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 = 3000;
ll num[maxn];
int n, ans[maxn];
void init() {
    num[0] = 0, num[1] = num[2] = 9;
    for (int i = 3; i < 20; i += 2)
        num[i] = num[i+1] = num[i-1] * 10;
}
int main() {
    init();
    while (scanf("%d", &n) && n) {
        int len = 1;
        while (n > num[len]) {
            n -= num[len];
            len++;
        }
        n--;
        int cnt = len / 2 + 1;
        while (n) {
            ans[cnt++] = n % 10;
            n /= 10;
        }
        for (int i = cnt; i <= len; i++)
            ans[i] = 0;
        ans[len]++;
        for (int i = 1; i <= len/2; i++)
            ans[i] = ans[len-i+1];
        for (int i = 1; i <= len; i++)
            printf("%d", ans[i]);
        printf("\n");
    }
    return 0;
}