USACO 1.2 Palindromic Squares (进制转换，回文)

/*
ID:twd30651
PROG:palsquare
LANG:C++
*/
#include<iostream>
#include<fstream>
#include<stdlib.h>
#include<string.h>
using namespace std;
int BASE;
char B[]={'0','1','2','3','4','5','6','7','8','9',
'A','B','C','D','E','F','G','H','I','J'};
int isPalindrome(char *s,int len)
{

for(int i=0,j=len-1;i<=j;i++,j--)
{
if(s[i]!=s[j])return 0;
}
if(len>0)return 1;
else return 0;
}
void getp(char *s,int num,int b)
{
int index=0;
while(num/b!=0)
{
s[index++]=B[num%b];
num=num/b;
}
s[index++]=B[num%b];;
s[index]='\0';
//    printf("%s\n",s);
}
int main(int argc,char *argv[])
{
freopen("palsquare.in","r",stdin);
freopen("palsquare.out","w",stdout);
scanf("%d",&BASE);
char s[20];
char t[20];
for(int i=1;i<=300;++i)
{
memset(s,0,sizeof(s));
memset(t,0,sizeof(t));
getp(s,i*i,BASE);
if(isPalindrome(s , strlen(s)))
{
getp(t,i,BASE);
int l=strlen(t);
for(int j=0;j<l/2;++j)
{
char tmp=t[j];
t[j]=t[l-j-1];
t[l-j-1]=tmp;
}

printf("%s %s\n",t,s);
}
}

return 0;
}