def test():
a = fibo(541)
a = str(a)
print is_pandigital(a[-9:])
print a[-9:]
b = fibo(2749)
b = str(b)
print is_pandigital(b[:9])
print b[:10]
def main1():
"""very slow"""
cnt = 329300
while True:
fib = fibo(cnt)
if is_pandigital(str(fib)[-9:]) and is_pandigital(str(fib)[:9]):
break
else:
print cnt
cnt += 1
print cnt
def gen_pandigitals(digit):
"""This function is very slow. expecially digit is over 5.
If digit is 9, this function return Memmory Error."""
pandigitals = []
below = 10 ** (digit-1)
above = 10 * below
for i in range(below, above):
if is_pandigital(i, s=digit):
pandigitals.append(i)
return pandigitals
def main():
start = default_timer()
max_ = 0
# A brute force approach is used, starting with 1 and multiplying
# the number by 1, 2 etc., concatenating the results, checking if
# it's 1 to 9 pandigital, and going to the next number when the
# concatenated result is greater than the greatest 9 digit pandigital
# value. The limit is set to 10000, since 1*10000=10000, 2*10000=20000 and
# concatenating this two numbers a 10-digit number is obtained.
for i in range(1, 10000):
n = 0
j = 1
while True:
tmp = i * j
n = n + tmp
j = j + 1
if n > max_ and is_pandigital(n, 9):
max_ = n
if i * j < 10:
n = n * 10
elif i * j < 100:
n = n * 100
elif i * j < 1000:
n = n * 1000
elif i * j < 10000:
n = n * 10000
elif i * j < 100000:
n = n * 100000
if n > 987654321:
break
end = default_timer()
print('Project Euler, Problem 38')
print('Answer: {}'.format(max_))
print('Elapsed time: {:.9f} seconds'.format(end - start))
def main():
start = default_timer()
# 8- and 9-digit pandigital numbers can't be prime, because
# 1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
# also divisible by 3, and therefore the whole number is divisible
# by 3. So we can start from the largest 7-digit pandigital number,
# until we find a prime.
i = 7654321
while (i > 0):
if is_pandigital(i, len(str(i))) and is_prime(i):
break
# Skipping the even numbers.
i = i - 2
end = default_timer()
print('Project Euler, Problem 41')
print('Answer: {}'.format(i))
print('Elapsed time: {:.9f} seconds'.format(end - start))
def main():
for n in xrange(9876, 9123, -1):
p = str(n*1) + str(n*2)
if is_pandigital(p):
print p
break
# http://blog.dreamshire.com/project-euler-104-solution/
from projecteuler import is_pandigital
def top_digits(n):
t = n * 0.20898764024997873 + (-0.3494850021680094)
t = int((pow(10, t - int(t) + 8)))
return t
fk, f0, f1 = 2, 1, 1
while not is_pandigital(f1) or not is_pandigital(top_digits(fk)):
f0, f1 = f1, (f1 + f0) % 10 ** 9
fk += 1
print "Project Euler 104 Solution =", fk
def main():
start = default_timer()
n = 0
# Initially I used a bigger array, but printing the resulting products
# shows that 10 values are sufficient.
products = zeros(10, int)
# To get a 1 to 9 pandigital concatenation of the two factors and product,
# we need to multiply a 1 digit number times a 4 digit numbers (the biggest
# one digit number 9 times the biggest 3 digit number 999 multiplied give
# 8991 and the total digit count is 8, which is not enough), or a 2 digit
# number times a 3 digit number (the smallest two different 3 digits number,
# 100 and 101, multiplied give 10100, and the total digit count is 11, which
# is too many). The outer loop starts at 2 because 1 times any number gives
# the same number, so its digit will be repeated and the result can't be
# pandigital. The nested loop starts from 1234 because it's the smallest
# 4-digit number with no repeated digits, and it ends at 4987 because it's
# the biggest number without repeated digits that multiplied by 2 gives a
# 4 digit number.
for i in range(2, 9):
for j in range(1234, 4987):
p = i * j
num_s = str(i) + str(j) + str(p)
if len(num_s) > 9:
break
num = int(num_s)
if is_pandigital(num, 9):
products[n] = p
n = n + 1
# The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
# the result can't be pandigital. The nested loop starts at 123 because
# it's the smallest 3-digit number with no digit repetitions and ends at
# 833, because 834*12 has 5 digits.
for i in range(12, 99):
for j in range(123, 834):
p = i * j
num_s = str(i) + str(j) + str(p)
if len(num_s) > 9:
break
num = int(num_s)
if is_pandigital(num, 9):
products[n] = p
n = n + 1
# Sort the found products to easily see if there are duplicates.
products = np.sort(products[:n])
sum_ = products[0]
for i in range(1, n):
if products[i] != products[i-1]:
sum_ = sum_ + products[i]
end = default_timer()
print('Project Euler, Problem 32')
print('Answer: {}'.format(sum_))
print('Elapsed time: {:.9f} seconds'.format(end - start))
#http://blog.dreamshire.com/project-euler-104-solution/
from projecteuler import is_pandigital
def top_digits(n):
t = n * 0.20898764024997873 + (-0.3494850021680094)
t = int((pow(10, t - int(t) + 8)))
return t
fk, f0, f1 = 2, 1, 1
while not is_pandigital(f1) or not is_pandigital(top_digits(fk)):
f0, f1 = f1, (f1+f0) % 10**9
fk += 1
print "Project Euler 104 Solution =", fk
def main2():
fn, f0, f1 = 2, 1, 1
while not is_pandigital(f1) or not is_pandigital(top_digits(fn)):
f0, f1 = f1, (f1 + f0) % 10 ** 9
fn += 1
print fn