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。