USACO section1.2 Palindromic Squares

给出一个数 n（十进制），求出 1-300 范围内所有平方（n 进制）为回文串的数，并打印。

/*
PROG: palsquare
LANG: C++
*/
# include <cstdio>
# include <cstring>

# define N 300

void strRev(char *s)
{
    char ch;
    int len = strlen(s), mid = len / 2;
    for (int i = 0; i < mid; ++i)
        ch = s[i], s[i] = s[len-1-i], s[len-1-i] = ch;
}

void to(int base, int x, char *s)
{
    int i = 0, t;
    while (x > 0)
    {
        t = x % base;
        if (t > 9) t += 'A'-10;
        else t += '0';
        s[i++] = t;
        x /= base;
    }
    s[i] = 0;
    strRev(s);
}

char isPal(char *s)
{
    int len = strlen(s), mid = len / 2;
    for (int i = 0; i < mid; ++i)
        if (s[i] != s[len-i-1]) return 0;
    return 1;
}

int main()
{
    int n;

    freopen("palsquare.in", "r", stdin);
    freopen("palsquare.out", "w", stdout);

    scanf("%d", &n);
    for (int i = 1; i <= N; ++i)
    {
        char a[10], b[20];
        to(n, i, a), to(n, i*i, b);
        if (isPal(b)) printf("%s %s\n", a, b);
    }

    fclose(stdin);
    fclose(stdout);

    return 0;
}

/**/