Palindrome Numbers

                                                                                 Palindrome Numbers
Time Limit: 1000MS         Memory Limit: 65536K
Total Submissions: 3948         Accepted: 1505
Description

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 010 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*10^9 ). 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
Source

Tehran 2003 Preliminary

题意：

输入N求第N个回文数字。

思路：

先求出要求的数位数，然后对半分，直接求一半的数，后一半根据左边可以求得。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;
#define maxn 20
ll cnt[20];
void pro()//根据规律预处理。
{
    cnt[1]=9;
    for(int i=2; i<=20; i++)
    {
        if(i%2!=0)
            cnt[i]=cnt[i-1]*10;
        else
            cnt[i]=cnt[i-1];
    }
}
int main()
{

    pro();

    ll n;
    while(cin>>n&&n)
    {
        int i=1;
        while(n>0)
        {

            n=n-cnt[i];

            i++;
        }
        i--;//要求的数的位数
//        cout<<i<<endl;
        n=n+cnt[i];

        ll left=1;
        for(ll j=1; j<(i+1)/2; j++)
            left*=10;
        ll res=left+n-1;
        left=res;
//        cout<<left<<endl;
//        cout<<res<<endl;
        if(i%2==1)
        {
            left=left/10;
        }//如果位数是奇数，最后一位就是左边中间的，右边数字直接舍掉。
        while(left)
        {
            ll  right=left%10;
            res=res*10+right;//相当于取和左边对称的数
            left/=10;
        }
        cout<<res<<endl;
    }
    return 0;
}