1078 Palindrom Numbers

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Palindrom Numbers

Time 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;
}