Palindromic Squares

Palindromic Squares

Rob Kolstad
Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.


Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters 'A', 'B', and so on to represent the digits 10, 11, and so on.


Print both the number and its square in base B.

PROGRAM NAME: palsquare

INPUT FORMAT

A single line with B, the base (specified in base 10).

SAMPLE INPUT (file palsquare.in)

10

OUTPUT FORMAT

Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself.

SAMPLE OUTPUT (file palsquare.out)

1 1
2 4
3 9
11 121
22 484
26 676
101 10201
111 12321
121 14641
202 40804
212 44944
264 69696

/*
ID: des_jas1
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string.h>
#include <stdlib.h>
//#define fin cin
//#define fout cout
using namespace std;

int N,count;
int after[20],tt[20];
const char w[11]={'A','B','C','D','E','F','G','H','I','J','K'};

bool Is_Palin()
{
	int i=0,j=count;
	for(;i<=j;i++,j--)
		if(after[i]!=after[j])
			return false;
	return true;

}

int main() 
{
	ofstream fout ("palsquare.out");
    ifstream fin ("palsquare.in");
	int i=1,tp,k;
	fin>>N;
	for(;;)
	{
		tp=i*i;
		if(i>300)
			break;
		count=0;
	memset(after,0,sizeof(after));
    while(tp>=N) //把平方转化成in base B
    {
        after[count]=tp%N; 
        tp=tp/N;
		count++;
    }
    after[count]=tp;
	k=i;
		if(Is_Palin())
		{
			int n=0;
			while(k>=N)//是回文就i把原来number也转换成base B
    {
        tt[n]=k%N; 
        k=k/N;
		n++;
    }
			tt[n]=k;
			int j;
			for(j=n;j>=0;j--)
			{
				if(tt[j]>=10)
				{
					fout<<w[tt[j]-10];//输出10进制以上的数要考虑用字母替换
				}
				else
					fout<<tt[j];
			}
			fout<<" ";
			for(j=count;j>=0;j--)
			{
				if(after[j]>=10)
				{
					fout<<w[after[j]-10];
				}
				else
					fout<<after[j];
			}
			fout<<endl;
		}
		i++;
	}
	fout.close();
	fin.close();
    return 0;
}