Find next largest palindrome number - c++

Question

I wrote the following program to find the nearest largest palindrome for each of the entered numbers. The first line gives input of how many integers are to be entered.
My problem is I've tried with test cases on geekforgeeks and spojtoolkit (also random numbers whose nearest palindrome I know) and am getting the right answer. SPOJ however says wrong answer. Can anyone suggest cases where my code may not work or possible corrections in my code.
using namespace std;
void evenmirror(char g[1000000],int n)
{
int k=n/2;
for(int i=n/2-1;i>=0;i--)
{
g[k]=g[i];
k++;
}
cout<<g;
}
void oddmirror(char g[1000000],int n)
{
int k=n/2+1;
for(int i=n/2-1;i>=0;i--)
{
g[k]=g[i]; k++;
}
cout<<g;
}
int main()
{
int n,i,j,m;
char g[1000000];
cin>>m;
for(int t=0;t<m;t++)
{
cin>>g;
n=strlen(g);
if(n==1 && g[0]!='9'){cout<<++g[0]; continue;}//single digits except9
int s=0;
int h=0;
if(g[0]=='9' && g[n-1]=='9')
{
for(i=0;i<n;i++)
{
if(g[i]=='9') {h++; }
}
if(h==n)
{
for(i=0;i<=n;i++)
{
if(i==0 || i==n) { cout<<1;}
else {cout<<0;}
}
s=1;
}
}
if(n%2==0 && s==0)
{
i=n/2-1; j=n/2;
while(g[i]==g[j] && i>0)
{
i--; j++;
if(i==0){break;}
}
//if palindrome
if(j==n-1 && g[j]==g[i])
{
if(g[n/2-1]<'9'){ g[n/2-1]++; }
else if(g[n/2]=='9'){(g[n/2-1])='0'; g[n/2-2]+=1;}
evenmirror(g,n);
}//if even string is not palindrome
else
{
if(g[i]<g[j])
{
if(g[n/2-1]<'9'){ g[n/2-1]++; }
else if(g[n/2]=='9'){ g[n/2-1]='0'; g[n/2-2]+=1;}
evenmirror(g,n);
}
else{ evenmirror(g,n); }
}
}
//if odd number string
else if(n%2!=0 && s==0)
{
i=(n/2)-1; j=(n/2)+1;
while(g[i]==g[j] && i>0)
{
i--; j++;
}
//if palindrome
if(j==n-1 && g[j]==g[i])
{
if(g[n/2]<'9'){ g[n/2]++; }
else if(g[n/2]=='9'){ g[n/2]='0'; g[n/2-1]+=1;}
oddmirror(g,n);
}//if odd not palindrome
else
{
char k=n/2+1;
if(g[i]<g[j])
{
if(g[n/2]<'9'){ g[n/2]++; }
else if(g[n/2]=='9'){(g[n/2])='0'; g[n/2-1]+=1;}
oddmirror(g,n);
}
else{ oddmirror(g,n); }
}
}
cout<<endl;
}
return 0;
}
Here is the question: http://www.spoj.com/problems/PALIN/

Answer 1

I didn't debug your code (you didn't ask for it and I think that it should be done by you anyway) but for the following input I got wrong answer:
Input: 9 (expected output 11)
Output: :
Good luck fixing it :)

Answer 2

Since this is an algorithm question, I'll try to keep it language neutral.
The simple way to find the next palindrome is to keep testing until you find one, stepping up one at a time. This is simply a loop, incrementing one at a time and testing.
The test for a palindrome is quite simple. I think you've overcomplicated it by splitting it into odd and even cases. You can either turn the number into a string and check the reverse() is the same, which is neat and tidy (because it's inside a call), but takes memory, or you can do it in place using two pointers. One pointer at the beginning of the string, and one at the end, each moving towards the middle, with the condition being they have to show the same character as the other, and the loop ending when they both point to the middle, or the have switched relative positions.