# ZOJ Problem Set - 1078 Palindrom Numbers

Palindrom Numbers

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a[20];
bool b[18];
int main()
{
int n,i,j,k;
while(scanf("%d",&n)&&n)
{
for(i=2;i<=16;i++)
b[i]=false;
for(i=2;i<=16;i++)//2~16进制
{
int m=n;
int j=0;
while(m)
{
a[++j]=m%i;
m/=i;
}
for(k=1;k<=j/2;k++)//判断是不是回文串
{
if(a[k]!=a[j-k+1])
break;
}
if(k>j/2)
b[i]=true;
}
for(i=2;i<=16;i++)
if(b[i]) break;
if(i>16)
printf("Number %d is not a palindrom\n",n);
else
{
printf("Number %d is palindrom in basis",n);
for(i=2;i<=16;i++)
if(b[i])
printf(" %d",i);
printf("\n");
}
}
return 0;
}

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