UVA

Description

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：首先能够知道的是长度为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，啊，起初没这么做，有BUG。后面再加回来
#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("
");
	}
	return 0;
}