Section 1.2 palsquare

这道题是要我们判断在B进制下，1<= n <= 300, n^2是否为回文数，如果是回文数的话就输出

该数及其平方在B进制下的形式。因为进制 2 <= B <= 20,所以先定义一个一个进制数组

         const char B[21] = "0123456789ABCDEFGHIJ";

 然后输入进制n，1->300求出其平方在n进制下的形式，判断是否为回文数，是的话再将数字

在n进制下的形式求出。

/*
ID:yucept21
LANG:C++
TASK:palsquare
*/

#include<cstdio>
#include<cstring>
#include<cstdlib>

const int N = 15;
const char B[21] = "0123456789ABCDEFGHIJ";
char a[N], b[N];
int n;

int main()
{

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

    scanf( "%d", &n);
    for( int i = 1; i <= 300; i ++)
    {
        memset( a, 0, sizeof a);
        memset( b, 0, sizeof b);
        int tt = i * i;
        int top = 0;
        while( tt != 0)
        {
            a[top ++] = B[tt % n];
            tt /= n;
        }
        int len1 = top;
        bool mark = true;
        for( int j = 0; j <= len1 / 2; j ++)
            if( a[j] != a[len1 - 1 - j]) {
                mark = false;
                break;
            }
        if( mark || len1 == 1)
        {
            top = 0;
            int k = i;
            while( k != 0)
            {
                b[top ++] = B[k % n];
                k /= n;
            }
            int len2 = top;
            for( int c = len2 - 1; c >= 0; c --)
                printf( "%c", b[c]);
            printf( " %s\n", a);
        }
    }
    return 0;
}