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V3.py
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V3.py
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# Solved by Ostap Baranov in preparation for the Russian Unified State Exam in CS, 2023.
#
# All tasks were developed by Krulov S. S. in 2023 and belongs to the © National Education Publishing, LLC.
import math
from turtle import *
from itertools import product, permutations
from functools import lru_cache
print("№2:") # ywzx
def columns(y, z, w, x):
return not(y <= (not(z <= w))) and ((not z) <= ((not w) == x))
for holes in product([0, 1], repeat=5):
table = [(1, holes[0], 1, 1), (holes[1], holes[2], 0, 0), (holes[3], 0, 0, holes[4])]; F = [0, 1, 1]
if len(table) == len(set(table)):
for answer2 in permutations('yzwx'):
print(*answer2, sep='') if [columns(**dict(zip(answer2, variables))) for variables in table] == F else None
print("№5:") # 12
for N in range(1, 100):
B = bin(N)[2:]
if len(B) % 2 == 0:
B = B[:len(B) // 2] + '1' + B[len(B) // 2:] # Не совсем логично, но пихаем в середину 1
R = int(B, 2)
if R >= 26:
print(N, "а само число такое", R)
break
print("№6:") # 1200
screensize(10000, 10000)
tracer(0)
hideturtle()
speed(10)
pensize(0.1)
color("black", "red")
scale = 100
counter6 = 0
up()
left(90)
forward(100 * scale)
right(90)
forward(100 * scale)
right(30)
down()
begin_fill()
for repeat in range(2):
forward(30 * scale)
right(90)
forward(40 * scale)
right(90)
end_fill()
up()
canvas = getcanvas()
for X in range(-250 * scale, 250 * scale, scale):
for Y in range(-250 * scale, 250 * scale, scale):
scanner = canvas.find_overlapping(X, Y, X, Y)
counter6 += 1 if len(scanner) == 1 and scanner[0] == 5 else 0
print(counter6)
update()
exitonclick()
print("№8") # 72
counter = 0
for let1, let2, let3, let4, let5, let6 in product("КНОРСЯ", repeat=6): # Считываем алфавит ИЗ ПРИМЕРА
word = let1 + let2 + let3 + let4 + let5 + let6
counter += 1 # Здесь мы просто нумеруем слова, счётчик не функциональный
if word.count('К') <= 3 and word.count('Я') == 2:
print(counter, word)
break
print("№12:") # 121222
string12 = '22' + '1' * 2024 + '22'
while '2111' in string12 or '1112' in string12:
string12 = string12.replace('111', '1', 1)
string12 = string12.replace('21', '12', 1) if '21' in string12 else string12.replace('12', '1', 1)
print(string12)
print("№14:") # 1071
bruh = 243**540 - 6 * 9**530 + 21 * 3**511 - 3 * 3**70 - 200
counter = 0
while bruh > 0:
if bruh % 9 == 8:
counter += 1
bruh //= 9
print(counter)
print("№15:") # 24
def treug(n, m, k):
return n + m > k and n + k > m and m + k > n # функция по смыслу, кукож тот ещё
for A in range(1, 1000):
while True: # Как в предыдущем варианте, плюсом НЕ МАКС изменили по информации из примечания
if all(not((treug(x, 11, 18) == (max(x, 5) <= 15)) and (treug(x, A, 5))) for x in range(1, 1000)):
print(A)
break
A -= 1 # Чтобы прога не уходила в отрицательные числа из-за особенностей ТРЕУГ функции
print("№16:") # 887040
@lru_cache(None)
def F(n):
if n < 3:
return 1
elif n > 2 and n % 2 != 0:
return F(n - 1) + F(n - 2)
elif n > 2 and n > 2 and n % 2 == 0:
return sum(F(i) for i in range(1, n)) # Преобразуем сигму по смыслу, ничего особенного
print(F(24))
print("№17") # 203 99820
counter = sums = 0
with open('17var03.txt') as file17:
sequence = [int(numbers) for numbers in file17]
maximum = max(sequence)
for elem1, elem2, elem3 in zip(sequence, sequence[1:], sequence[2:]):
last_digits = str(elem1 % 10) + str(elem2 % 10) + str(elem3 % 10)
if last_digits.count('0') == 1 and elem1 + elem2 + elem3 < maximum:
counter += 1
sums = max(sums, elem1 + elem2 + elem3)
print(counter, sums)
print("№19:") # 76
def heap19(rocks1, positions):
if rocks1 >= 153 or positions > 3:
return positions == 3
elif positions % 2 == 0:
return heap19(rocks1 + 1, positions + 1) or heap19(rocks1 * 2, positions + 1)
else:
return heap19(rocks1 + 1, positions + 1) and heap19(rocks1 * 2, positions + 1)
for answer19 in range(1, 152 + 1):
print(answer19) if heap19(answer19, 1) else None
print("№20:") # 38 75
def heap20(rocks1, positions):
if rocks1 >= 153 or positions > 4:
return positions == 4
elif positions % 2 != 0:
return heap20(rocks1 + 1, positions + 1) or heap20(rocks1 * 2, positions + 1)
else:
return heap20(rocks1 + 1, positions + 1) and heap20(rocks1 * 2, positions + 1)
for answer20 in range(1, 152 + 1):
print(answer20) if heap20(answer20, 1) else None
print("№21:") # 74
def heap21(rocks1, positions):
if rocks1 >= 153 or positions > 5:
return positions == 3 or positions == 5
elif positions % 2 == 0:
return heap21(rocks1 + 1, positions + 1) or heap21(rocks1 * 2, positions + 1)
else:
return heap21(rocks1 + 1, positions + 1) and heap21(rocks1 * 2, positions + 1)
for answer21 in range(1, 152 + 1):
print(answer21) if heap21(answer21, 1) else None
print("№23:") # 1620
func23 = lambda start, end, exception: func23(start - 1, end, exception) + func23(start // 2, end, exception) if start > end and start != exception else start == end
print(func23(50, 20, 10) * func23(20, 1, 10)) # Ради одного "не равно" буквами сорим, но что поделать
print("№24:") # 55
with open('24var03.txt') as file24:
f = file24.read().strip().replace('AB', '_') # АВ - пара, не выйдет сравнения. Представим чем-то одним
array = list()
min_len = float('inf')
for index in range(len(f)):
if f[index] == '_':
array.append(index)
if len(array) == 21:
min_len = min(index - array[0] + 1, min_len)
array = array[1:]
print(min_len + 21) # Символов - пара, нужно ещё столько же (НЕ умножить на 2, это другое)
print("№25:") # Ответ верный
digits = '0123456789' # Чтобы не спамить, выносим все цифры
for length in range(4):
for asterisk in product(digits, repeat=length):
for question_mark in digits: # просто пробуем по цифре
mask = int(f"32{''.join(asterisk)}21{question_mark}4")
print(mask, mask // 2049) if mask % 2049 == 0 else None