• Nov 18, 2022 •AustinLeath
0 likes • 6 views
primes=[] products=[] def prime(num): if num > 1: for i in range(2,num): if (num % i) == 0: return False else: primes.append(num) return True for n in range(30,1000): if len(primes) >= 20: break; else: prime(n) for previous, current in zip(primes[::2], primes[1::2]): products.append(previous * current) print (products)
• Sep 9, 2023 •AustinLeath
0 likes • 28 views
print("test")
• Feb 26, 2023 •wabdelh
0 likes • 4 views
#You are given a two-digit integer n. Return the sum of its digits. #Example #For n = 29 the output should be solution (n) = 11 def solution(n): return (n//10 + n%10)
• Jun 26, 2025 •AustinLeath
0 likes • 2 views
def format_timestamp(timestamp_epoch): """ Convert epoch timestamp to formatted datetime string without using datetime package. Args: timestamp_epoch (int/float): Unix epoch timestamp (seconds since 1970-01-01 00:00:00 UTC) Returns: str: Formatted datetime string in 'YYYY-MM-DD HH:MM:SS' format """ # Constants for time calculations SECONDS_PER_DAY = 86400 SECONDS_PER_HOUR = 3600 SECONDS_PER_MINUTE = 60 # Handle negative timestamps and convert to integer timestamp = int(timestamp_epoch) # Calculate days since epoch and remaining seconds days_since_epoch = timestamp // SECONDS_PER_DAY remaining_seconds = timestamp % SECONDS_PER_DAY # Calculate hours, minutes, seconds hours = remaining_seconds // SECONDS_PER_HOUR remaining_seconds %= SECONDS_PER_HOUR minutes = remaining_seconds // SECONDS_PER_MINUTE seconds = remaining_seconds % SECONDS_PER_MINUTE # Calculate date (simplified, ignoring leap seconds) year = 1970 days = days_since_epoch while days >= 365: is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) days_in_year = 366 if is_leap else 365 if days >= days_in_year: days -= days_in_year year += 1 # Month lengths (non-leap year for simplicity, adjusted later for leap years) month_lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): month_lengths[1] = 29 month = 0 while days >= month_lengths[month]: days -= month_lengths[month] month += 1 # Convert to 1-based indexing for month and day month += 1 day = days + 1 # Format the output string return f"{year:04d}-{month:02d}-{day:02d} {hours:02d}:{minutes:02d}:{seconds:02d}" # Example timestamp (Unix epoch seconds) timestamp = 1697054700 formatted_date = format_timestamp(timestamp) print(formatted_date + " UTC") # Output: 2023-10-11 18:45:00
0 likes • 14 views
# question3.py from itertools import product V='∀' E='∃' def tt(f,n) : xss=product((0,1),repeat=n) print('function:',f.__name__) for xs in xss : print(*xs,':',int(f(*xs))) print('') # this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r) def parta(p,q,r) : a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or not r) b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r) c= a and b return c def partb(p,q,r) : a=(p or q and r) and (p or not q or not r) and (p or not q or not r)and (p or q or not r) b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r) c= a and b return c print("part A:") tt(parta,3) print("part B:") tt(partb,3)
• Jun 16, 2024 •lagiath
print('hello, world')