Python
# Базовый
for a in range(1, 1500):
if all(((x > 56) or (y >= x) or (3*x - y < a))
for x in range(10000) for y in range(10000)):
print(a)
break
# Поразрядная конъюнкция
for a in range(1, 1000):
if all((x & 47 == 0) or ((x & 13 == 0) <= (x & a != 0))
for x in range(1000)):
print(a); break
# ДЕЛ
for a in range(1, 1000):
if all(((x % 17 != 0) or (x % 12 != 0)) <= (x % a != 0)
for x in range(1000)):
print(a); break Способ 2
# Отрезки
def f(x):
b = 66 <= x <= 75
c = 71 <= x <= 85
a = a1 <= x <= a2
return (not (a)) <= (b == c)
ox = [y for x in (66,75,71,85) for y in (x, x+0.01, x-0.01)]
m = []
for a1 in ox:
for a2 in ox:
if a2 > a1 and all(f(x) == 1 for x in ox):
m.append(a2 - a1)
print(min(m)) # max — если просят макс. длину
# Тип Крылова 2025
def F(b, c, x):
A = 3 <= x <= 60
B = x in b
C = x in c
return C <= (A and (not B))
for y in range(1, 10000):
B = [i for i in range(2, 177) if 177 % i == 0]
C = [i for i in range(2, y) if y % i == 0]
if len(C) > 0 and all(F(B, C, x) for x in range(-10000, 10000)):
print(y)
# Множества
p = [5, 10, 15, 20, 25, 30]
q = [15, 18, 21, 24, 27, 30]
a = list(range(1000))
for x in range(1000):
if (not((x in p) <= (not(x in q))) and (x in a)) == False:
a.remove(x)
print(a)📚 Теория
Логические выражения и кванторы. Проверка на множестве значений.