# print('{0} has {1} elements.'.format(greatest,num))
def main(limit):
chainLength = [1]
longest = 0
gen = 0
for i in range(2,limit+1):
chain = 0
while (i not in range(1,len(chainLength)+1)):
if i % 2 == 0:
i = int(i/2)
chain += 1
else:
i = (3*i + 1)
chain += 1
chain += chainLength[i-1]
chainLength.append(chain)
for i in range(0,len(chainLength)):
if (chainLength[i] > longest):
longest = chainLength[i]
gen = i+1
return str(gen) + " generates a chain of length " + str(longest)
# print(main(1000000), analytics.lap(), analytics.maxMem())
print(Main2(MAX), analytics.lap(), analytics.maxMem())
multiply by #14 and divide by #1 and then you don’t need to remultiply all the numbers every time. you’re doing 2 operations instead of 12.
store the largest one so far in a variable, and compare new one to that and replace if new one is larger"""
def string_slicer(string, n):
Max_prod = 1
Max_prod_str = ''
i = 0
while i < (len(string) - n):
sliding_window = string[i:i + n]
# window = str(string[i + start:(i+start) +n])
# print("window", sliding_window)
window_prod = 1
if "0" not in sliding_window:
for j in sliding_window:
window_prod *= int(j)
if window_prod >= Max_prod:
Max_prod = window_prod
Max_prod_str = sliding_window
i += 1
return Max_prod
# print(product_getter_5000(l))
# print(product_getter_5000(string_slice(number, 4)))
# print(product_getter_5000(string_slice(number, 13)))
# print(string_slicer(number, 13))
print(string_slicer(number, 13), analytics.lap(), analytics.maxMem())
print(product_getter_5000(string_slice(number, 13)), analytics.lap(),
analytics.maxMem())
Diag = []
for diag in list_of_lists:
if len(diag) > 3:
i = 0
while i < (len(diag) - n):
sliding_window = diag[i:i + n]
window_prod = 1
if "00" not in sliding_window:
for j in sliding_window:
window_prod *= int(j)
if window_prod >= prod:
prod = window_prod
prod_numbs = sliding_window
Diag = diag[:]
i += 1
return prod, prod_numbs, Diag
def main(matrix, n):
A = matrix
window = n
lists = []
prod_r, prod_num_r, row = row_checker(rows(A), window)
prod_c, prod_num_c, col = col_checker(cols(A), window)
prod_d, prod_num_d, diag = diags_checker(diags(A), window)
return max(prod_r, prod_c, prod_d)
print(main(A, 4), analytics.lap(), analytics.maxMem())
# filename = "primes"
test = 600851475143
test1 = 83
# print(len(prime_tests.primes))
# The prime factors of 13195 are 5, 7, 13 and 29.
# What is the largest prime factor of the number 600851475143 ?
factor_list = []
def main(n):
pf = set([])
while n%2==0:
pf.add(2)
n //= 2
for i in range(3,int(n**(.5)),2):
while n % i == 0:
pf.add(i)
n //= i
if n > 1:
pf.add(n)
LPF = sorted(list(pf))
print(LPF)
if LPF == n:
return LPF
print(main(test), analytics.lap(), analytics.maxMem())