906. Super Palindromes (Hard)

嗯…暂且打表吧…

思路,打表

Don’t know the reason but not interested in this problem, so let’s calculate a table to break it.

class Solution:
def superpalindromesInRange(self, left: str, right: str) -> int:
L, R = int(left), int(right)
ans = 0

for i in self.table:
if i < L:
continue
elif i <= R:
ans += 1
else:
break

return ans

table = [
1,
4,
9,
121,
484,
10201,
12321,
14641,
40804,
44944,
1002001,
1234321,
4008004,
100020001,
102030201,
104060401,
121242121,
123454321,
125686521,
400080004,
404090404,
10000200001,
10221412201,
12102420121,
12345654321,
40000800004,
1000002000001,
1002003002001,
1004006004001,
1020304030201,
1022325232201,
1024348434201,
1210024200121,
1212225222121,
1214428244121,
1232346432321,
1234567654321,
4000008000004,
4004009004004,
100000020000001,
100220141022001,
102012040210201,
102234363432201,
121000242000121,
121242363242121,
123212464212321,
123456787654321,
400000080000004,
10000000200000001,
10002000300020001,
10004000600040001,
10020210401202001,
10022212521222001,
10024214841242001,
10201020402010201,
10203040504030201,
10205060806050201,
10221432623412201,
10223454745432201,
12100002420000121,
12102202520220121,
12104402820440121,
12122232623222121,
12124434743442121,
12321024642012321,
12323244744232321,
12343456865434321,
12345678987654321,
40000000800000004,
40004000900040004
]