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

Sample Output

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 ;

 }

巴特西

UVa 12050 - Palindrome Numbers (回文数)

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