 别名]
def permutePrimes(prime, character):
# Replaces each instance of character in prime with 0-9 and
# returns a list of each resulting number that is also prime
digits = [x for x in range(10)]
digits = list(filter(lambda x: x != character, digits))
answers = [prime]
for x in digits:
testnum = replaceDigit(prime, character, x)
if Euler.isPrime(testnum) == True:
if len(str(testnum)) == len(str(prime)):
answers.append(testnum)
return answers
counter = 0
answer = 0
permutations = []
while answer == 0:
counter = Euler.nextPrime(counter)
for x in range(10):
if digitCount(counter,x) > 1:
if len(permutePrimes(counter, x)) > 7:
permutations = permutePrimes(counter, x)
answer = permutations[0]
print("The answer is " + str(answer))
print("The primes are: " + str(permutations))
t2 = time.clock()
print("Execution time = " + str(t2-t1)[:5] + " seconds")
示例3: reversed
# 需要导入模块: import Euler [as 别名]
# 或者: from Euler import nextPrime [as 别名]
# Problem 46: Goldbach's Other Conjecture
# Find the smallest odd composite number that cannot be written as
# the sum of a prime and twice a square.
import time, Euler
t1 = time.clock()
primes = [2,3,5,7,11]
candidate = 9
answer = 0
while True:
if candidate == primes[-1]:
primes.append(Euler.nextPrime(candidate))
candidate += 2
else:
answer_found = False
goldbach = False
for prime in reversed(primes[:-1]):
counter = 1
while goldbach == False:
result = (prime + (2 * (counter**2)))
if result == candidate:
goldbach = True
candidate += 2
elif result < candidate:
counter += 1
else:
break
if goldbach == True:
break示例4: int
# 需要导入模块: import Euler [as 别名]
# 或者: from Euler import nextPrime [as 别名]
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
large_prime_counter +=1
if large_prime_counter not in primes:
primes[large_prime_counter] = Euler.nextPrime(primes[large_prime_counter-1])
large = primes[large_prime_counter]
# Deal with the case where composite is too big
else:
small_prime_counter -= 1
large_prime_counter = small_prime_counter +1
primes[small_prime_counter] = Euler.previous_prime(primes[small_prime_counter+1])
small = primes[small_prime_counter]
large = primes[large_prime_counter]
min_ratio = 3
answer = 0
for key, value in composites.items():
n = key
phin = value[2]