UVA - 12050-Palindrome Numbers

12050 - Palindrome Numbers

Time limit: 3.000 seconds

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

题意就是把第i个回文数输出来。

长度为k的回文数有9*10(k-1)个，依次计算，得出n为长度多少的回文数，然后得出长度多少的第几个回文数，n计算完要-1.

1-9 9

10-99 9

100-999 90

1000-9999 90

然后900 900 9000 9000。。。

这个题好像又不会了。。。

傻子(╯‵□′)╯︵┻━┻，用C语言交了10几次，格式错误，都不知道看一下用什么交的吗？(▼ヘ▼#)

代码:

#include<stdio.h>
const int N=*1e5;
int n[N],i;
int h,m[N];
void gg(){
    n[]=,n[]=n[]=;
    for(i=;i<;i+=)
        n[i]=n[i+]=n[i-]*;
}
int main(){
    gg();
    while(~scanf("%d",&h)&&h){
        int len=;
        while(h>n[len]){
            h-=n[len];
            len++;
        }
        h--;
        int cnt=len/+;
        while(h!=){
            m[cnt++]=h%;
            h/=;
        }
        for(i=cnt;i<=len;i++)
            m[i]=;
            m[len]++;
        for(i=;i<=len/;i++){
            m[i]=m[len-i+];
        }
        for(i=;i<=len;i++)
            printf("%d",m[i]);
        printf("\n");
    }
    return ;
}