Solution of ZOJ 1078 Palindrome Number

Statement of the Problem

We say that a number is a palindrom if it is the sane when read from left to right or from right to left. For example, the number 75457 is a palindrom.

Of course, the property depends on the basis in which is number is represented. The number 17 is not a palindrom in base 10, but its representation in base 2 (10001) is a palindrom.

The objective of this problem is to verify if a set of given numbers are palindroms in any basis from 2 to 16.

Input Format

Several integer numbers comprise the input. Each number 0 < n < 50000 is given in decimal basis in a separate line. The input ends with a zero.

Output Format

Your program must print the message Number i is palindrom in basis where I is the given number, followed by the basis where the representation of the number is a palindrom. If the number is not a palindrom in any basis between 2 and 16, your program must print the message Number i is not palindrom.

Sample Input

17
19
0

Sample Output

Number 17 is palindrom in basis 2 4 16
Number 19 is not a palindrom

Source: South America 2001

 

下面是我的实现代码：

 

 #include <cstdio> #include <iostream> #include <vector> using namespace std; bool is_palin_num( int n, int base ) { bool res = true; vector<int> m; while( n != 0 ) { m.push_back( n % base ); n /= base; } for( int i = 0; i < m.size(); ++i ) { if( m[i] != m[m.size()-i-1] ) { res = false; break; } } return res; } int main() { int n, base; bool flag = false; vector<int> palin_base; while( scanf( "%d", &n ) != EOF && n != 0 ) { for( base = 2; base <= 16; ++base ) if( is_palin_num( n, base ) ) palin_base.push_back( base ); if( palin_base.empty() ) { printf( "Number %d is not a palindrom", n ); } else { printf( "Number %d is palindrom in basis", n ); for( int i = 0; i < palin_base.size(); ++i ) printf( " %d", palin_base[i] ); } palin_base.clear(); printf("/n"); } return 0; }