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2081. Sum of kMirror Numbers
Description
A kmirror number is a positive integer without leading zeros that reads the same both forward and backward in base10 as well as in basek.
 For example,
9
is a 2mirror number. The representation of9
in base10 and base2 are9
and1001
respectively, which read the same both forward and backward.  On the contrary,
4
is not a 2mirror number. The representation of4
in base2 is100
, which does not read the same both forward and backward.
Given the base k
and the number n
, return the sum of the n
smallest kmirror numbers.
Example 1:
Input: k = 2, n = 5 Output: 25 Explanation: The 5 smallest 2mirror numbers and their representations in base2 are listed as follows: base10 base2 1 1 3 11 5 101 7 111 9 1001 Their sum = 1 + 3 + 5 + 7 + 9 = 25.
Example 2:
Input: k = 3, n = 7 Output: 499 Explanation: The 7 smallest 3mirror numbers are and their representations in base3 are listed as follows: base10 base3 1 1 2 2 4 11 8 22 121 11111 151 12121 212 21212 Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499.
Example 3:
Input: k = 7, n = 17 Output: 20379000 Explanation: The 17 smallest 7mirror numbers are: 1, 2, 3, 4, 5, 6, 8, 121, 171, 242, 292, 16561, 65656, 2137312, 4602064, 6597956, 6958596
Constraints:
2 <= k <= 9
1 <= n <= 30
Solutions

class Solution { public long kMirror(int k, int n) { long ans = 0; for (int l = 1;; ++l) { int x = (int) Math.pow(10, (l  1) / 2); int y = (int) Math.pow(10, (l + 1) / 2); for (int i = x; i < y; i++) { long v = i; for (int j = l % 2 == 0 ? i : i / 10; j > 0; j /= 10) { v = v * 10 + j % 10; } String ss = Long.toString(v, k); if (check(ss.toCharArray())) { ans += v; if (n == 0) { return ans; } } } } } private boolean check(char[] c) { for (int i = 0, j = c.length  1; i < j; i++, j) { if (c[i] != c[j]) { return false; } } return true; } }