yuda110

#소수구하는 모듈인 primesieve로 일단 구함. 그중에 고름
#순환소수인지 판별할 때는 순열조합 모듈인 itertools.permutations를 사용

import primesieve

def is_circular_prime(num):
   for 검사해야할num in permutations(str(num)):
      검사해야할num = int(''.join(검사해야할num))
      if 검사해야할num == 소수:
         result = True

yuda110

from itertools import permutations
from collections import deque
import primesieve

#def is_prime(num):
#    result = False
#    for i in range(2, num):
#        if num > 2 and num % i == 0 :
#            return False
#        else :

yuda110

#python3
from math import factorial

def get_factorial_num(range_from, range_to):
    result_sum = 0
    for num in range(range_from, range_to):
        list_num = list(str(num))
        sum = 0
        for i in list_num :
            sum = sum + factorial(int(i))

yuda110

#python3
import math

square_arr = []
for i in range(0, 10) :
    square_arr.append(i ** 5)

def sum_out_squares_of_five(*args) :
    result = 0
    for a in args:

yuda110

#python3
def get_square_set(a_range_to, b_range_to) :
    square_list = set()
    for a in range(2, a_range_to+1) :
        for b in range(2, b_range_to+1) :
            square_list.add(a**b)
    return len(square_list)

print(get_square_set(100, 100))

yuda110

#python3
#1. 일단 규칙부터 찾고
#2. 차근차근 더함

def sum_one_square(first_num, nth) :
    result_sum = 0
    for i in range(1, 5) :
        next_num = first_num + 2*i*nth
        result_sum += next_num
    return result_sum, next_num

yuda110

#python3
# n2 + an + b (단 | a | < 1000, | b | < 1000)
# 연속된 n에 대해 가장 많은 소수를 만들어내는 2차식을 찾아 a*b를 구해라
import cProfile

def is_prime_num(num) :
    if num <= 1  : return False
    for i in range(2, int(num**0.5) + 1):
        if num % i==0:
            return False

yuda110

#python3
import re

def cal_unit_fraction(denominator) :
    return re.findall('.(\d+)', str(1/denominator))

def cal_recurring_cycle(deno) :
    num_list = cal_unit_fraction(deno)
    int(num_list[0])
    for num in num_list :

yuda110

#python3
def cal_fibo(digit) :
    num1 = 1
    num2 = 1
    seq = 3
    while True :
        num3 = num1 + num2
        num1 = num2
        num2 = num3
        if len(str(num3)) >= digit :

yuda110

#python3
import itertools

def cal_lexicographic_permutation(range_to) :
    seq = 1
    num = '0123456789'
    num_list = itertools.permutations(num)
    for lexi_num in num_list :
        if seq == range_to :
            return "".join(lexi_num)