2017ecjtu-summer training #1 UVA 12050

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  数字回文串从小到大排序，输出第n个回文数字

不会写此题啊，膜一下网上大佬代码，没有全看懂，只看懂一部分，没有注释--_--!，他日再破吧。。。

AC代码

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <queue>
#include <vector>
#include <algorithm>
#define maxn 3000
using namespace std;
typedef long long ll;
ll num[maxn];
int n, ans[maxn];
 
void init()
{
    num[] = , num[] = num[] = ;
    for (int i = ; i < ; i += )
        num[i] = num[i+] = num[i-] * ;           //预处理i位的回文数的个数，9,9,90,90,900....
}
 
int main()
{
    init();
    while (scanf("%d", &n) && n)
    {
        int len = ;
        while (n > num[len])
        {
            n -= num[len];                           //判断第n个回文数字有多少位，len
            len++;
        }
        n--;                                          
 
        int cnt = len /  + ;                       //写出回文数字后半边
        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+];                //前半边由后半边复制过来
        for (int i = ; i <= len; i++)
            printf("%d", ans[i]);
        printf("\n");
    }
    return ;
}