UVa 12050 - Palindrome Numbers (回文数)

A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionally numbers can of course be ordered in size. The first few palindrome numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, ...

The number 10 is not a palindrome (even though you could write it as 010) but a zero as leading digit is not allowed.

Input

The input consists of a series of lines with each line containing one integer value i (1 ≤ i ≤ 2 ∗ 109 ). This integer value i indicates the index of the palindrome number that is to be written to the output, where index 1 stands for the first palindrome number (1), index 2 stands for the second palindrome number (2) and so on. The input is terminated by a line containing ‘0’.

Output

For each line of input (except the last one) exactly one line of output containing a single (decimal) integer value is to be produced. For each input value i the i-th palindrome number is to be written to the output.

Sample Input

1

12

24

0

Sample Output

1

33

151

题意：求第n个回文串
思路：首先可以知道的是长度为k的回文串个数有9*10^(k-1),那么依次计算，得出n是长度为多少的串，然后就得到是长度为多少的第几个的回文串了，有个细节注意的是，
n计算完后要-1

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 typedef long long ll;
 using namespace std;
 const int maxn = ;
 ll num[maxn];
 int n, ans[maxn];
 void init() {
     num[] = , num[] = num[] = ;
     for (int i = ; i < ; i += )
         num[i] = num[i+] = num[i-] * ;
 }
 int main() {
     init();
     while (scanf("%d", &n) && n) {
         int len = ;
         while (n > num[len]) {
             n -= num[len];
             len++;
         }
         n--;
         int cnt = len /  + ;
         while (n) {
             ans[cnt++] = n % ;
             n /= ;
         }
         for (int i = cnt; i <= len; i++)
             ans[i] = ;
         ans[len]++;
         for (int i = ; i <= len/; i++)
             ans[i] = ans[len-i+];
         for (int i = ; i <= len; i++)
             printf("%d", ans[i]);
         printf("\n");
     }
     return ;
 }