1(1001.111)
201000,110)
3(111.110)
4(1000, 101)
51010,101)
6(1011,110)
7(1100,110)
8(1101, 101)
(1100.101)
10(1001.11)
11(110,11)
12(101,10)
13(11,11)
14100.11)
15(101,100)
16(100.111)
17(10,1000)
18(100,1001)
19(101,1000)
20110.110)
21(111,111)
22(1001,111)
по информатике типо надо обратное решение сделать) вобще не ебу
[0, 0, "a", 1]
[1, 1, "b", 2]
[1, 2, "bb", 3]
[2, 2, "b", 4]
[1, 3, "bbb", 5]
[2, 3, "bb", 6]
[3, 3, "b", 7]
[1, 4, "", 8]
[2, 4, "bbb", 9]
[3, 4, "bb", 10]
[4, 4, "b", 11]
[1, 5, "", 12]
[2, 5, "", 13]
[3, 5, "bbb", 14]
[4, 5, "bb", 15]
[5, 5, "b", 16]
[0, 6, "aa", 17]
[6, 6, "a", 18]
[6, 7, "aa", 19]
[7, 7, "a", 20]
[6, 8, "aaa", 21]
[7, 8, "aa", 22]
[8, 8, "a", 23]
[5, 9, "baaab", 24]
[9, 9, "b", 25]
s = "aaaab"
n = 0
for r in 0...s.size
for l in 0..r
t = s[l..r]
if t.reverse == t
n += 1
p [l,r,t,n]
end
end
end