UVALive 2889 Palindrome Numbers

Question：
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
题意大意：将回文数从大到小排列，1，2，3，4，5，6，7，8，9，11，22，，，，99，，，输入一个数n，让你输出第n大的回文数是多少。
思路：仔细找你会发现一位数和两位数的回文数都只有9个，三位数和四位数的回文数有90个，五位数和六位数的回文数有900个，以此类推。
(http://acm.hust.edu.cn/vjudge/contest/121559#problem/H)

#include <iostream>#include <cstdio>using namespace std;int main(){    int n,i,j,t1,t2,t3;    while (scanf("%d",&n),n)    {        i=9;j=1;        while (1)        {            if(n<=i)            {                t1=j+n-1;   //此部分数位奇数位，设位数为n，先算出前面部分n/2+1位                cout<<t1;                t2=t1/10;                while(t2)                {                    cout<<t2%10; // 再算出后面的n/2位                    t2/=10;                }                break;            }            n-=i;            if(n<=i)            {                t1=j+n-1;  //此部分数位偶数位，设位数为n，先算出前面部分n/2位                cout<<t1;                t2=t1;                while(t2)                {                    cout<<t2%10;  // 再算出后面的n/2位                    t2/=10;                }                break;            }            n-=i;            i*=10;            j*=10;        }        cout<<endl;    }    return 0;}

体会：做题要善于总结规律，发现规律，千万不要嫌麻烦，多算几组可能就发现规律了。