usaco Palindromic Squares

/*
ID: modengd1
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
char leter[20]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J'};
//进制转换，将x转化为base进制的数的字符串表示形式存入out数组
void changetheBase(int x,int base,char out[60])
{
stack<char> S;
int i=0;
while(x>0)
{
S.push(leter[x%base]);
x/=base;
}
for(i=0;!S.empty();i++)
{
out[i]=S.top();
S.pop();
}
out[i]=0;
}
//检测一个以'\0'结尾的字符串是不是回文串
bool isPalindromic(char input[60])
{
int temp[60];
int len=strlen(input);
for(int i=0;i<len;i++)
{
temp[len-i-1]=input[i];
}
for(int i=0;i<len;i++)
{
if(input[i]!=temp[i])
return false;
}
return true;
}

int main()
{
freopen("palsquare.in","r",stdin);
freopen("palsquare.out","w",stdout);
int square[301],base;

scanf("%d",&base);
for(int i=1;i<=300;i++)
{
char out1[60],out2[60];
changetheBase(i,base,out1);
changetheBase(i*i,base,out2);
if(isPalindromic(out2))
{
cout<<out1<<' '<<out2<<endl;
}
}
return 0;
}