Python build_int示例

示例#1

文件： solve.py 项目： Deddryk/projecteulerv2

def substring_divisible(lst):
    if not build_int(lst[1:4]) % 2 == 0:
        return False
    if not build_int(lst[2:5]) % 3 == 0:
        return False
    if not build_int(lst[3:6]) % 5 == 0:
        return False
    if not build_int(lst[4:7]) % 7 == 0:
        return False
    if not build_int(lst[5:8]) % 11 == 0:
        return False
    if not build_int(lst[6:9]) % 13 == 0:
        return False
    if not build_int(lst[7:10]) % 17 == 0:
        return False
    return True

示例#2

def substring_divisible(lst):
    if not build_int(lst[1:4]) % 2 == 0:
        return False
    if not build_int(lst[2:5]) % 3 == 0:
        return False
    if not build_int(lst[3:6]) % 5 == 0:
        return False
    if not build_int(lst[4:7]) % 7 == 0:
        return False
    if not build_int(lst[5:8]) % 11 == 0:
        return False
    if not build_int(lst[6:9]) % 13 == 0:
        return False
    if not build_int(lst[7:10]) % 17 == 0:
        return False
    return True

示例#3

文件： solve.py 项目： Deddryk/projecteulerv2

Solution to problem 43.

"""

from utils.misc import build_int, prev_perm

def substring_divisible(lst):
    if not build_int(lst[1:4]) % 2 == 0:
        return False
    if not build_int(lst[2:5]) % 3 == 0:
        return False
    if not build_int(lst[3:6]) % 5 == 0:
        return False
    if not build_int(lst[4:7]) % 7 == 0:
        return False
    if not build_int(lst[5:8]) % 11 == 0:
        return False
    if not build_int(lst[6:9]) % 13 == 0:
        return False
    if not build_int(lst[7:10]) % 17 == 0:
        return False
    return True

sum = 0
perm = range(9, -1, -1)
while perm[0] != 0:
    if substring_divisible(perm):
        sum += build_int(perm)
    perm = prev_perm(perm)
print sum

示例#4

文件： solve.py 项目： Deddryk/projecteulerv2

# Author: Deddryk

"""
Solution to problem 41.

This is solved by checking every permutation of the digits 1..n
for n in 9..2 for primality.

"""

from utils.primes import probable_prime
from utils.misc import prev_perm, build_int

x = range(9, 0, -1)

for num_digits in range(9, 1, -1):
    x = range(num_digits, 0, -1)
    while x != range(1, num_digits + 1):
        if probable_prime(build_int(x)):
            print(build_int(x))
            exit()
        x = prev_perm(x)

示例#5

"""

from utils.misc import build_int, prev_perm


def substring_divisible(lst):
    if not build_int(lst[1:4]) % 2 == 0:
        return False
    if not build_int(lst[2:5]) % 3 == 0:
        return False
    if not build_int(lst[3:6]) % 5 == 0:
        return False
    if not build_int(lst[4:7]) % 7 == 0:
        return False
    if not build_int(lst[5:8]) % 11 == 0:
        return False
    if not build_int(lst[6:9]) % 13 == 0:
        return False
    if not build_int(lst[7:10]) % 17 == 0:
        return False
    return True


sum = 0
perm = range(9, -1, -1)
while perm[0] != 0:
    if substring_divisible(perm):
        sum += build_int(perm)
    perm = prev_perm(perm)
print sum

示例#6

# Author: Deddryk
"""
Solution to problem 41.

This is solved by checking every permutation of the digits 1..n
for n in 9..2 for primality.

"""

from utils.primes import probable_prime
from utils.misc import prev_perm, build_int

x = range(9, 0, -1)

for num_digits in range(9, 1, -1):
    x = range(num_digits, 0, -1)
    while x != range(1, num_digits + 1):
        if probable_prime(build_int(x)):
            print(build_int(x))
            exit()
        x = prev_perm(x)