Решить 3 номер 12 везде ! 50 . программа питон. нужно еще схемы сделать!

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])      

from itertools import product

i = open('input.txt', 'r')

o = open('output.txt', 'w+')

l = int(i.

alpha = "245"

d = []

for s in product(alpha, repeat=l):

    d.append([" ".join(

for q in d[1: ]:

    s = " ".join(q)

    if int(s.count('2')) > 2:

        continue

    o.write(" ".join(q) + '\n')

i.close()

o.close()

код к : «перевод кода с питона на паскаль»

var

n,c,i,ost,b,t,d: integer;

a,k: string;

begin

readln(n);

c: =1;

for i: =1 to n do

c: =c*3;

c: =c-1;

for i: =1 to c do

begin

k: ='';

b: =i;

t: =0;

while b> 0 do

begin

ost: =b mod 3;

if ost=0 then inc(t);

d: =(-sqr(ost)+5*ost+4)div 2;

str(d,a);

k: =a+k;

b: =b div 3;

end;

while length(k)do

begin

k: ='2'+k;

inc(t);

end;

if t< =2 then writeln(k);

end;

end.