USACO Palindromic Squares (数制转换)

题目大意：找出1到300的数中其平方在b进制下是回文的数，并打印这些数数及其平方在b进制下的表示。

/*
ID: lijian42
LANG: C++
TASK: palsquare
*/
#include <stdio.h>
#define LEN 20
#define N 300
char t[20]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J'};
char s1[LEN],s2[LEN];
int cnt1,cnt2;
int b;
void tran(int k,char s[],int &cnt)
{
    cnt=0;
    while(k)
    {
        s[cnt++]=t[k%b];
        k/=b;
    }
}
bool judge(char s[],int cnt)
{
    int i,j;
    for(i=0,j=cnt-1;i<j;i++,j--)
    {
        if(s[i]!=s[j])  return false;
    }
    return true;
}
void print()
{
    int i;
    for(i=cnt2-1;i>=0;i--)  printf("%c",s2[i]);
    printf(" ");
    for(i=cnt1-1;i>=0;i--)   printf("%c",s1[i]);
    puts("");
}
void solve()
{
    int i;
    for(i=1;i<=N;i++)
    {
        tran(i*i,s1,cnt1);
        if(judge(s1,cnt1))
        {
            tran(i,s2,cnt2);
            print();
        }
    }
}
int main()
{
    freopen("palsquare.in","r",stdin);
    freopen("palsquare.out","w",stdout);
    while(~scanf("%d",&b))
    {
        solve();
    }
    return 0;
}