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;

}

/**/