spoj The Next Palindrome

题意：比给出的数大的最小回文数。

先用前n/2长对称到后面，如果没变大，在中间加1，进位，再对称。

//#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include <stack>
using namespace std;
typedef long long lon;
const lon SZ=,INF=0x7FFFFFFF;

void work(string &str)
{
    int bs=;
    for(int i=;i<str.size()/;++i)
    {
        if(str[i]>str[str.size()--i])bs=;
        else if(str[i]<str[str.size()--i])bs=-;
        str[str.size()--i]=str[i];
    }
    if(bs<=)
    {
        int c=;
        str[(str.size()-)/]+=;
        for(int i=(str.size()-)/;i>=;--i)
        {
            str[i]+=c;
            if(str[i]>'')str[i]='',c=;
            else c=;
        }
        if(c)
        {
            str=""+str;
        }
        for(int i=;i<str.size()/;++i)
        {
            str[str.size()--i]=str[i];
        }
    }
}

int main()
{
    std::ios::sync_with_stdio();
    //freopen("d:\\1.txt","w",stdout);
    lon casenum;
    cin>>casenum;
    for(lon time=;time<=casenum;++time)
    //for(;scanf("%d",&n)!=EOF;)
    {
        string str;
        cin>>str;
        work(str);
        cout<<str<<endl;
    }
    return ;
}