USACO:Palindromic Squares;回文判断+进制转化

/*
TASK:palsquare
LANG:C++
ID:huibaochen
*/
/*一题：给你一个数n代表进制，算出从1到300的数的平方转化为n进制后是否进回文序列，是的话输出这个是的n进制和这个数的平方的n进制*/
#include <iostream>
#include <stdio.h>
#include <string.h>
int m, n, len, num, r;
char str[1000 +5], ch[100 + 5];

using namespace std;

void yes(int m)
{
    num = m * m;
    len = 0;
    while(num){
        r = num % n;
        num = num / n;
        if(r >= 0 && r <= 9)
            str[len++] = r + 48;
        else
            str[len++] = r + 'A' - 10;
    }
    for(int i = 0; i < len / 2; i++){
        if(str[i] != str[len - 1 - i])
            return;
    }
    len = 0;
    while(m){
       r = m % n;
       m = m / n;
       if(r >= 0 && r <= 9)
            ch[len++] = r + 48;
       else
            ch[len++] = r + 'A' - 10;
    }
    for(int i = 0; i < len / 2; i++){
        swap(ch[i], ch[len - i - 1]);
    }
    printf("%s %s\n", ch, str);

}

int main()
{
    freopen ("palsquare.in", "r", stdin);
    freopen ("palsquare.out", "w", stdout);
    scanf("%d", &n);
    for(int i = 1; i <= 300; i++){
        yes(i);
    }
    return 0;
}