Ответы на вопрос:
ответ:я не уверена в ответах(давно не решала такие задачи)
16
[29, 31]
[28, 30]
12
[21, 23]
[20, 22]
Объяснение:
def moves(heap):
a,b=heap
return[(a+1,b),(2*a,b),(a,b+1),(a,b*2)]
table={(k,s):0 for k in range(501)for s in range(501)}
for x in table:
if any(sum(t)>=69 for t in moves(x)):
table[x]=1
for x in table:
if table[x]==0 and all(table[t]==1 for t in moves(x)):
table[x]=2
for x in table:
if table[x]==0 and any(table[t]==2 for t in moves(x)):
table[x]=3
for x in table:
if table[x]==0 and all(table[t]==3 or table[t]==1 for t in moves(x)):
table[x]=4
print(min([s for s in range(1,118)if any(table[t]==1 for t in moves((5,s)))]))
print([s for s in range(1,118)if table[(5,s)]==3])
print([s for s in range(1,118)if table[(5,s)]==4])
def moves(heap):
a,b=heap
return[(a+1,b),(2*a,b),(a,b+1),(a,b*2)]
table={(k,s):0 for k in range(501)for s in range(501)}
for x in table:
if any(sum(t)>=53 for t in moves(x)):
table[x]=1
for x in table:
if table[x]==0 and all(table[t]==1 for t in moves(x)):
table[x]=2
for x in table:
if table[x]==0 and any(table[t]==2 for t in moves(x)):
table[x]=3
for x in table:
if table[x]==0 and all(table[t]==3 or table[t]==1 for t in moves(x)):
table[x]=4
print(min([s for s in range(1,118)if any(table[t]==1 for t in moves((5,s)))]))
print([s for s in range(1,118)if table[(5,s)]==3])
print([s for s in range(1,118)if table[(5,s)]==4])
Популярно: Информатика
-
77777201877777712.08.2022 13:03
-
fa7b3ar10.12.2022 17:41
-
кит21725.01.2020 10:11
-
Yanaaa1324.06.2022 03:46
-
kovmihser22810.08.2021 16:26
-
dzubanuksofia14.02.2023 00:10
-
srySyutoduofyi19.02.2021 17:31
-
alekseyblohinov29.03.2021 03:10
-
everlast11206.01.2023 20:10
-
obilan22820.01.2023 12:16