题目链接：http://acm.swust.edu.cn/problem/797/
Time limit(ms): 1000 Memory limit(kb): 10000
Description
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.
Input
A single line with B, the base (specified in base 10).
Output
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 Input
| 10 |
Sample Output
|
1 1
2 4
3 9
11 121
22 484
26 676
101 10201
111 12321
121 14641
202 40804
212 44944
264 69696
|
Hint
题目大意：一个n代表进制(n<=20)，输出十进制数300以内的此时进制下的所有这个数的平方是回文数的所有数和它的平方
解题思路：直接转换判断就是了，判读回文可以利用回文数性质，在转换进制的数组中直接判断，具体的看代码
代码如下：
#include<iostream>
using namespace std;
int n;
char str[] = "0123456789ABCDEFGHIJ";
void out(int x){
int a[], i = ;
while (x){
a[++i] = x%n;
x /= n;
}
for (; i; cout << str[a[i--]]);
}
void find(int x){
int a[], i = , s = x*x, t;
while (s){
a[++i] = s%n;
s /= n;
}
t = i;
for (; i&&a[i] == a[t - i + ]; --i); {
if (!i){
out(x), cout << ' ';
out(x*x), cout << endl;
}
}
}
int main()
{
cin >> n;
for (int i = ; i <= ; i++)
find(i);
return ;
}