1078 Palindrom Numbers

Palindrom NumbersTime limit: 1 Seconds Memory limit: 32768K
Total Submit: 5044 Accepted Submit: 2193 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

Problem Source: South America 2001

http://acm.zju.edu.cn/show_problem.php?pid=1078


#include <iostream>


using namespace std;




int convert(int n, int list[], int basis); //转换进制




int main()




...{


int num[1000];


bool flag1, flag2;


int basis[17], counter = 1;


int n, len;




cin >> n;


while ( n != 0 )




...{


flag1 = 1;


counter = 1;


for ( int i = 2; i < 17; i++ )




...{


len = convert(n, num, i)+1;


flag2 = true;


for ( int j = 0; j < len/2; j++ )




...{


if ( num[j] != num[len-j-1] )




...{


flag2 = false; //flag2标志该数在该进制下是否回文数


break;


}


}


if ( flag2 )




...{


flag1 = 0;


basis[counter] = i;


counter++;


}


}




if ( !flag1 ) //flag1 标志是否有解






...{


cout << "Number " << n << " is palindrom in basis";


for ( int k = 1; k < counter; k++ )




...{


cout << " " << basis[k];


}


cout << endl;


}


else




...{


cout << "Number " << n << " is not a palindrom" << endl;


}






cin >> n;


}




return 0;


}




int convert(int n, int list[], int basis)




...{


int total = 0;




while ( n >= basis )




...{


list[total] = n % basis;


total++;


n /= basis;


}




list[total] = n;




return total;


}