## primes numbers finder

• mo_ak

0 likes • Mar 12, 2021

Python

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#question1.pydef rose(n) :if n==0 :yield []else :for k in range(0,n) :for l in rose(k) :for r in rose(n-1-k) :yield [l]+[r]+[r]def start(n) :for x in rose(n) :print(x) #basically I am printing x for each rose(n) fileprint("starting program: \n")start(2) # here is where I call the start function

import itertoolsimport stringimport timedef guess_password(real):chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuationattempts = 0for password_length in range(1, 9):for guess in itertools.product(chars, repeat=password_length):startTime = time.time()attempts += 1guess = ''.join(guess)if guess == real:return 'password is {}. found in {} guesses.'.format(guess, attempts)loopTime = (time.time() - startTime);print(guess, attempts, loopTime)print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")val = input("Enter a password you want to crack that is 9 characters or below: ")print(guess_password(val.lower()))

def sum_of_powers(end, power = 2, start = 1):return sum([(i) ** power for i in range(start, end + 1)])sum_of_powers(10) # 385sum_of_powers(10, 3) # 3025sum_of_powers(10, 3, 5) # 2925

def to_roman_numeral(num):lookup = [(1000, 'M'),(900, 'CM'),(500, 'D'),(400, 'CD'),(100, 'C'),(90, 'XC'),(50, 'L'),(40, 'XL'),(10, 'X'),(9, 'IX'),(5, 'V'),(4, 'IV'),(1, 'I'),]res = ''for (n, roman) in lookup:(d, num) = divmod(num, n)res += roman * dreturn resto_roman_numeral(3) # 'III'to_roman_numeral(11) # 'XI'to_roman_numeral(1998) # 'MCMXCVIII'

# Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.# For example, if n is 10, the output should be “2, 3, 5, 7”. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”.# Python program to print all primes smaller than or equal to# n using Sieve of Eratosthenesdef SieveOfEratosthenes(n):# Create a boolean array "prime[0..n]" and initialize# all entries it as true. A value in prime[i] will# finally be false if i is Not a prime, else true.prime = [True for i in range(n + 1)]p = 2while (p * p <= n):# If prime[p] is not changed, then it is a primeif (prime[p] == True):# Update all multiples of pfor i in range(p * 2, n + 1, p):prime[i] = Falsep += 1prime[0]= Falseprime[1]= False# Print all prime numbersfor p in range(n + 1):if prime[p]:print (p)# driver programif __name__=='__main__':n = 30print("Following are the prime numbers smaller")print("than or equal to ", n)print("than or equal to ", n)SieveOfEratosthenes(n)

class Solution(object):def floodFill(self, image, sr, sc, newColor):R, C = len(image), len(image[0])color = image[sr][sc]if color == newColor: return imagedef dfs(r, c):if image[r][c] == color:image[r][c] = newColorif r >= 1: dfs(r-1, c)if r+1 < R: dfs(r+1, c)if c >= 1: dfs(r, c-1)if c+1 < C: dfs(r, c+1)dfs(sr, sc)return image