def test_is_prime(self):
print('is_prime')
self.assertTrue(is_prime(5))
self.assertFalse(is_prime(8))
self.assertFalse(is_prime(0))
self.assertFalse(is_prime(-1))
self.assertFalse(is_prime(-3))
from myfunc import is_prime
sum = 0
count = 0
for n in range(9,1000000):
trunc = True
i = n
if count == 11:
break
while len(str(i)) != 0:
if not is_prime(i):
trunc = False
break
if len(str(i)) == 1:
break
i = int(str(i)[1:])
i = n
while len(str(i)) != 0:
if not is_prime(i):
trunc = False
break
if len(str(i)) == 1:
break
i = int(str(i)[:-1])
if trunc:
print n
sum += n
count += 1
print sum
print count
import math
from myfunc import is_prime
n = 600851475143
for i in range(3,int(math.sqrt(n)),2):
if (n % i == 0):
if (is_prime(i)):
print i
from myfunc import is_prime
from itertools import permutations
max = 0
#Since all 9-pandigital numbers will sum to 45, and therefore are divisible by 3
#And the samething goes for 8 digit pandigital numbers, as they sum to 36
#We only need to go up to 7 digits
for i in permutations(['1','2','3','4','5','6','7']):
num = ''.join(i)
if (is_prime(int(num))):
if (num > max):
max = num
print max
from myfunc import is_prime, prime_list
bestcount = 0
bestproduct = 0
for n in xrange(0,200):
print "n=",n
for a in xrange(-1000,1000):
count = 0
for b in prime_list(1000):
if (((n**2)+(a*n)+b) > 0):
if (is_prime((n**2)+(a*n)+b)):
count += 1
if (count > bestcount):
bestcount, bestproduct = count, a*b
print bestproduct
def test_is_prime(self):
print("测试is_prime方法")
#测试用例以test_开头；使用unittest提供的断言方法；
self.assertTrue(is_prime(5))
self.assertTrue(is_prime(5))
import myfunc
count = 0
for i in myfunc.prime_list():
if i > 1000000:
break
tmp = myfunc.shift_digits(str(i), 1)
circ = True
while int(tmp) != i:
if not myfunc.is_prime(int(tmp)):
circ = False
tmp = myfunc.shift_digits(tmp, 1)
if circ:
count += 1
print count
