UVa12050

12050 Palindrome Numbers
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：

       将回文数从小到大依次排列，输入一个正整数n，需要输出第n小的回文数。

分析：

       开一个数组存储长度为i的回文数一共有几个，这样就可以根据输入的n先确定要输出的回文数的长度len，并确定第n小的回文数是长度为len的回文数中的第几个，然后依次求出每一位的数即可得到最终的答案。

#include <cstdio>
#define MAX_BIT 20
// ans数组记录答案，下标从1开始
int bit_num[MAX_BIT + 2],n,ans[MAX_BIT + 2];
void set_bitnum(){
    bit_num[0] = 0;
    bit_num[1] = bit_num[2] = 9;
    for(int i = 3 ; i <= MAX_BIT ; i += 2)
        bit_num[i] = bit_num[i + 1] = 10 * bit_num[i - 1];
}
int calc_bit(int& n){
    int bitnum = 0;
    while(n > bit_num[bitnum])
        n -= bit_num[bitnum],bitnum++;
    return bitnum;
}
int main(){
    set_bitnum();
    while(scanf("%d",&n) == 1 && n){
        int len = calc_bit(n);
        //printf("len: %d
",len);
        int index = len / 2 + 1;
        n = n - 1; // 需要从零开始排列
        while(n)
            ans[index++] = n % 10,n /= 10;
        for(int i = index ; i <= len ; i++)
            ans[i] = 0;
        ans[len] += 1;
        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]);
        puts("");
    }
    return 0;
}