[Swust OJ 797]--Palindromic Squares(回文数水题)

 

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：

 1 #include<iostream>

 2 using namespace std;

 3 int n;

 4 char str[] = "0123456789ABCDEFGHIJ";

 5 void out(int x){

 6     int a[17], i = 0;

 7     while (x){

 8         a[++i] = x%n;

 9         x /= n;

10     }

11     for (; i; cout << str[a[i--]]);

12 }

13 void find(int x){

14     int a[17], i = 0, s = x*x, t;

15     while (s){

16         a[++i] = s%n;

17         s /= n;

18     }

19     t = i;

20     for (; i&&a[i] == a[t - i + 1]; --i); {

21         if (!i){

22             out(x), cout << ' ';

23             out(x*x), cout << endl;

24         }

25     }

26 }

27 int main()

28 {

29     cin >> n;

30     for (int i = 1; i <= 300; i++)

31         find(i);

32     return 0;

33 }

View Code