poj 2402 Palindrome Numbers

#include <iostream>
#include <string>
#include <math.h>
using namespace std;
__int64 list[20]={0,9,9,90,90,900,900,9000,9000,90000,90000,900000,900000,9000000,9000000,
90000000,90000000,900000000,900000000,9000000000};
//list[i]表示位数为i的满足回文数要求共有list[i]个
int main()
{
    __int64 n,sum;
    int i;
    while(cin>>n&&n)
    {
        sum=0;
        for(i=1;i<=19;++i)
        {
            if(sum+list[i]>=n)        //i表示位数
                break;
            sum+=list[i];
        }
        string str(i,'0');
        str[0]=str[i-1]='1';
        n=n-sum-1;
        if(i==1)
        {
            str[0]+=n;
        }
        else
        {
            int index=-1,s=(int)pow(10.0,(i-2)/2+(i-2)%2);
            for( ;s>=1 && n!=0 ;s/=10 )
            {
                index++;
                if(n<s)
                    continue;
                str[index]+=n/s;        //不能写成=n/s,因为最外层的str[0],str[i-1]起始值是1
                str[i-1-index]=str[index];
                n=n%s;
            }
        }
        cout<<str<<endl;    
    }
    return 0;
}