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.
A single line with B, the base (specified in base 10).
10
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. NOTE WELL THAT BOTH INTEGERS ARE IN BASE B!
1 1 2 4 3 9 11 121 22 484 26 676 101 10201 111 12321 121 14641 202 40804 212 44944 264 69696
题解 ：简单题 进制转换，然后判断是否是回文数
/*
ID: cxq_xia1
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int B;
char trans[20]={‘0‘,‘1‘,‘2‘,‘3‘,‘4‘,‘5‘,‘6‘,‘7‘,‘8‘,‘9‘,‘A‘,‘B‘,‘C‘,‘D‘,‘E‘,‘F‘,‘G‘,‘H‘,‘I‘,‘J‘};
char ans1[25],ans2[50];
int cntLen1,cntLen2;
bool isPalsquare(char a[],int n)
{
for(int i=0;i<n;i++)
{
if(a[i]!=a[n-1-i])
return false;
}
return true;
}
void output()
{
for(int i=cntLen1-1;i>=0;i--)
cout << ans1[i];
cout << " ";
for(int i=cntLen2-1;i>=0;i--)
cout << ans2[i];
cout << endl;
}
int main()
{
freopen("palsquare.in","r",stdin);
freopen("palsquare.out","w",stdout);
cin >> B;
for(int i=1;i<=300;i++)
{
cntLen1=0;cntLen2=0;
memset(ans1,0,sizeof(ans1));
memset(ans2,0,sizeof(ans2));
int tmp1,tmp2;
tmp1=i;
while(tmp1!=0)
{
ans1[cntLen1++]=trans[tmp1%B];
tmp1/=B;
}
tmp2=i*i;
while(tmp2!=0)
{
ans2[cntLen2++]=trans[tmp2%B]
;
tmp2/=B;
}
if(isPalsquare(ans2,cntLen2))
{
output();
}
}
return 0;
}
原文：http://www.cnblogs.com/WillsCheng/p/4776924.html